Question 14 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\big(\frac\pi2-\text{x}\big)\sin\text{x}-2\cos\text{x}}{\big(\frac\pi2-\text{x}\big)+\cot\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\big(\frac\pi2-\text{x}\big)\sin\text{x}-2\cos\text{x}}{\big(\frac\pi2-\text{x}\big)+\cot\text{x}}$ If $\text{x}\rightarrow\frac{\pi}{2},$ then $\frac\pi2-\text{x}\rightarrow0$ let $\frac\pi2-\text{x}=\text{y}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big(\text{y}\sin\big(\frac{\pi}{2}-\text{y}\big)-2\cos\big(\frac{\pi}{2}-\text{y}\big)\Big)}{\text{y}+\cot\big(\frac\pi2-\text{y}\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\Big(\frac{\text{y}\cos\text{y}-2\sin\text{y}}{1+\tan\text{y}}\Big)$ $=\lim\limits_{\text{y}\rightarrow{0}}\Bigg(\frac{\cos\text{y}-2\frac{\sin\text{y}}{\text{y}}}{1+\frac{\tan\text{y}}{\text{y}}}\Bigg)$ $=\frac{\lim\limits_{\text{y}\rightarrow{0}}\cos\text{y}-2\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}}{1+\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan\text{y}}{\text{y}}}$ $\Big[\because\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}=1,\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}=1\Big]$ $=\frac{1-2}{1+1}=\frac{-1}{2}$ $=-\frac12$
View full question & answer→Question 24 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}}$
Answer$\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\big(\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}\big)}\times\frac{\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{(6\text{x}-5)-(4\text{x}+5)}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{2\text{x}-10}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{2(\text{x}-5)}$ $=\frac{\sqrt{6(5)-5}+\sqrt{4(5)+5}}{2}$ $=\frac{\sqrt{25}+\sqrt{25}}{2}$ $=\frac{5+5}{2}=5$
View full question & answer→Question 34 Marks
Evaluate the following limit: $\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$
Answer$\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2(\sin\text{a}\cos\text{h})-\text{a}^2\sin\text{a}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{a}^2+2\text{ah}+\text{h}\big)(\sin\text{a}\cos\text{h})-\text{a}^2\sin\text{a}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}^2\sin\text{a}(\cos-1)+2\text{ah}\sin\text{a}\cos\text{h}+\text{h}^2\sin\text{a}\cos\text{h}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}^2\sin\text{a}(\cos-1)}{\text{h}}+\lim\limits_{\text{h}\rightarrow0}\frac{2\text{ah}\sin\text{a}\cos\text{h}}{\text{h}}\\\ +\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2\sin\text{a}\cos\text{h}}{\text{h}}+\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin\text{a}\cos\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{a}^2\sin\text{a}\sin^2\big(\frac{\text{h}}{2}\big)}{\frac{\text{h}}{2}}+2\text{a}\sin\text{a}+0+\text{a}^2\cos\text{a}$ $=0+2\text{a}\sin\text{a}+\text{a}^2\cos\text{a}$ $=2\text{a}\sin\text{a}+\text{a}^2\cos\text{a}$
View full question & answer→Question 44 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\{\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\text{x}\}}{\cos^2\beta\text{x}-\cos^2\alpha\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\{\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\text{x}\}}{\cos^2\beta\text{x}-\cos^2\alpha\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\Big\{2\sin\frac{(\alpha+\beta+\alpha-\beta)}{2}\times\cos\frac{(\alpha+\beta-\alpha+\beta)}{2}\times+2\sin\alpha\cos\alpha\text{x}\Big\}}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big\{2\sin\alpha\text{x}\cos\beta\text{x}+2\sin\alpha\text{x}\cos\alpha\text{x}\big\}}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}(\cos\beta\text{x}+\cos\alpha\text{x})}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{(\cos\beta\text{x}-\cos\alpha\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{\Big(1-2\sin^2\big(\frac{\beta\text{x}}{2}\big)-1+2\sin^2\big(\frac{\alpha\text{x}}{2}\big)\Big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{2\sin^2\big(\frac{\alpha\text{x}}{2}\big)-2\sin^2\big(\frac{\beta\text{x}}{2}\big)}$ $=\frac{2\alpha}{\alpha^2-\beta^2}$
View full question & answer→Question 54 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{3}+\sqrt{2}\big)}{\text{x}^2-6}$
Answer$\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{3}+\sqrt{2}\big)}{\text{x}^2-6}$$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\Big(\sqrt{\big(\sqrt{3}+\sqrt{2}\big)^2}\Big)}{\text{x}^2-6}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{5+2\sqrt{6}}\big)}{\text{x}^2-6}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{5+2\sqrt{6}}\big)}{\text{x}^2-6}\times\frac{\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)}{\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{{7+2\text{x}}-{7-2\sqrt{10}}}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{5+2\text{x}-5-2\sqrt{6}}{\big(\text{x}^2-6\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{2\big(\text{x}-\sqrt{6}\big)}{\big(\text{x}^2-\sqrt{6}\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{10}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{2}{\big(\text{x}+\sqrt{6}\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)\Big)}$
$=\frac{2}{\big(2\sqrt{6}\big)\Big(2\sqrt{5+2\sqrt{6}}\Big)}$
$=\frac{1}{\big(2\sqrt{6}\big)\Big(\sqrt{5+2\sqrt{6}}\Big)}$
$=\frac{1}{\big(2\sqrt{6}\big)\Big(\sqrt{3}+\sqrt{2}\Big)}$
$=\frac{\big(\sqrt{3}-\sqrt{2}\big)}{\big(2\sqrt{6}\big)}$
View full question & answer→Question 64 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\text{cosec}^2\text{x}}{1-\cot\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\text{cosec}^2\text{x}}{1-\cot\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\big(1+\cot^2\text{x}\big)}{1-\cot\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-1-\cot^2\text{x}}{1-\cot\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{1-\cot^2\text{x}}{1-\cot\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{(1-\cot\text{x})(1+\cot\text{x})}{(1-\cot\text{x})}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}(1+\cot\text{x})$ $=1+\cot\frac{\pi}{4}$ $=1+1$ $=2$
View full question & answer→Question 74 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}\big)}{\text{x}}\times\frac{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{(1+3\text{x})-(1-3\text{x})}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{6\text{x}}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{6}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$ $=\frac{6}{\sqrt{1}+\sqrt{1}}$ $=\frac62$ $=3$
View full question & answer→Question 84 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\pi}\frac{1+\cos\text{x}}{\tan^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\pi}\frac{1+\cos\text{x}}{\tan^2\text{x}}$ As$\text{ x}\rightarrow\pi,\text{x}-\pi\rightarrow0,$let $\text{ x }-\pi=\text{y}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{1+\cos(\pi+\text{y})}{\tan^2(\pi+\text{y})}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{1-\cos\text{y}}{\tan^2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{2\sin^2\frac{\text{y}}{2}}{\tan^2\text{y}}$ $=\frac{\lim\limits_{\text{y}\rightarrow0}2\sin^2\frac{\text{y}}{2}}{{\lim\limits_{\text{y}\rightarrow0}\tan^2\text{y}}}$ $=\frac{2\Bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{\text{y}^2}{4}}{\bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\bigg)\times\text{y}^2}$ $=\frac{2\times1\times\frac{\text{y}^2}{4}}{1\times\text{y}^2}$ $\begin{bmatrix}\therefore\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\\\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1 \end{bmatrix}$ $=2\times1\times\frac{1}{4}$ $=\frac{1}{2}$
View full question & answer→Question 94 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}2\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{4}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}2\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{4}\big)}=\lim\limits_{\text{h}\rightarrow{0}}\frac{1\sin\big(\frac{\pi+\text{h}}{2}\big)}{\cos\big(\frac{\pi+\text{h}}{2}\big)\big(\cos\big(\frac{\pi+\text{h}}{2}\big)-\sin\big(\frac{\pi+\text{h}}{2}\big)\big)}$ $=\lim\limits_{\text{h}\rightarrow{0}}\frac{1-\cos\big(\frac{\text{h}}{2}\big)}{-\sin\big(\frac{\text{h}}{2}\big)\big(\frac{1}{\sqrt{2}}\cos\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\sin\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\sin\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\cos\big(\frac{\text{h}}{4}\big)\big)}$ $=\lim\limits_{\text{h}\rightarrow{0}}\frac{1-\cos\big(\frac{\text{h}}{2}\big)}{\sqrt{2}\sin\big(\frac{\text{h}}{2}\big)\sin\big(\frac{\text{h}}{4}\big)}$ $=\lim\limits_{\text{h}\rightarrow{0}}\frac{2-\sin^2\big(\frac{\text{h}}{4}\big)}{\sqrt{2}\sin\big(\frac{\text{h}}{2}\big)\sin\big(\frac{\text{h}}{4}\big)}$ $=\sqrt{2}\lim\limits_{\text{h}\rightarrow{0}}\frac{\sin\big(\frac{\text{h}}{4}\big)}{\sin\big(\frac{\text{h}}{2}\big)}$ $=\sqrt{2}\lim\limits_{\text{h}\rightarrow{0}}\frac{\frac{\sin\big(\frac{\text{h}}{4}\big)}{\big(\frac{\text{h}}{4}\big)}\times\big(\frac{\text{h}}{4}\big)}{\frac{\sin\big(\frac{\text{h}}{2}\big)}{\big(\frac{\text{h}}{2}\big)}\times\big(\frac{\text{h}}{2}\big)}$ $=\sqrt{2}\times\frac{\frac14}{\frac12}$ $=\frac{1}{\sqrt{2}}$
View full question & answer→Question 104 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{5}+\sqrt{2}\big)}{\text{x}^2-10}$
Answer$\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{5}+\sqrt{2}\big)}{\text{x}^2-10}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\Big(\sqrt{\big(\sqrt{5}+\sqrt{2}\big)^2}\Big)}{\text{x}^2-10}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{7+2\sqrt{10}}\big)}{\text{x}^2-10}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{7+2\sqrt{10}}\big)}{\text{x}^2-10}\times\frac{\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)}{\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{{7+2\text{x}}-{7-2\sqrt{10}}}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{2\big(\text{x}-\sqrt{10}\big)}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{2}{\big(\text{x}-\sqrt{10}\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$ $=\frac{2}{\big(\sqrt{10}+\sqrt{10}\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$ $=\frac{2}{\big(2\sqrt{10}\big)\Big(2\sqrt{7+2\sqrt{10}}\Big)}$ $=\frac{1}{\big(2\sqrt{10}\big)\Big(\sqrt{7+2\sqrt{10}}\Big)}$ $=\frac{1}{\big(2\sqrt{10}\big)\Big(\sqrt{5}+\sqrt{2}\Big)}$ $=\frac{\big(\sqrt{5}-\sqrt{2}\big)}{\big(6\sqrt{10}\big)}$
View full question & answer→Question 114 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}-2}{\text{x}^2-\text{x}}-\frac{1}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$
Answer$\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}-2}{\text{x}^2-\text{x}}-\frac{1}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$$=\lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1)}{\text{x}\big(\text{x}^2-3\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}\big(\text{x}^2-3\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}\big(\text{x}^2-1\text{x}-2\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}(\text{x}-1)(\text{x}-2)}\Bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{(\text{x}-2)^2-1}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2+4-4\text{x}-1}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2-4\text{x}+3}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2-4\text{x}+3}{\text{x}(\text{x}-2)(\text{x}-1)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{\text{x}^2-\text{x}-3\text{x}+3}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{\text{x}(\text{x}-1)-3(\text{x}-1)}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{(\text{x}-3)(\text{x}-1)}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\frac{(1-3)}{1(1-2)}$
$=\frac{-2}{-1}$
$=2$
View full question & answer→Question 124 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-\text{x}-6}{\text{x}^3+3\text{x}^2+\text{x}-3}$
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-\text{x}-6}{\text{x}^3+3\text{x}^2+\text{x}-3}$Now $\text{x}^2-\text{x}-6$
$=\text{x}^2-3\text{x}+2\text{x}-6$
$=\text{x}(\text{x}-3)+2(\text{x}-3)$
$=(\text{x}+2)(\text{x}-3)\ \cdots(\text{i})$
Dividing $\text{x}^3-3\text{x}^2+\text{x}-3\text{ by }(\text{x}-3), \text{ we get}$
Thus (x - 3) is a factor of $\text{x}^3-3\text{x}^2+\text{x}-3\ \cdots(\text{ii})$
Substituting (i) and (ii) in the given expression
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}+2)(\text{x}-3)}{\big(\text{x}^2+1\big)(\text{x}-3)}$
$=\frac{\text{x}+2}{\text{x}^2+1}=\frac{3+2}{9+1}=\frac{5}{10}$
$=\frac12$ View full question & answer→Question 134 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}$ $=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}\times\frac{\sqrt{2+\cos\text{x}}+1}{{\sqrt{2+\cos\text{x}}+1}}$ $=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{(2+\cos\text{x})-1}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{1+\cos\text{x}}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}$ Let $\pi-\text{x}=\text{y},\text{x}\rightarrow\pi,\text{y}\rightarrow0$ $\Rightarrow\lim\limits_{\text{x}\rightarrow{\pi}}\frac{1+\cos\text{x}}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}=\lim\limits_{\text{y}\rightarrow0}\frac{1+\cos\text{x}(\pi-\text{y})}{\text{y}^2\big(\sqrt{2+\cos(\pi-\text{y})+1}\big)}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{1-\cos\text{y}}{\text{y}^2\sqrt{2-\cos\text{y}+1}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\frac{2\sin^2\text{y}}{2}}{\text{y}^2\sqrt{2-\cos\text{y}}+1}$ $=2\lim\limits_{\text{y}\rightarrow0}\Bigg(\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{1}{4}\frac{1}{\sqrt{2-\cos\text{y}}+1}$ $=2\times\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{2}\Big)^2\times\frac{1}{4}\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos\text{y}}+1}$ $=2\times1\times\frac{1}{4}\times\frac{1}{\sqrt{2-\cos\text{0}}+1}$ $=2\times1\times\frac{1}{4}\times\frac{1}{\sqrt{2-1}+1}$ $=2\times1\times\frac{1}{4}\times\frac{1}{{1}+1}$ $=2\times1\times\frac{1}{4}\times\frac{1}{{2}}$ $=\frac{1}{4}$
View full question & answer→Question 144 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$ Dividing $\text{x}^4-3\text{x}^3+2\text{ by }\text{x}^3-5\text{x}^2+3\text{x}+1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ Dividing $\text{x}^3-5\text{x}^2+3\text{x}+1\text{ by }\text{x}-1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{(\text{x}-1)(\text{x}^2-4\text{x}-1)}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}}{\big(\text{x}^2-4\text{x}-1\big)}$ $=1+2+\frac{7}{(1-4-1)}$ $=3-\frac{7}{4}$ $=\frac{12-7}{4}$ $=\frac{5}{4}$ View full question & answer→Question 154 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\cos\text{x}-\sin\text{x}}{\big(\frac\pi4-\text{x}\big)(\cos\text{x}+\sin\text{x})}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\cos\text{x}-\sin\text{x}}{\big(\frac\pi4-\text{x}\big)(\cos\text{x}+\sin\text{x})}$$\Rightarrow\text{x}\rightarrow\frac{\pi}{4},$ then $\frac\pi4-\text{x}\rightarrow0$ let $\frac\pi4-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cos\big(\frac\pi4+\text{y}\big)-\sin\big(\frac\pi4+\text{y}\big)}{-\text{y}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big[\big(\cos\frac\pi4\cos\text{y}-\sin\frac\pi4\sin\text{y}\big)-\big(\sin\frac\pi4\cos\text{y}+\cos\frac\pi4\sin\text{y}\big)\Big]}{-\text{y}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big[\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}-\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}\Big]}{-\text{y}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{-2\sin\text{y}}{\sqrt{2}}}{-\text{y}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\Big(\frac{\sin\text{y}}{\text{y}}\Big)\times\frac{1}{\lim\limits_{\text{y}\rightarrow{0}}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\sqrt{2}\times1\times\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}=\sqrt{2}\times\frac{1}{\frac{2}{\sqrt{2}}}$
$=\frac{\sqrt{2}\times\sqrt{2}}{2}=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^7-2\text{x}^5+1}{\text{x}^3-3\text{x}^2+2}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^7-2\text{x}^5+1}{\text{x}^3-3\text{x}^2+2}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\big(\text{x}^6+\text{x}^5-\text{x}^4-\text{x}^3-\text{x}^2-\text{x}-1\big)}{(\text{x}-1)\big(\text{x}^2-2\text{x}-2\big)}$ $= \lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^6+\text{x}^5-\text{x}^4-\text{x}^3-\text{x}^2-\text{x}-1\big)}{\big(\text{x}^2-2\text{x}-2\big)}$ $=\frac{(1+1-1-1-1-1-1)}{(1-2-2)}$ $=\frac{-3}{-3}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-3\text{x}^2+9\text{x}-2\text{ by }\text{x}^3-\text{x}-6$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9}{\text{x}^3-\text{x}-6}=\lim\limits_{\text{x}\rightarrow2}1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-2\text{x}-6\text{x+4}}{\text{x}^3-\text{x}-6}$ $\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}-1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-\text{x}-6\text{ by}\text{ x}-2$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{(\text{x}-2)(\text{x}^2+2\text{x}+3)}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)}{(\text{x}^2-2\text{x}+3)}$ $=1+\frac{3\times2-2}{2^2+2\times2+3}$ $=1+\frac{4}{11}$ $=\frac{15}{11}$ View full question & answer→Question 184 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}\big(3\sin\text{x}-4\sin^3\text{x}\big)}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{4\sin^3\text{x}}{\text{x}^3}$ $=4\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)^3$ $=4\times1$ $=4$
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Evaluate the following limit: If $\lim\limits_{\text{x}\rightarrow0}{\text{ kx cosec x}}=\lim\limits_{\text{x}\rightarrow0}\text{ x cosec kx,}{}$ find k.
Answer$\lim\limits_{\text{x}\rightarrow0}{\text{ kx cosec x}}=\lim\limits_{\text{x}\rightarrow0}\text{ x cosec kx,}{}$ $\lim\limits_{\text{x}\rightarrow0}\text{ kx}\frac{1}{\sin\text{x}}=\lim\limits_{\text{x}\rightarrow0}\text{x}\frac{1}{\sin\text{kx}}$ $\text{k}\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\text{x}}{\sin\text{x}}\Big)=\frac{1}{\text{k}}\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\text{kx}}{\sin\text{kx}}\Big)$ $\text{k}=\frac{1}{\text{k}}$ $\text{k}^2=1$ $\text{k}=\pm1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\text{x}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}\big)}{\text{x}}\times\frac{\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}^2\big)-\big(1-\text{x}^2\big)}{\text{x}\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}^2}{\text{x}\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}^2}\big)}$
$=\frac{2\times0}{\big(\sqrt{1}+\sqrt{1}\big)}$
$=\frac{2}{2}\times0$
$=0$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{2-\text{x}}-\sqrt{2+\text{x}}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{2-\text{x}}-\sqrt{2+\text{x}}}{\text{x}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}{\text{x}\times\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(2-\text{x})-(2+\text{x})}{\text{x}\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-2\text{x}}{\text{x}\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\frac{-2}{\sqrt{2}+\sqrt{2}}$
$=\frac{-2}{2\sqrt{2}}$
$=\frac{-1}{\sqrt{2}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}+4\sin3\text{x}}{4\sin2\text{x}+7\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}+4\sin3\text{x}}{4\sin2\text{x}+7\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{5+\frac{4\sin3\text{x}}{\text{x}}}{\frac{4\sin2\text{x}}{\text{x}}+7}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}5+4\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3}{4\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2+7}$ $=\frac{5+4\times1\times3}{4\times2+7}$ $=\frac{5+12}{8+7}$ $=\frac{17}{15}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\text{cosec}^2\text{x}-2}{\cot\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\text{cosec}^2\text{x}-2}{\cot\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\cot^2\text{x}+1-2}{\cot\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\cot^2\text{x}-1}{\cot\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{(\cot\text{x}-1)(\cot\text{x}+1)}{\cot\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}(\cot\text{x}+1)$ $=\cot\frac{\pi}{4}+1$ $=1+1$ $=2$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}}-1\big)}{\text{x}}\times\frac{\big(\sqrt{1+\text{x}}+1\big)}{\big(\sqrt{1+\text{x}}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x}-1)}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{1+\text{x}}+1\big)}$ $=\frac{1}{\sqrt{1}+1}=\frac12$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{3}}}\frac{\sqrt{3}-\tan\text{x}}{\pi-3\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{3}}}\frac{\sqrt{3}-\tan\text{x}}{\pi-3\text{x}}$ If $\text{x}\rightarrow\frac{\pi}{3},\frac{\pi}{3}-\text{x}\rightarrow0,\pi-3\text{x}\rightarrow0$ Let $\frac\pi3-\text{x}=\text{y}$ they y → 0 $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{3}-\tan\big(\frac{\pi}{3}-\text{y}\big)}{3\big(\frac{\pi}{3}-\text{x}\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\begin{pmatrix}\frac{\bigg(\sqrt{3}-\frac{\tan\frac\pi3-\tan\text{y}}{1+\tan\frac\pi3.\tan\text{y}}\bigg)}{3\text{y}}\end{pmatrix}$ $=\lim\limits_{\text{y}\rightarrow{0}}\begin{pmatrix}\frac{\Big(\sqrt{3}-\frac{\sqrt{3}-\tan\text{y}}{1+\sqrt{3}\tan\text{y}}\Big)}{3\text{y}}\end{pmatrix}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big(\sqrt{3}-\tan\text{y}-\sqrt{3}+\tan\text{y}\big)}{3\big(1+\sqrt{3}\tan\text{y}\big)\text{y}}$ $=\frac{4}{3}\times\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\Big(1+\sqrt{3}\frac{\tan\text{y}}{\text{y}}\times\text{y}\Big)}$ $=\frac{4\times1}{3}\times\frac{1}{1+0}$ $=\frac{4}{3}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow7}\frac{4-\sqrt{9+\text{x}}}{1-\sqrt{8-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow7}\frac{4-\sqrt{9+\text{x}}}{1-\sqrt{8-\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow7}\frac{\big(4-\sqrt{9+\text{x}}\big)}{1-\sqrt{8-\text{x}}}\times\frac{\big(4+\sqrt{9+\text{x}}\big)}{\big(4+\sqrt{9+\text{x}}\big)}\times\frac{\big(1+\sqrt{8-\text{x}}\big)}{\big(\sqrt{1+\sqrt{8+\text{x}}}\big)}$ $=\lim\limits_{\text{x}\rightarrow7}\frac{\big((4)^2-\big(\sqrt{9+\text{x}}\big)^2\big)}{\big((1)^2-\big(\sqrt{8-\text{x}}\big)^2\big)}\times\frac{1+\sqrt{8-\text{x}}}{4+\sqrt{9+\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow7}\frac{(16-9-\text{x})\times\big(1+\sqrt{8-\text{x}}\big)}{(1-8+\text{x})\times\big(4+\sqrt{9+\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow7}\frac{7-\text{x}}{(-7)(7-\text{x})}\frac{\big(1+\sqrt{8-\text{x}}\big)}{\big(4+\sqrt{9+\text{x}}\big)}$ $=\frac{1}{(-1)}\times\frac{(1+1)}{(4+4)}=\frac{-1}{4}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\cos2\text{x}-\cos8\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\cos2\text{x}-\cos8\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{-2\sin\big(\frac{2\text{x}+8\text{x}}{2}\big)\sin\big(\frac{2\text{x}-8\text{x}}{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin^2\text{x}}{\sin5\text{x}\times\sin(-3\text{x})}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}\sin^2\text{x}}{-\big(\lim\limits_{\text{x}\rightarrow0}\sin5\text{x}\big)\big(-\lim\limits_{\text{x}\rightarrow0}\sin3\text{x}\big)}$ $=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}^2}{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}}{5\text{x}}\times5\text{x}\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\big)\times3\text{x}}$ $=\frac{1\times\text{x}^2}{1\times5\text{x}\times1\times3\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=\frac{\text{x}^2}{15\text{x}^2}$ $=\frac{1}{15}$
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Evaluate the following limit: If $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-1}{\text{x}-1},$ find all possible value of a.
AnswerIf $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-1}{\text{x}-1}\ \cdots{\text{(i})}$ $\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}$ $=3(\text{a})^{3-1}$ $=3\text{a}^{2}\ \cdots{\text{(ii})}$ $\text{R.H.S}=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^4-1}{\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^4-1}{\text{x}-1}$ $=4(1)^{4-1}$ $=4\ \cdots{(\text{iii})}$ Substituting (ii) and (iii) in (i), $3\text{a}^8=4$ $\Rightarrow\text{a}^{2}=\frac43$ $\Rightarrow\text{a}=\pm\frac{2}{\sqrt{3}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^2-\text{x}-2}{\big(\text{x}^2+\text{x}\big)+\sin(\text{x}+1)}$
Answer$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^2-\text{x}-2}{\big(\text{x}^2+\text{x}\big)+\sin(\text{x}+1)}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{(\text{x}-2)(\text{x}+2)}{\text{x}(\text{x}+1)+\sin(\text{x}+1)}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\frac{\text{x}(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}+\frac{\sin(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\frac{\text{x}}{\text{x}-2}+\frac{\sin(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{(\text{x}-2)}\Bigg(\frac{1}{\text{x}+\frac{\sin(\text{x}+1)}{\text{x}+1}}\Bigg)$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\text{x}-2}\times\frac{1}{\lim\limits_{\text{x}\rightarrow-1}(\text{x})+\lim\limits_{\text{x}\rightarrow-0}\sin\frac{\text{x}+1}{\text{x}+1}}$ $=\Big(\frac{1}{-1-2}\Big)\times\frac{1}{(-1)+1}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac{1}{0}$ $\Big[\because\frac10=\infty\Big]$ $=\infty$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\infty}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}\sqrt{\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}\sqrt{\text{x}+2}$ $=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big[\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big]\times\frac{\big[\sqrt{\text{x}+1}+\sqrt{\text{x}}\big]}{\big[\sqrt{\text{x}+1}+\sqrt{\text{x}}\big]}\times\frac{\sqrt{\text{x}+2}\times\sqrt{\text{x}+2}}{\sqrt{\text{x}+2}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{(\text{x}+1-\text{x})}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\times\frac{(\text{x}+2)}{\sqrt{\text{x}+2}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{1(\text{x}+2)}{\text{x}\big(\sqrt{\text{x}+1}+\sqrt{{\text{x}}}\big)\big(\sqrt{\text{x}+2}\big)}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{x}\Big(1+\frac{2}{\text{x}}\Big)}{\sqrt{\text{x}}\bigg(\sqrt{1+\frac{1}{\text{x}}}+1\bigg)\bigg(\sqrt{1+\frac{2}{\text{x}}}\bigg)\sqrt{\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\Big(1+\frac{2}{\text{x}}\Big)}{\bigg(\sqrt{1+\frac{1}{\text{x}}}+\sqrt{1}\bigg)\bigg(\sqrt{1+\frac{2}{\text{x}}}\bigg){}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{(1+0)}{(1+1)\times1}=\frac12$
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Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\text{n}\sin\Big(\frac{\pi}{4\text{n}}\Big)\cos\Big(\frac{\pi}{4\text{n}}\Big)$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\text{n}\sin\Big(\frac{\pi}{4\text{n}}\Big)\cos\Big(\frac{\pi}{4\text{n}}\Big)$ $=\lim\limits_{\text{n}\rightarrow\infty}2\Big(\text{n}\sin\frac{\pi}{4\text{n}}\cos\frac{\pi}{4\text{n}}\Big)\times\frac12$ $=\lim\limits_{\text{n}\rightarrow\infty}\text{n}\times\sin\frac{\pi}{2\text{n}}\times\frac12$ $\text{n}\rightarrow\infty,$ then $\frac{1}{\text{n}}\rightarrow0, $ let $\frac{1}{\text{n}}=\text{y}$ $=\frac12\lim\limits_{\frac{1}{\text{n}}\rightarrow\infty}\frac{1}{\text{y}}\sin\Big(\frac{\pi}{2}\Big)\Big(\frac{1}{\text{n}}\Big)$ $=\frac12\lim\limits_{{\text{y}}\rightarrow\infty}\frac{\sin\big(\frac\pi2\big)\text{y}}{\text{y}}$ $=\frac12\Bigg(\lim\limits_{{\text{y}}\rightarrow\infty}\frac{\sin\big(\frac{\pi\text{y}}2\big)}{\frac{\pi\text{y}}{2}}\Bigg)\times\frac\pi2$ $=\frac12\times1\times\frac\pi2$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac\pi4$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+3\Big)\Big(\text{x}^{\frac{1}{3}}-3\Big)}{\text{x}-27}$
Answer$\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+3\Big)\Big(\text{x}^{\frac{1}{3}}-3\Big)}{\text{x}-27}$ $=\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+9\Big)}{\text{x}-27}$ $=\lim\limits_{\text{x}\rightarrow{27}}\frac{\text{x}^\frac{2}{3}-27^{\frac{2}{3}}}{\text{x}-27}$ Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$ $=\frac23(27)^{\frac23-1}$ $=\frac23(27)^{\frac{-1}{3}}$ $=\frac23\times\frac{1}{(27)^{\frac{1}{3}}}$ $=\frac{2}{3}\times\frac{1}{3}$ $=\frac29$
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Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}{\Big(\frac{1}{\text{n}^2}+\frac{2}{\text{n}^2}+\frac{3}{\text{n}^2}+\ \cdots+\frac{\text{n}-1}{\text{n}^2}}{}\Big)$
Answer$\lim\limits_{\text{n}\rightarrow\infty}{\Big(\frac{1}{\text{n}^2}+\frac{2}{\text{n}^2}+\frac{3}{\text{n}^2}+\ \cdots+\frac{\text{n}-1}{\text{n}^2}}{}\Big)$$=\lim\limits_{\text{n}\rightarrow\infty}\Big(\frac{1+2+3+\ \cdots+(\text{n}-1)}{{n}^2}\Big)$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\text{n}+1\big)(\text{n})}{2\times\text{n}^2}$ $\Big[1+2+3+\ \cdots+({\text{n}}-1)=\frac{({\text{n}}-1)({\text{n}})}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{n}^2-\text{n}}{2\text{n}^2}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\lim\limits_{\text{n}\rightarrow{\infty}}\frac{1-\frac{1}{\text{n}}}{2}$
$=\frac{1-0}{2}=\frac12$
$=\frac{1}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\cot^2\text{x}-3}{\text{cosec x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\cot^2\text{x}-3}{\text{cosec x}-2}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec}^2\text{x}-1\big)-3}{\text{cosec x}-2}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec}^2\text{x}-4\big)}{\text{cosec x}-2}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec }\text{x}-2\big)\big(\text{cosec }\text{x}+2\big)}{\text{cosec x}-2}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\text{cosec }\text{x}+2$ $=\text{cosec}\frac{\pi}{6}+2$ $=2+2$ $=4$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big\{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big\}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big\{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big\}$ $=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big[\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big]\times\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}\big)}{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2+1}\big)}$ $=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\frac{\text{x}\big(\text{x}^2+1-\text{x}^2+1\big)}{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}(2)}{\text{x}\Big(\sqrt{1+\frac{1}{\text{x}^2}}+\sqrt{1-\frac{1}{\text{x}^2}}\Big)}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{2}{\sqrt{1+\frac{1}{\text{x}^2}}+\sqrt{1-\frac{1}{\text{x}^2}}}$ $=\frac{2}{2}=1$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1+\cos\text{x}}{\tan^2\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1+\cos\text{x}}{\tan^2\text{ x}}$ $\Rightarrow\text{x}\rightarrow{\pi},\text{x}-{\pi}\rightarrow0,$ let $\text{y}=\text{x}-\pi$ $\Rightarrow\lim\limits_{\text{x}-{\pi}\rightarrow{{0}}}\frac{1+\cos\text{x}}{\tan^2\text{x}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos(\pi+\text{y})}{\tan^2(\pi+\text{y})}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos\text{y}}{\tan^2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\tan^2\text{y}}$ $=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{\text{y}^2}{4}\times\frac{1}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan\text{y}}{\text{y}}\Big)^2\times\text{y}^2}$ $=2\times1\times\frac{\text{y}^2}{4}\times\frac{1}{1\times\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$ $=\frac{1}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\big(\sqrt{1+\text{x}}-\sqrt{1-\text{x}}\big)}\times\frac{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{\big(\sqrt{1+\text{x}}\big)^2-\big(\sqrt{1-\text{x}}\big)^2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{1+\text{x}-1+\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{2\text{x}}$ $=\frac12\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{\text{x}}\Big)\text{x}$ $=\frac{1}{2}\lim\limits_{\text{x}\rightarrow0}{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}$ $=\frac{1}{2}\big(\sqrt{1}+\sqrt{1}\big)$ $=\frac{1}{2}(1+1)=\frac22$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\pi}\frac{\sqrt{5+\cos\text{x}-2}}{(\pi-\text{x})^2}$
Answer$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sqrt{5+\cos\text{x}-2}}{(\pi-\text{x})^2}$ $⇒ \text{x} → \pi,$ then $\pi-\text{x}\rightarrow0,$ let $\pi-\text{x}=\text{y}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5+\cos(\pi-\text{y})}-2}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5-\cos\text{y}}-2}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5-\cos\text{y}}-2}{\text{y}^2}\times\frac{\big(\sqrt{5-\cos\text{y}}+2\big)}{\big(\sqrt{5-\cos\text{y}}+2\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{(5-\cos\text{y}-4)}{\text{y}^2\big(\sqrt{5-\cos\text{y}}+2\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{5-\cos\text{y}}+2\big)}$ $=2\times\frac14\times\frac{1}{\sqrt{4}+2}=2\times\frac14\times\frac14$ $=\frac18$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\sqrt{\text{x}}-\sqrt{2}}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\sqrt{\text{x}}-\sqrt{2}}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)\big(\sqrt{\text{x}}+\sqrt{2}\big)}{\big(\sqrt{\text{x}}-\sqrt{2}\big)\big(\sqrt{\text{x}}+\sqrt{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)\big(\sqrt{\text{x}}+\sqrt{2}\big)}{(\text{x}-2)}$ $=\sqrt{2}+\sqrt{2}$ $=2\sqrt{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos2\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\tan\text{x}}{2\sin^2\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\frac{\tan\text{x}}{\text{x}}}{\frac{2\sin^2\text{x}}{\text{x}^2}}$ $=\frac{\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}}{2\lim\limits_{\text{x} \rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$ $=\frac{1}{2\times1}$ $=\frac{1}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-1}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{1-2\sin^2\big(\frac{\text{ax}}{2}\big)-1+2\sin^2\big(\frac{\text{bx}}{2}\big)}{1-2\sin^2\big(\frac{\text{cx}}{2}\big)-1}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin^2\big(\frac{\text{ax}}{2}\big)+2\sin^2\big(\frac{\text{bx}}{2}\big)}{-2\sin^2\big(\frac{\text{cx}}{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{-\sin^2\big(\frac{\text{ax}}{2\text{ax}}\big)4\text{a}^2\text{x}^2+\sin\big(\frac{\text{bx}}{2}\big)4\text{b}^2\text{x}^2}{-\sin^2\big(\frac{\text{cx}}{2}\big)4\text{c}^2\text{x}^2}$ $=\frac{-\text{a}^2+\text{b}^2}{-\text{c}^2}$ $=\frac{\text{a}^2-\text{b}^2}{\text{c}^2}$
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Evaluate the following limit: If $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x}),$ find all possible value of a.
AnswerIf $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x})\ \cdots{\text{(i})}$ $\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}$ $=9(\text{a})^{9-1}$ $=9\text{a}^{8}\ \cdots{\text{(ii})}$ $\text{R.H.S}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x})$ $=4+5=9\ \cdots{\text{(iii})}$ Substituting (ii) and (iii) in (i) $9\text{a}^8=9$ $\Rightarrow\text{a}^{8}=1$ $\Rightarrow\text{a}^4=1$ $\text{a}^2=1$ $\Rightarrow\text{a} = 1\text{ and a} = -1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin2\pi\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin2\pi\text{ x}}$ ⇒ x → 1, then x - 1 →0, let x - 1 = y $=\lim\limits_{(\text{x}-1)\rightarrow{0}}\frac{(1-\text{x})(1+\text{x})}{\sin2\pi\text{ x}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{-\text{y}(1+\text{y}+1)}{\sin2\pi(\text{y}+1)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin(2\pi\text {y}+2\pi)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin2\pi\text {y}}$ $=-\lim\limits_{\text{y}\rightarrow{0}}(\text{y}+2)\times\frac{\text{y}}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\sin\frac{2\pi\text {y}}{\text {y}\times2\pi}\Big)\times2\pi\text {y}}$ $=-2\times\frac{1}{1\times2\pi}$ $=-\frac{1}{\pi}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}}{2\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}}{2\text{x}^2}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}\big)\times\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}+\text{x}^2\big)-(\text{x}+1)}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $=\frac{1}{2\big(\sqrt{1}+\sqrt{1}\big)}$ $=\frac{1}{2\times2}$ $=\frac14$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$ $\Rightarrow\text{x}\rightarrow\frac{\pi}{4},\text{x}-\frac{\pi}{4}\rightarrow0,$ let $\text{x}-\frac{\pi}{4}=\text{y}$ $\Rightarrow\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{x}}=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\frac{\Big(1-\sin2\big(\text{y}+\frac{\pi}{4}\big)\Big)}{1+\cos4\big(\text{y}+\frac{\pi}{4}\big)}$ $=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\Bigg(\frac{1-\sin\big(\frac\pi2+2\text{y}\big)}{1+\cos(\pi+4\text{y})}\Bigg)$ $=\lim\limits_{\text{x}-\frac\pi4\rightarrow{0}}\frac{1-\cos2\text{y}}{1-\cos4\text{y}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\text{y}}{2\sin^22\text{y}}$ $=\frac{\lim\limits_{\text{y}\rightarrow{0}}\sin^2\text{y}}{\lim\limits_{\text{y}\rightarrow{0}}\sin^22\text{y}}$ $=\frac{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}\Big)^2\times\text{y}^2}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\text{y}^2}$ $=\frac{1\times\text{y}^2}{1\times4\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac14$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$ $=\frac{2\lim\limits_{\text{x}\rightarrow0}\sin^2\text{x}+\lim\limits_{\text{x}\rightarrow0}\tan^2\text{x}}{\lim\limits_{\text{x}\rightarrow0}\text{x}\sin\text{x}}$ $=\frac{\Big(2\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}^2\Big)+\big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}\big)^2\times\text{x}^2}{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)\times\text{x}^2}$ $=\frac{\big(2\times1\times\text{x}^2\big)+\big(1\times\text{x}^2\big)}{\big(1\times\text{x}^2\big)}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ and }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=\frac{3\text{x}^2}{\text{x}^2}$ $=3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{\sqrt{2+\cos\text{x}-1}}{(\pi-\text{x})^2}$
Answer$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{\sqrt{2+\cos\text{x}-1}}{(\pi-\text{x})^2}$ $\text{x}\rightarrow{\pi},$ then $\text{x}-{\pi}\rightarrow0,$ let $\text{x}-\pi=\text{y}$ $\Rightarrow\lim\limits_{{\text{x}\rightarrow{\pi}}}\frac{\big(\sqrt{2+\cos\text{x}}-1\big)}{(\pi-\text{x})^2}=\lim\limits_{{\text{x}-{\pi}}\rightarrow0}\frac{\sqrt{2+\cos(\text{x})}-1}{(-1)^2(\text{x}-\pi)^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2+\cos(\pi+\text{y})}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2-\cos\text{y}}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\big(\sqrt{2-\cos\text{y}}-1\big)\big(\sqrt{2-\cos\text{y}}+1\big)}{\text{y}^2\big(\sqrt{2\cos\text{y}}+1\big)}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{(2\cos\text{y}-1)}{\big(\sqrt{2-\cos\text{y}+1}\big)\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{(1-\cos\text{y})}{\big(\sqrt{2-\cos\text{y}}+1\big)\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=2\lim\limits_{\text{y}\rightarrow0}\Bigg(\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos0+1}}$ $=2\times\frac14\times\frac{1}{\sqrt{2}-1+1}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac{1}{4}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-\sqrt{\text{x}}}{\sqrt{\text{x}}-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-\sqrt{\text{x}}}{\sqrt{\text{x}}-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^2-\sqrt{\text{x}}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}{\big(\sqrt{\text{x}}-1\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-\text{x}}{\big(\sqrt{\text{x}-1}\big)\big({\text{x}^2}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\text{x}^3-1\big)}{\big(\sqrt{\text{x}-1}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}(\text{x}-1)\big(\text{x}^2+1+\text{x}\big)}{\big(\sqrt{\text{x}-1}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\sqrt{\text{x}+1}\big)\big({\text{x}^2+1+\text{x}}\big)}{\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\frac{1(1+1)(1+1+1)}{1+1}$
$=\frac62$
$=3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{ x}-3}{\text{cosec}\text{ x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{x}-3}{\text{cosec }\text{x}-2}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{x}-4}{\text{cosec }\text{x}-2}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{(\text{cosec }\text{x}-2)(\text{cosec x}+2)}{(\text{cosec }\text{x}-2)}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}(\text{cosec }\text{x}+2)$ $=\text{cosec}\frac{\pi}{6}+2$ $=2+2=4$
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Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \dots+\text{n}^3}{(\text{n}-1)^4}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+3^3\ \dots+\text{n}^3}{(\text{n}-1)^4}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big[\frac12(\text{n})(\text{n}+1)\Big]^2}{(\text{n}-1)^4}$ $\bigg[1^3+2^3+3^3+\ \cdots+\text{n}^3=\Big(\frac12\text{n}(\text{n}+1)\Big)^2\bigg]$ $=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\frac{\frac14\text{n}^2\big(\text{n}^2+1+2\text{n}\big)}{(\text{n}-1)^4}\Bigg)$ $=\frac14\lim\limits_{\text{n}\rightarrow\infty}\bigg(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{(\text{n}-1)^2(\text{n}-1)^2}\bigg)$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\Big(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{\text{n}^4+\text{n}^2-2\text{n}^3+\text{n}^2+1-2\text{n}-2\text{n}^3-2\text{n}+4\text{n}^2}\Big)$ $=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}\Big)}{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}+\frac{1}{\text{n}^2}+\frac{1}{\text{n}^4}-\frac{2}{\text{n}^3}-\frac{2}{\text{n}}-\frac{2}{\text{n}^3}+\frac{4}{\text{n}^2}\Big)}$ $=\frac{1}{4}\Big(\frac14\Big)$ $=\frac14$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{cosec x}-\cot\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{cosec x}-\cot\text{x}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big)\times\frac{1}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1}{\sin\text{x}}\Big(\frac{1-\cos\text{x}}{\text{x}}\Big)\Big)$ $=\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{1}{\sin\text{x}}\bigg(\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}}\bigg)\bigg)$ $=2\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{1}{\frac{\sin\text{x}}{\text{x}}}\times\text{x}\bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)\times\frac{\text{x}}{4}\Bigg)$ $=2\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{1}{\frac{\sin\text{x}}{\text{x}}}\bigg)\times\frac{1}{\text{x}}\times\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)\times\frac{\text{x}}{4}$ $=2\times\frac{1}{\text{x}}\times\frac{\text{x}}{4}$ $=\frac{1}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-\cos\text{dx}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-\cos\text{dx}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{{\Big(-2\sin\big(\frac{\text{a}+\text{b}}{2}\big)\times\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\Big)}}{-2\sin\big(\frac{\text{c}+\text{d}}{2}\big)\times\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}{\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}}{\lim\limits_{\text{x}\rightarrow0}{\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}}$ $=\frac{\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}{\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}\times\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\Bigg)\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}}{\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}}\times\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}\Bigg)}{\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}{\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}\times\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}\Bigg)\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}{\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}\times\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}\Bigg)}$ $=\frac{(\text{a}+\text{b})(\text{a}-\text{b})}{(\text{c}+\text{d})(\text{c}-\text{d})}$ $\Big[\because\ \lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}$$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\frac{\text{x}^\frac{3}{4}-\text{a}^\frac{3}{4}}{\text{x}-\text{a}}}$ [Dividing numerator and denominator by x - a]
$=\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\frac{3}{4}-\text{a}^\frac{3}{4}}{\text{x}-\text{a}}}$
Applying the formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$ in numerator and $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{m}}-\text{a}^{\text{m}}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator respectively
Here, $\text{n}=\frac{2}{3},\text{m}=\frac{3}{4}$
$\Rightarrow\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{x}}}\frac{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}{\text{x}-\text{a}}}=\frac{\frac{2}{3}(\text{a})^{\frac{2}{3}-1}}{\frac{3}{4}(\text{a})^{\frac{3}{4}-1}}$
$=\frac{8}{9}\text{a}^{\frac{-1}{3}+\frac{1}{4}}$
$=\frac89\text{a}^{\frac{-1}{12}}$
View full question & answer→Question 544 Marks
Find $\lim\limits_{\text{x}\rightarrow-\frac{5}{2}}{[\text{x}]}.$
Answer$\lim\limits_{\text{x}\rightarrow-\frac{5}{2}}{[\text{x}]}$ $\lim\limits_{\text{x}\rightarrow\frac{5}{2}}{[\text{x}]}=\Big[\frac{5}{2}\Big],$ $=[2.5]=2$ [By definition of geatest integer function] $\Rightarrow\ \lim\limits_{\text{x}\rightarrow\frac{5}{2}}{[\text{x}]}=2$
View full question & answer→Question 554 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{a}\sin\text{x}-\text{x}\sin\text{a})}{\text{ax}(\text{x}-\text{a})}$ If t = x - a Then, as x → a, t → 0 $\therefore\ \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{a}\sin\text{x}-\text{x}\sin\text{a})}{\text{ax}(\text{x}-\text{a})}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\big(\text{a}\sin(\text{t}+\text{a})-(\text{t}+\text{a})\sin\text{a}\big)}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\text{a}\sin\text{t}\cos\text{a}+\text{a}\sin\text{a}\cot\text{t}-\text{t}\sin\text{a}-\text{a}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{a}}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\text{a}\sin\text{t}\cos\text{a}+\text{a}\sin\text{a}(\cos\text{t}-1)-\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\cos\text{a}+\text{a}\sin\text{a}\Big(2\sin^2\big(\frac{\text{t}}{2}\big)\Big)-\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\text{t}\cos\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}+\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\text{a}\Big(2\sin^2\big(\frac{\text{t}}{2}\big)\Big)}{\text{a}(\text{t}+\text{a})\text{t}}-\lim\limits_{\text{t}\rightarrow0}\frac{\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\frac{\text{a}\cos\text{a}}{\text{a}^2}+0-\frac{\sin\text{a}}{\text{a}^2}$ $=\frac{\text{a}\cos\text{a}-\sin\text{a}}{\text{a}^2}$
View full question & answer→Question 564 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{3+\text{x}}-\sqrt{5-\text{x}}\big)}{\big(\text{x}^2-1\big)}\times\frac{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(3+\text{x})-(5-\text{x})}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2+2\text{x}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{2}{(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\frac{2}{(1+1)\big(\sqrt{3+1}+\sqrt{5-1}\big)}=\frac{2}{(2)(2+2)}$
$=\frac14$
View full question & answer→Question 574 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}^\circ-\sin2\text{x}^\circ}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}^\circ-\sin2\text{x}^\circ}{\text{x}^3}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}-\sin\frac{2\pi\text{x}}{180}}{\text{x}^3}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}-2\sin\frac{\pi\text{x}}{180}\cos\frac{\pi\text{x}}{180}}{\text{x}^3}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}\big(2\sin^2\frac{\pi\text{x}}{360}\big)}{\text{x}^3}$ $=4\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\text{x}}\bigg)\times\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\text{x}}\bigg)\times\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\text{x}}\bigg)$ $=4\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\frac{\pi\text{x}}{180}}\times\frac{\pi}{180}\Bigg)\times\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\frac{\pi\text{x}}{360}}\times\frac{\pi}{360}\Bigg)\times\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\frac{\pi\text{x}}{360}}\times\frac{\pi}{360}\Bigg)$ $=4\times\frac{\pi}{180}\times\frac{\pi}{360}\times\frac{\pi}{360}$ $=\Big(\frac{\pi}{180}\Big)^3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\text{x}^2+3\text{x}-6}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\text{x}^2+3\text{x}-6}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\big[\text{x}^2+\text{x}-2\big]}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\big[\text{x}^2+2\text{x}-\text{x}-2\big]}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3(\text{x}+2)(\text{x}-1)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3(\text{x}+2)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)}{3(\text{x}+2)\big(\sqrt{\text{x}}+1\big)}$ $=\frac{(2-3)}{3(1+2)(1+1)}$ $=\frac{-1}{3\times3\times2}$ $=-\frac{1}{18}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}\times\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{\big({\text{x}^2+1}-{5}\big)}{(\text{x}-2)\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)(\text{x}-2)}{(\text{x}-1)\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)}{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$ $=\frac{(2+2)}{\sqrt{4+1}+\sqrt{5}}$ $=\frac{4}{2\sqrt{5}}=\frac{2}{\sqrt{5}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)^2}$ $\Rightarrow\ \text{x}\rightarrow\frac{\pi}{2},$ then $\frac{\pi}{2}-\text{x}\rightarrow0,$ let $\frac{\pi}{2}-\text{x}=\text{y}$ $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}\rightarrow0}=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2-\sin\big(\frac{\pi}{2}-\text{y}\big)}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2-\cos\text{y}}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big(\sqrt{2-\cos\text{y}}-1\big)}{\text{y}^2}\times\frac{\big(\sqrt{2-\cos\text{y}}+1\big)}{\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big({2-\cos\text{y}}-1\big)}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{{1-\cos\text{y}}}{\big(\sqrt{2-\cos\text{y}}+1\big)\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=2\times\Bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)\times\frac14\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos\text{y}}+1}$ $=2\times1\times\frac14\times\frac{1}{1+1}=\frac14$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}},\text{x}>1$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}\times\frac{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}\times\frac{\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^-1}}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big[\big(\text{x}^2-1\big)-(\text{x}-1)\big]\times\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^2-\text{x}\big)\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}(\text{x}-1)\sqrt{\text{x}^2-1}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}{(\text{x}+1)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}-1\big)}$ $=\frac{\sqrt{2}}{2\big(\sqrt{2}-1\big)}$ $=\frac{\sqrt{2}}{2\times\big(\sqrt{2}-1\big)}\times\frac{\sqrt{2}+1}{\sqrt{2}+1}$ $=\frac{\sqrt{2}+1}{\sqrt{2}}$
View full question & answer→Question 624 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}-\sin2\text{x}}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}-\sin2\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin2\text{x}}{\cos2\text{x}}-\sin2\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}\big(\frac{1}{\cos2\text{x}}-1\big)}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(1-\cos2\text{x})}{\text{x}^2\cos2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}\big(2\sin^2\text{x}\big)}{\text{x}^3\cos2\text{x}}$ $=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{x}}\big)\Big(\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{\text{x}^2}\Big)}{\big(\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}\big)}$ $=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{x}}\times2\big)\Big(\lim\limits_{\text{x}\rightarrow0}2\big(\frac{\sin\text{x}}{\text{x}}\big)^2\Big)}{\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}}$ $=\frac{(2\times1)(2\times1)}{1}$ $=4$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^3+\text{x}^2+4\text{x}+12}{\text{x}^3+3\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^3+\text{x}^2+4\text{x}+12}{\text{x}^3+3\text{x}+2}$ $=\lim\limits_{\text{x}\rightarrow-2}\frac{(\text{x}+2)\big(\text{x}^2+\text{x}+6\big)}{(\text{x}+2)\big(\text{x}^2-2\text{x}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^2+\text{x}+6}{\text{x}^2-2\text{x}+1}$ $=\frac{(-2)^2-(-2)+6}{(-2)^2-2(-2)+1}$ $=\frac{4+2+6}{4+4+1}$ $=\frac{12}{9}$ $=\frac{4}{3}$
View full question & answer→Question 644 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^{3}+1}{\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^{3}+1}{\text{x}+1}$$=\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^3-(-1)^3}{\text{x}-(-1)}$ [Dividing numerator and denominator by x - 1]
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
Here, n = 3, a = -1
$\Rightarrow\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3-(-1)^3}{\text{x}-(-1)}=\text{na}^{\text{n}-1}$
$=3(-1)^{3-1}$
$=3(-1)^2$
$=3$
View full question & answer→Question 654 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}}{\text{x}\sqrt{\text{a}^2+\text{ax}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}}{\text{x}\sqrt{\text{a}^2+\text{ax}}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}\big)}{\text{x}\sqrt{\text{a}^2+\text{ax}}}\times\frac{\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}{\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a}+\text{x})-\text{a}}{\text{x}\sqrt{\text{a}^2+\text{ax}}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{x}\sqrt{\text{a}^2+\text{ax}}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{\text{a}^2+\text{ax}}\big)\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\frac{1}{\text{a}\big(\sqrt{2\text{a}}\big)}$
$=\frac{1}{2\text{a}\sqrt{\text{a}}}$
View full question & answer→Question 664 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}+7\text{x}}{4\text{x}+\sin2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}+7\text{x}}{4\text{x}+\sin2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\frac{\sin3\text{x}}{\text{x}}+\frac{7\text{x}}{\text{x}}\big)}{\big(\frac{4\text{x}}{\text{x}}+\frac{\sin2\text{x}}{\text{x}}\big)}$ $=\frac{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3\Big)+7}{4+\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\Big)\times2}$ $=\frac{3+7}{4+2}$ $=\frac{10}{6}$ $=\frac53$
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Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\frac{\sin\big(\frac{\text{a}}{2^{\text{n}}}\big)}{\sin\big(\frac{\text{b}}{2^{\text{n}}}\big)}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{\sin\big(\frac{\text{a}}{2^{\text{n}}}\big)}{\sin\big(\frac{\text{b}}{2^{\text{n}}}\big)}$ $\text{n}\rightarrow\infty,$ then $\frac{1}{\text{n}}=\text{h} \rightarrow0$ $=\frac{\lim\limits_{\text{h}\rightarrow\infty}\sin\Bigg(\frac{\text{a}}{2^{\frac{1}{\text{h}}}}\Bigg)}{\lim\limits_{\text{h}\rightarrow\infty}\sin\Bigg(\frac{\text{b}}{2^{\frac{1}{\text{h}}}}\Bigg)}$ $=\frac{\begin{pmatrix}\lim\limits_{\text{h}\rightarrow\infty}\frac{\sin\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}\end{pmatrix}}{\begin{pmatrix}\lim\limits_{\text{h}\rightarrow\infty}\frac{\sin\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{b}}{2^{\frac{1}{\text{h}}}}\end{pmatrix}}$ $=\frac{1\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{1\times\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}=\frac{\text{a}}{\text{b}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1-\cos2\text{x}}{1-\cos\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^2\text{x}}{2\sin^2\frac{\text{x}}{2}}$ $=\frac{\lim\limits_{\text{x} \rightarrow0}(\sin\text{x})^2}{\lim\limits_{\text{x} \rightarrow0}\big(\sin\frac{\text{x}}{2}\big)^2}$ $=\frac{\lim\limits_{\text{x} \rightarrow0}\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)^2}{\lim\limits_{\text{x} \rightarrow0}\big(\sin\frac{\text{x}}{2}\big)^2}$ $=4\lim\limits_{\text{x} \rightarrow0}\cos^2\frac{\text{x}}{2}$ $=4\times1$ $=4$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}$ $\Rightarrow\text{x}\rightarrow1,\text{x}-1\rightarrow0,$ let $\text{x}-1=\text{y}\Rightarrow\text{y}\rightarrow0$ $=\lim\limits_{\text{x}\rightarrow{0}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}=\lim\limits_{\text{x}-1\rightarrow{0}}\frac{(1-\text{x})(1+\text{x})}{\sin\pi\text{ x}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{(-\text{y})(1+\text{y}+1)}{\sin\pi(\text{y}+1)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin\pi(\text{y}+1)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{-\sin\pi(\text{y}+1)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\frac{\sin\pi\text{y}}{2}}$ $=\frac{\lim\limits_{\text{y}\rightarrow{0}}\text{y}(\text{y}+2)}{\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\pi\text{y}}{\pi\text{y}}\times\pi\text{y}}$ $=\frac{2}{\pi}$
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Evaluate the following limit: Evaluate: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1.2+2.3+3.4+\ \cdots+\text{n}(\text{n}+1)}{\text{n}^3}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1.2+2.3+3.4+\ \cdots+\text{n}(\text{n}+1)}{\text{n}^3}$ $=\ \lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}}{\text{n}^3}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{n}(\text{n}+1)\Big[\frac{(2\text{n}+1)+3}{6}\Big]}{\text{n}^3}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(\text{n}+1)(2\text{n}+4)}{6}}{\text{n}^3}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(1+\frac{1}{\text{n}}\Big)\Big(2+\frac{4}{\text{n}}\Big)}{6}$ $=\frac{1\times2}{6}$ $=\frac{1}{3}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$As $\text{x}\rightarrow\frac{\pi}{4},\frac\pi4-\text{x}\rightarrow0,$ let $\frac{\pi}{4}-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\cos\big(\frac\pi4-\text{y}\big)-\sin\big(\frac\pi4-\text{y}\big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\big[\cos\frac\pi4\cos\text{y}+\sin\frac{\pi}{4}\sin\text{y}+\sin\frac\pi4\cos\text{y}-\cos\frac\pi4\sin\text{y}\big]}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\Big(\frac{\cos\text{y}}{\sqrt{2}}+\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}\Big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{2\cos\text{y}}{\sqrt{2}}}{\text{y}^2}=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\sqrt{2}\cos\text{y}}{\text{y}^2}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\frac{(1-\cos\text{y})}{\text{y}^2}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\frac{\text{y}^2}{4}}\times\frac14$
$=\sqrt{2}\times2\times\frac14\bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\bigg)^2$
$=\sqrt{2}\times2\times\frac14\times1$
$=\frac{1}{\sqrt{2}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$$=\lim\limits_{\text{x} \rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}-\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\text{x}^2}{3\text{x}^2}$
$=\lim\limits_{\text{x} \rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^2-\frac{2}{3}\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}$
$=1-\frac{2}{3}\times1$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=1-\frac23$
$=\frac{3-2}{3}$
$=\frac{1}{3}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}}{\text{x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}}{\text{x}-2}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{\big(\sqrt{1+4\text{x}}-\sqrt{5+2\text{x}}\big)}{(\text{x}-2)}\times\frac{\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}{\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(1+4\text{x})-(5+2\text{x})}{(\text{x}-2)\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{-4+2\text{x}}{(\text{x}-2)\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$ $=\frac{2}{\sqrt{1+8}+\sqrt{5+4}}=\frac{2}{\sqrt{9}+\sqrt{9}}$ $=\frac{2}{6}=\frac13$
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Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{x}^4+7\text{x}^3+46\text{x}+\text{a}}{\text{x}^4+6},$ where a is a non-zero real number.
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{x}^4+7\text{x}^3+46\text{x}+\text{a}}{\text{x}^4+6}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{1+\frac{7}{\text{x}}+\frac{46}{\text{x}^3}+\frac{\text{a}}{\text{x}^4}}{1+\frac{6}{\text{x}^4}}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}(1-\text{x})\tan\Big(\frac{\pi\text{x}}{2}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow1}(1-\text{x})\tan\Big(\frac{\pi\text{x}}{2}\Big)$ When x → 1, x - 1 → 0, let x - 1 = y, then y → 0 $=\lim\limits_{{(\text{x}-1)\rightarrow0}}-(\text{x}-1)\tan\frac{\pi\text{x}}{2}$ $=-\lim\limits_{\text{y}\rightarrow0}\text{y}\tan\frac{\text{y}}{2}(\text{y}+1)$ $=-\lim\limits_{\text{y}\rightarrow0}\text{y}\times\tan\Big(\frac{\pi}{2}+\frac{\pi}{2}\text{y}\Big)$ $=\lim\limits_{\text{y}\rightarrow0}\text{y}\times\cot\frac{\pi}{2}\text{y}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\text{y}}{\tan\frac{\pi\text{y}}{2}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\frac{\pi\text{y}}{2}\times\frac{2}{\pi}}{\tan\frac{\pi\text{y}}{2}}$ $=\frac2\pi$
View full question & answer→Question 764 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-64}{\text{x}^2-16}$
Answer$\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-64}{\text{x}^2-16}$$=\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-4^3}{\text{x}^2-4^2}$
$=\lim\limits_{\text{x}\rightarrow{4}}\frac{\frac{\text{x}^{3}-4^3}{\text{x}-4}}{\frac{\text{x}^2-4^2}{\text{x}-4}}$ [Dividing numerator and denominator by x - 4]
$=\frac{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^3-4^3}{\text{x}-4}}{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-4}{\text{x}-4}}$
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$ in numerator and $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{m}}-\text{a}^{\text{m}}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator
$\Rightarrow\text{n}=3,\text{m}=2$
$\Rightarrow\frac{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^3-4^3}{\text{x}-4}}{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-4}{\text{x}-4}}=\frac{3(4)^{3-1}}{2(4)^{2-1}}=\frac{3(4)^2}{2(4)}$
$=6$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin(3+\text{x})-\sin(3-\text{x})}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(3+\text{x})-\sin(3-\text{x})}{\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\cos\big(\frac{3+\text{x}+3-\text{x}}{2}\big)\sin\big(\frac{3+\text{x}-3+\text{x}}{2}\big)}{\text{x}}$ $=2\lim\limits_{\text{x} \rightarrow0}\frac{\cos3.\sin\text{x}}{\text{x}}$ $=2\cos3\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}$ $=2\cos3\times1$ $=2\cos3$
View full question & answer→Question 784 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{\text{x}^2-8\text{x}}+\text{x}\big)$
Answer$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{\text{x}^2-8\text{x}}+\text{x}\big)$ $=\lim\limits_{\text{y}\rightarrow\infty}\Big(\sqrt{\text{y}^2+8\text{y}}-\text{y}\Big),$ where y = -x on rationalising $=\lim\limits_{\text{y}\rightarrow\infty}\frac{\big(\sqrt{\text{y}^2+8\text{y}}-\text{y}\big)\big(\sqrt{\text{y}^2+8\text{y}}+\text{y}\big)}{\big(\sqrt{\text{y}^2+8\text{y}}+\text{y}\big)}$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{\text{y}^2+8\text{y}+\text{y}^2}{\sqrt{\text{y}^2+8\text{y}}+\text{y}}$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{8\text{y}}{\text{y}\sqrt{1+\frac{8}{\text{y}}}+\text{y}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{8}{\sqrt{1+\frac{8}{\text{y}}}+1}$ $=\frac{8}{1+1}=\frac{8}{2}$ $=4$
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Evaluate the following limit: $\lim\limits_{\text{h}\rightarrow0}\frac{\text{(a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$
Answer$\lim\limits_{\text{h}\rightarrow0}\frac{\text{(a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{a}^2+\text{h}^2+2\text{ah}\big)\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}^2\sin(\text{a}+\text{h})+\text{h}^2\sin(\text{a}+\text{h})+2\text{ah}\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\Big[\frac{\text{a}^2(\sin(\text{a}+\text{h})-\sin\text{a})}{\text{h}}+\frac{\text{h}^2\sin(\text{a}+\text{h})}{\text{h}(\text{a}+\text{h})}\times(\text{a}+\text{h})+\frac{2\text{ah}}{\text{h}}(\sin(\text{a}+\text{h}))\Big]$ $=\Bigg[\text{a}^2\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{a}+\text{h}+\text{a}}{2}\big)\sin\big(\frac{\text{a}+\text{h}-\text{a}}{2}\big)}{\text{h}}\Bigg]+[0]+2\text{a}\lim\limits_{\text{h}\rightarrow0}\sin(\text{a}+\text{h})$ $=\Big(2\text{a}^2\cos\text{a}\times\frac12\Big)+(2\text{a}\sin\text{a})$ $=\text{a}^2\cos\text{a}+2\text{a}\sin\text{a}$
View full question & answer→Question 804 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow4}\frac{2-\sqrt{\text{x}}}{4-\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow4}\frac{2-\sqrt{\text{x}}}{4-\text{x}}$$\lim\limits_{\text{x}\rightarrow4}\frac{\big(2-\sqrt{\text{x}}\big)\big(2+\sqrt{\text{x}}\big)}{(4-\text{x})\big(2+\sqrt{\text{x}}\big)}$
$\lim\limits_{\text{x}\rightarrow4}\frac{(2)^2-\big(\sqrt{\text{x}}\big)^2}{(4-\text{x})\big(2+\sqrt{\text{x}}\big)}$
$\lim\limits_{\text{x}\rightarrow4}\frac{(4-\text{x})}{(4-\text{x})\big(2+\sqrt{\text{x}}\big)}$
$=\frac{1}{2+\sqrt{4}}$
$=\frac{1}{2+2}$
$=\frac14$
View full question & answer→Question 814 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^2\text{x}}$
AnswerGiven that: $\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2-(1+\cos\text{x})}{\sin^{2}\text{x}\big[\sqrt{2}+\sqrt{1+\cos\text{x}}\big]}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1+\cos\text{x}}{\sin^{2}\text{x}\Big[\sqrt{2}+\sqrt{1+\cos\text{x}}\Big]}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^{2}\frac{\text{x}}{2}}{(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2})^{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^{2}\frac{\text{x}}{2}}{4\sin^{2}\frac{\text{x}}{2}\cos^{2}\frac{\text{x}}{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2}{4\cos^{2}\frac{\text{x}}{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$ Taking limit, we get: $=\frac{2}{4\cos^{2}0}\times\frac{1}{(\sqrt{2}+\sqrt{2})}$ $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2\sqrt{2}}=\frac{1}{4\sqrt{2}}$ Hence, the required answer is $\frac{1}{4\sqrt{2}}.$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{1}}\frac{1+\cos\pi\text{x}}{(1-\text{x})^2}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1+\cos\pi\text{x}}{(1-\text{x})^2}$ $\Rightarrow\text{x}\rightarrow1,\text{x}-1\rightarrow0,$ let $\text{x}-1=\text{y}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos\pi(\text{y}+1)}{(-\text{y})^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos(\pi+\pi\text{y})}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos(\pi\text{y})}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2-\sin^2\frac{\pi\text{y}}{2}}{\text{y}^2}$ $=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\pi\text{y}}{2}}{\frac{\pi\text{y}}{2}}\Bigg)^2\times\frac{\pi^2}{4}$ $=2\times1\times\frac{\pi^2}{4}$ $=\frac{\pi^2}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\sqrt{\text{x}}-\sin\sqrt{\text{a}}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\sqrt{\text{x}}-\sin\sqrt{\text{a}}}{\text{x}-\text{a}}$$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\sqrt{\text{x}}-\sin\sqrt{\text{a}}}{\big(\sqrt{\text{x}}-\sqrt{\text{a}}\big)\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{2\sin\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)\cos\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)}{\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)\Big({\sqrt{\text{x}}-\sqrt{\text{a}}}\Big)}$
$=2\begin{pmatrix}\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}}{\Big(\frac{{\sqrt{\text{x}}+\sqrt{\text{a}}}}{2}\Big)}\end{pmatrix}\times\frac12\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\cos\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}$
$=2\times1\times\frac{1}{2}\times\cos\sqrt{\text{a}}\times\frac{1}{2\sqrt{\text{a}}}$
$=\frac{\cos\sqrt{\text{a}}}{2\sqrt{\text{a}}}$
View full question & answer→Question 844 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-2\sin\text{x}\cos\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}(1-\cos\text{x})}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}(1-\cos\text{x})}{\text{x}^3}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(1-\cos^2\text{x}\big)}{\text{x}^3(1+\cos\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\sin^2\text{x}\big)}{\text{x}^3(1+\cos\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^3\text{x}}{\text{x}^3(1+\cos\text{x})}$ $=2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin\text{x}}{\text{x}}\Big)^3\times\lim\limits_{\text{x}\rightarrow0}\frac{1}{(1+\cos\text{x})}$ $=2\times1\times\frac{1}{(1+1)}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{\cos\text{x}}-\sqrt{\sin\text{x}}}{\text{x}-\frac{\pi}{4}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{\cos\text{x}}-\sqrt{\sin\text{x}}}{\text{x}-\frac{\pi}{4}}$$=\lim\limits_{{\text{x}-{\frac\pi4\rightarrow0}}}\frac{\big(\sqrt{\cos\text{x}}-\sqrt{\sin\text{x}}\big)}{\big(\text{x}-\frac{\pi}{4}\big)}\times\frac{\big(\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}\big)}{\big(\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}\big)}$
$=\lim\limits_{{\text{x}-{\frac\pi4\rightarrow0}}}\frac{\big(\sqrt{\cos\text{x}}-\sqrt{\sin\text{x}}\big)}{\big(\text{x}-\frac{\pi}{4}\big)\big(\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}\big)}$
As $\text{x}\rightarrow\frac{\pi}{4}\Rightarrow\text{x}-\frac\pi4\rightarrow0\Rightarrow$let $\text{x}-\frac\pi4=\text{y}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\Big(\cos\big(\frac\pi4+\text{y}\big)-\sin\big(\frac\pi4+\text{y}\big)\Big)}{\text{y}\Big(\sqrt{\cos\big(\frac\pi4+\text{y}\big)}+\Big(\sqrt{\sin\big(\frac\pi4+\text{y}\big)}\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\big(\cos\frac\pi4\cos\text{y}-\sin\frac{\pi}{4}\sin\text{y}\big)-\big(\sin\frac\pi4\cos\text{y}-\cos\frac{\pi}{4}\sin\text{y}\big)}{\text{y}\Big(\sqrt{\cos\big(\frac\pi4+\text{y}\big)}+\Big(\sqrt{\sin\big(\frac\pi4+\text{y}\big)}\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\Big(\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}-\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}\Big)}{\text{y}\Big(\sqrt{\cos\big(\frac\pi4+\text{y}\big)}+\Big(\sqrt{\sin\big(\frac\pi4+\text{y}\big)}\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\Big(-2\frac{\sin\text{y}}{\sqrt{2}}\Big)}{\text{y}\Big(\sqrt{\cos\big(\frac\pi4+\text{y}\big)}+\Big(\sqrt{\sin\big(\frac\pi4+\text{y}\big)}\Big)}$
$=-\sqrt{2}\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}\Big)\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{\cos\big(\text{y}+\frac\pi4}\big)+\lim\limits_{\text{y}\rightarrow0}\sqrt{\sin\big(\text{y}+\frac{\pi}{4}\big)}}$
$=-\sqrt{2}\times1\times\frac{1}{\sqrt{\cos\frac{\pi}{4}}+\sqrt{\sin\frac\pi4}}$
$=-\sqrt{2}\times\frac{1}{\big(\frac{1}{\sqrt{2}}\big)^{\frac{1}{2}}+\big(\frac{1}{\sqrt{2}}\big)^{\frac{1}{2}}}$ $\Big[\because\cos\frac\pi4=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\Big]$
$=\frac{-\sqrt{2}}{\big(\frac{1}{\sqrt{2}}\big)^{\frac{1}{2}}+(1+1)^{\frac{1}{2}}}$
$=\frac{-\sqrt{2}}{\sqrt{2}\big(\frac{1}{\sqrt{2}}\big)^{\frac{1}{2}}}$
$=-\frac{1}{2^{\frac14}}$
View full question & answer→Question 864 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}{(\text{x}-1)(\text{x}+1)}\times\frac{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{((3+\text{x})-(5-\text{x}))}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2+2\text{x}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2(\text{x}-1)}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2}{(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\frac{2}{(1+1)\big(\sqrt{3+1}+\sqrt{5-1}\big)}$
$=\frac{2}{(2)(2+2)}$
$=\frac14$
View full question & answer→Question 874 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^2-\text{x}-2}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^2-\text{x}-2}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(\text{x}+1)}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{1}{\frac{\text{x}}{\text{x}+1}+\frac{\sin(\text{x}-2)}{(\text{x}-2)(\text{x}+1)}}$ $=\lim\limits_{\text{x}\rightarrow2}(\text{x}+1)\Bigg(\frac{1}{\text{x}+\frac{\sin(\text{x}-2)}{\text{x}-2}}\Bigg)$ $=\lim\limits_{\text{x}\rightarrow2}(\text{x}+1)\frac{1}{\lim\limits_{\text{x}\rightarrow2}(\text{x})+\lim\limits_{\text{x}\rightarrow2}\frac{\sin(\text{x}-2)}{\text{x}-2}}$ $=(2+1)\times\frac{1}{(2)\lim\limits_{\text{x}\rightarrow2\rightarrow0}\frac{\sin(\text{x}-2)}{(\text{x}-2)}}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=3\times\frac{1}{2+1}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\text{ax}}{\text{x}}+\text{b}}{\text{a}+\frac{\sin\text{bx}}{\text{bx}}}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}\times\text{a}+\text{b}}{\text{a}+\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{bx}}{\text{bx}}\times\text{b}}$ $=\frac{\text{a}+\text{b}}{\text{a}+\text{b}}$ $=1$
View full question & answer→Question 894 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{{\text{x}-1}}{\sqrt{\text{x}^2+3}-2}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{{\text{x}-1}}{\sqrt{\text{x}^2+3}-2}$$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\times\big(\sqrt{\text{x}^2+3}+2\big)}{\big(\sqrt{\text{x}^2+3}-2\big)\big(\sqrt{\text{x}^2+3}+2\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\big(\sqrt{\text{x}^2+3}+2\big)}{\big({\text{x}^2+3}-4\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\big(\sqrt{\text{x}^2+3}+2\big)}{\big({\text{x}^2-1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2+3}+2}{{\text{x}}+1}$
Putting the value x = 1
$\Rightarrow\frac{\sqrt{1+3}+2}{1+1}$
$=\frac{2+2}{2}$
$=\frac{4}{2}=2$
View full question & answer→Question 904 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{3+\text{x}}-1}{2-\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{3+\text{x}}-1}{2-\text{x}}$$=\lim\limits_{\text{x}\rightarrow2}\frac{\big(\sqrt{3-\text{x}}-1\big)}{2-\text{x}}\times\frac{\big(\sqrt{3-\text{x}}+1\big)}{\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(3-\text{x})-1}{(2-\text{x})\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(2-\text{x})}{(2-\text{x})\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{1}{\big(\sqrt{3-\text{x}}+1\big)}$
$=\frac{1}{\sqrt{3-2}+1}=\frac{1}{1+1}$
$=\frac{1}{2}$
View full question & answer→Question 914 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{{\text{x}}-\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{{\text{x}}-\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{({\text{x}}-\text{a})\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}{\big(\sqrt{\text{x}}-\sqrt{\text{a}}\big)\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}-\text{a})-\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}{(\text{x}-\text{a})}$
$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)$
$=2\sqrt{\text{a}}$
View full question & answer→Question 924 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}+2)^{\frac{3}{2}}-(\text{a}+2)^{\frac{3}{2}}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}+2)^{\frac{3}{2}}-(\text{a}+2)^{\frac{3}{2}}}{\text{x}-\text{a}}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}+2)^{\frac{3}{2}}-(\text{a}+2)^{\frac{3}{2}}}{(\text{x}+2)-(\text{a}+2)}$ Let x + 2 = y, a + 2 = b $\Rightarrow\lim\limits_{({\text{x}+2)}\rightarrow{(\text{a}+2)}}\frac{(\text{y})^\frac{3}{2}-(\text{b})^\frac{3}{2}}{(\text{y})-(\text{b})},$ $\Big[\text{Using formula}\ \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}\Big]$ $=\frac{3}{2}(\text{b})^{\frac{3}{2}-1}$ $=\frac{3}{2}(\text{a}+2)^{\frac{3}{2}-1}$ $=\frac{3}{2}(\text{a}+2)^{\frac{1}{2}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\sqrt{\text{x}}-\cos\sqrt{\text{a}}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\sqrt{\text{x}}-\cos\sqrt{\text{a}}}{\text{x}-\text{a}}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{-2\sin\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)\times\sin\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}{\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)\times\Big({\sqrt{\text{x}}-\sqrt{\text{a}}}\Big)}$ $=-2\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)\times\lim\limits_{\text{x}\rightarrow{\text{a}}}\sin\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)\times\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}\times\frac12$ $=-2\sin\sqrt{\text{a}}\times1\times\frac{1}{2\sqrt{\text{a}}}\times\frac12$ $=-\frac{1}{2\sqrt{\text{a}}}\sin\sqrt{\text{a}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\sqrt{\text{x}-2}-\sqrt{4-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\sqrt{\text{x}-2}-\sqrt{4-\text{x}}}$$=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\big(\sqrt{\text{x}-2}-\sqrt{4-\text{x}}\big)}\times\frac{\big(\sqrt{\text{x}}-2+\sqrt{4-\text{x}}\big)}{\big(\sqrt{\text{x}}-2+\sqrt{4-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{(\text{x}-2)-(4-\text{x})}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{\text{x}-2-4+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{2(\text{x}-3)}$
$=\frac{1}{2}\lim\limits_{\text{x}\rightarrow3}\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)$
$=\frac{1}{2}\big(\sqrt{3-2}+\sqrt{4-\text{x}}\big)$
$=\frac{1}{2}\big(\sqrt{1}+\sqrt{1}\big)$
$=\frac{1}{2}(1+1)=\frac22$
$=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}^2+\text{cx}}-\text{x}}{}$
Answer$\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}^2+\text{cx}}-\text{x}}{}$$=\lim\limits_{\text{x}\rightarrow{\infty}}\Bigg(\big(\sqrt{\text{x}^2+\text{cx}}-\text{x}\big)\frac{\big(\sqrt{\text{x}^2+\text{cx}}+\text{x}\big)}{\sqrt{\text{x}^2+\text{cx}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\big(\text{x}^2+\text{cx}-\text{x}^2\big)}{\sqrt{\text{x}^2+\text{cx}+\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{cx}}{\sqrt{\text{x}^2+\text{cx}+\text{x}}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{c}}{\sqrt{1+\frac{\text{c}}{\text{x}}+1}}$
$=\frac{\text{c}}{1+1}=\frac{\text{c}}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{5\text{x}-4}-\sqrt{\text{x}}\big)}{\text{x}-1}\times\frac{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{((5\text{x}-4)-\text{x)}}{(\text{x}-1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$ $=4\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)}{(\text{x}-1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$ $=4\lim\limits_{\text{x}\rightarrow1}\frac{1}{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$ $=4\times\frac{1}{\sqrt{5-4}+\sqrt{1}}$ $=4\times\frac{1}{\sqrt{5-4}+\sqrt{1}}$ $=4\times\frac{1}{\sqrt{1}+\sqrt{1}}$ $=\frac{4}{2}=2$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}$Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1},$ here, $\text{n}=\frac27$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}=\frac{2}{7}(\text{a})^{\frac{2}{7}-1}$
$=\frac{2}{7}\text{a}^{\frac{-5}{7}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^3-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^3-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{5\text{x}-4}-\sqrt{\text{x}}\big)}{(\text{x}-1)\big(\text{x}^2+1+1\big)}\times\frac{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(5\text{x}-4-\text{x})}{(\text{x}-1)\big(\text{x}^2+1+\text{x}\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4(\text{x}-1)}{(\text{x}-1)\big(\text{x}^2+\text{x}+1\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\frac{4}{(1+1+1)\big(\sqrt{5-4}+\sqrt{1}\big)}$
$=\frac{4}{(3)(1+1)}$
$=\frac{4}{3\times2}=\frac23$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos4\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos4\text{x}}{\text{x}^2}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^22\text{x}}{\text{x}^2}$ $=2\lim\limits_{\text{x} \rightarrow0}\Big(\frac{\sin2\text{x}}{\text{x}}\Big)^2$ $=2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin2\text{x}}{2\text{x}}\Big)^2\times(2)^2$ $=2\times1\times4$ $=8$
View full question & answer→Question 1004 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^{\text{n}}-3^\text{n}}{\text{x}-\text{3}}=108,$ find the value of n.
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^{\text{n}}-3^\text{n}}{\text{x}-\text{3}}=108$$\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{3}}\frac{\text{x}^{\text{n}}-3^{\text{n}}}{\text{x}-3}$
Applying the formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
Here, n = n, a = 3
$\Rightarrow{\lim\limits_{\text{x}\rightarrow{3}}}\frac{\text{x}^{\text{n}}-3^\text{n}}{\text{x}-3}=\text{n}(3)^{\text{n}-1}$
It is given that $\text{n}(3)^{\text{n}-1}=108$
$\Rightarrow\text{n}(3)^{\text{n}-1}=2\times2\times3\times3\times3$
$=(2)^2\times(3)^3$
$=4(3)^{4-1}$
$\Rightarrow\text{n}=4$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}^2+\text{x}^2}-\text{a}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}^2+\text{x}^2}-\text{a}}{\text{x}^2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{\text{a}^2+\text{x}^2}-\text{a}\big)}{\text{x}^2}\times\frac{\big(\sqrt{\text{a}^2+\text{x}^2}+\text{a}\big)}{\big(\sqrt{\text{a}^2+\text{x}^2}+\text{a}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{a}^2+\text{x}^2-\text{a}^2\big)}{\text{x}^2\sqrt{\text{a}^2+\text{x}^2}+\text{a}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{\text{x}^2\big(\sqrt{\text{a}^2+\text{x}^2}+\text{a}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\sqrt{\text{a}^2+\text{x}^2+\text{a}}}$ $=\frac{1}{\text{a}+\text{a}}$ $=\frac{1}{2\text{a}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=9,$ find all possible value of a.
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=9\ \cdots{\text{(i)}}$ $\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}$ $=9(\text{a})^{9-1}$ $=9\text{a}^{8}$ It is given that $9^\text{a}=9$ [From (i)] $\Rightarrow\text{a}^{8}=\frac99=1$ $\Rightarrow\text{a}^4=1$ $\text{a}^2=1$ $\text{a}=\pm1$ $\Rightarrow \text{a} = 1 \text{ and a} = -1$
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Evaluate the following limit: $\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}},\text{x}\ne0$
Answer$\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}}$$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}\big)}{\text{h}}\times\frac{\big(\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big)}{\big(\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h}-\text{x})}{\text{h}\big(\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big)}$
$=\frac{1}{\sqrt{\text{x}}+\sqrt{\text{x}}}$
$=\frac{1}{2\sqrt{\text{x}}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\frac{\text{a}+\text{x}+\text{a}-\text{x}}{2}\big)\cos\big(\frac{\text{a}+\text{x}-\text{a}+\text{x}}{2}\big)-\sin\text{a}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}(\cos\text{x}-1)}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin\text{a}(1-\cos\text{x})}{\text{x}\sin\text{x}}$ $=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}\frac{\text{x}}{2}}{\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)\text{x}}$ $=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}\frac{\text{x}}{2}}{\big(\cos\frac{\text{x}}{2}\big)\text{x}}$ $=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{\tan\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\times\frac12$ $=-2\sin\text{a}\times1\times\frac12$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=-\sin\text{a}$
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Evaluate the following limit: Evaluate: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1^4+2^4+3^4+\ \cdots+\text{n}^4}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \cdots+\text{n}^3}{\text{n}^5}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^4+2^4+3^4+\ \cdots+\text{n}^4}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \cdots+\text{n}^3}{\text{n}^5}$ $=\ \lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(1+\text{n})(1+2\text{n})\big(-1+3\text{n}+3\text{n}^2\big)}{30}}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)^2}{\text{n}^5}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\frac{1}{\text{n}}+1\big)\big(\frac{1}{\text{n}}+2\big)\big(-\frac{1}{\text{n}^2}+\frac{3}{\text{n}}+3\big)}{30}\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}^5}\Bigg(\frac{\text{n}^2\big(\text{n}^2+2\text{n}+1\big)}{4}\Bigg)$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\frac{1}{\text{n}}+1\big)\big(\frac{1}{\text{n}}+2\big)\big(-\frac{1}{\text{n}^2}+\frac{3}{\text{n}}+3\big)}{30}\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(\frac{1}{\text{n}}+\frac{2}{\text{n}^2}+\frac{1}{\text{n}^3}\Big)}{4}$ $=\frac{1\times2\times3}{30}-0$ $=\frac15$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{1+\text{cosec}^3\text{x}}{\cot^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{1+\text{cosec}^3\text{x}}{\cot^2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{(1+\text{cosec}\text{x})\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{ x}\big)}{\big(\text{cosec}^2\text{x}-1\big)}$ $=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{(\text{cosec}\text{x}+1)\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{x}\big)}{(\text{cosecx}-1)\big(\text{cosecx+1}\big)}$ $=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{ x}\big)}{(\text{cosec }\text{x}-1)}$ $=\frac{1+\text{cosec}^2\frac{3\pi}{2}-\text{cosec}\frac{3\pi}{2}}{\text{cosec}\frac{3\pi}{2}-1}$ $=\frac{1+(-1)^2-(-1)}{(-1)-1}$ $\Big[\therefore\text{cosec}\frac{3\pi}{2}=-1\Big]$ $=\frac{1+1+1}{-2}$ $=\frac{-3}{2}$
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Evaluate the following limits: $\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}}{\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}}{\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}}$ Rationalising the numerator and the denominator: $=\lim\limits_{\text{x}\rightarrow\infty}\Bigg[\frac{\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)}{\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)}\times\frac{\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)}{\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)}\times\frac{\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)}{\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)}\Bigg]$ $=\lim\limits_{\text{x}\rightarrow\infty}\Bigg[\frac{\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)}{\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)\big(\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}\big)\big(\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}\big)}\Bigg]$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{\big(\text{x}^2+\text{a}^2\big)-\big(\text{x}^2+\text{b}^2\big)}{\big(\text{x}^2+\text{c}^2\big)-\big(\text{x}^2+\text{d}^2\big)}\times\bigg(\frac{\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}}\bigg)$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\Big(\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}\Big)\bigg(\frac{\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}}\bigg)$ Dividing the numerator and the denominator by x: $=\lim\limits_{\text{x}\rightarrow{\infty}}\Big(\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}\Big)\begin{pmatrix}\frac{\sqrt{1+\frac{\text{a}^2}{\text{x}^2}}+\sqrt{1+\frac{\text{a}^2}{\text{x}^2}}}{\sqrt{1+\frac{1}{\text{a}^2}}+\sqrt{1+\frac{\text{b}^2}{\text{a}^2}}}\end{pmatrix}$ $\text{As x}\rightarrow\infty,\frac{1}{\text{x}},\frac{1}{\text{x}^2}\rightarrow0$ $=\Big(\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}\Big)\Big(\frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}\Big)$ $=\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(\cos3\text{x}-\cos\text{x})}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(\cos3\text{x}-\cos\text{x})}{\text{x}^3}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}\Big(-2\sin\big(\frac{3\text{x}+\text{x}}{2}\big)\sin\big(\frac{3\text{x}-\text{x}}{2}\big)\Big)}{\text{x}^3}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}(-2\sin2\text{x}\sin\text{x})}{\text{x}^3}$ $=\frac{-2\lim\limits_{\text{x} \rightarrow0}\sin2\text{x}\times\lim\limits_{\text{x} \rightarrow0}\sin2\text{x}\times\lim\limits_{\text{x} \rightarrow0}\sin\text{x}}{\text{x}^3}$ $=-2\Big(\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\Big)\times\Big(2\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\Big)\times\Big(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)$ $=-2(1\times2)\times(2)\times(1)$ $=-8$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{4\text{x}^2-7\text{x}}+2\text{x}\big)$
Answer$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{4\text{x}^2-7\text{x}}+2\text{x}\big)$ Substitute y = -x $=\lim\limits_{\text{y}\rightarrow\infty}\Big(\sqrt{4\text{y}^2+7\text{y}}-2\text{y}\Big)$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{\Big(\sqrt{4\text{y}^2+7\text{y}}-2\text{y}\Big)\Big(\sqrt{4\text{y}^2+7\text{y}}+2\text{y}\Big)}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{\Big(4\text{y}^2+7\text{y}+4\text{y}^2\big)}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{(7\text{y})}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{7}{\sqrt{4+\frac{7}{\text{y}}}+2}$ $=\frac{7}{2+2}=\frac{7}{4}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^2-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^2-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{5\text{x}-4}-\sqrt{\text{x}}\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{((5\text{x}-4)-\text{x})}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4(\text{x}-1)}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4}{(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\frac{4}{(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{1}\big)}$
$=\frac{4}{2(1+1)}$
$=\frac{4}{4}=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}-2\text{x}}{3\text{x}-\sin^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}-2\text{x}}{3\text{x}-\sin^2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\tan3\text{x}}{\text{x}}-\frac{\tan2\text{x}}{\text{x}}}{\frac{3\text{x}}{\text{x}}-\frac{\sin^2\text{x}}{\text{x}}}$ $=\frac{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\times3\Big)-\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}}{2\text{x}}\times2\Big)}{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{3\text{x}}{2}\Big)-\lim\limits_{\text{x}\rightarrow0}\frac{(\sin\text{x})^2}{\text{x}}}$ $=\frac{3-2}{3-\big(\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=\frac{3-2}{3-0}=\frac13$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=\frac13$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2+1-\cos\text{x}}{\text{x}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2+1-\cos\text{x}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^2+2\sin^2\frac{\text{x}}{2}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^2\Bigg[1+2\bigg(\frac{\sin\frac{\text{x}}{2}}{\text{x}}\bigg)^2\Bigg]}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1+2\Bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)^2\times\frac{1}{4}}{\frac{\sin\text{x}}{\text{x}}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1+\lim\limits_{\text{x} \rightarrow0}2\Bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)^2\times\frac{1}{4}}{\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}}$ $\frac{1+2\times1\times\frac{1}{4}}{1}=\frac{1+\frac{1}{2}}{1}$ $=\frac{3}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x})^{{6}}-1}{(1+\text{x})^2-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x})^{{6}}-1}{(1+\text{x})^2-1}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(1+\text{x})^{{{6}}}-1^6}{1+\text{x}-1}}{\frac{(1+\text{x})^2-1^2}{1+\text{x}-1}}$
$\Rightarrow \text{Let} 1 + \text{x} = \text{y}, \text{as x} → 0, \text{y} → 1$
$=\frac{\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^6-1^6}{\text{y}-1}}{\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^2-1}{\text{y}-1}}$
$=\frac{6(1)^{6-1}}{2(1)^{2-1}}$ $\Big[\text{Using formula} \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{a}-\text{a}}=\text{na}^{\text{n}-1}\Big]$
$=\frac62$
$=3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{\pi}{8}}\frac{\cot4\text{x}-\cos4\text{x}}{(\pi-8\text{x})^3}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{8}}\frac{\cot4\text{x}-\cos4\text{x}}{(\pi-8\text{x})^3}$ When $\text{x}\rightarrow\frac\pi8,\frac\pi8-\text{x}\rightarrow0,$ let $\frac{\pi}{8}-\text{x}=\text{y}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cot4\big(\frac\pi8-\text{y}\big)-\cos4\big(\frac\pi8-\text{y}\big)}{(8)^3\text{y}^3}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cot4\big(\frac\pi2-4\text{y}\big)-\cos4\big(\frac\pi2-4\text{y}\big)}{(8)^3\text{y}^3}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan4\text{y}-\sin4\text{y}}{(8)^2\text{y}^3}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{\sin4\text{y}}{\cos4\text{y}}-\sin4\text{y}}{8^3\text{y}^3}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}-\sin4\text{y}\cos4\text{y}}{\cos4\text{y}\times\text{y}^3\times8^3}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}\big(2\sin^22\text{y}\big)}{\cos4\text{y}\times\text{y}^3\times8^3}$ $=\frac{2}{8^3}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}}{\text{y}}\times\frac{\sin^22\text{y}}{\text{y}^2}\times\frac{1}{\cos4\text{y}}$ $=\frac{2}{8^3}\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}}{4\text{y}}\times4\Big)\times\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\cos4\text{y}}$ $=\frac{2}{8^2}(1\times4)\times(1)\times4\times\frac11$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\cos\theta=1\Big]$ $=\frac{2\times4\times4}{8\times8\times8}$ $\frac{1}{16}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos5\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos5\text{x}}{\text{x}^2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(-2\sin\big(\frac{3\text{x}+5\text{x}}{2}\big)\sin\big(\frac{3\text{x}-5\text{x}}{2}\big)\Big)}{\text{x}^2}$ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{-2\sin4\text{x}\sin(-\text{x})}{\text{x}^2}\Big)$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin4\text{x}\sin\text{x}}{\text{x}^2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin4\text{x}}{\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$ $=2\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin4\text{x}}{4\text{x}}\times4\Big)\times\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=8$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{5}{7}}-\text{a}^{\frac{5}{7}}}{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{5}{7}}-\text{a}^{\frac{5}{7}}}{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}$$\Rightarrow\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\frac{\text{x}^{\frac{5}{7}}-\text{a}^{\frac{5}{7}}}{\text{x}-\text{a}}}{\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}}$ [Dividing numrator and denominator by x - a]
$=\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{5}{7}}-\text{a}^{\frac{5}{7}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}}$
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
Here, $\text{n}=\frac57$ is numerator and applying $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\text{m}-\text{a}^\text{m}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator, where $\text{m}=\frac27$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\frac{\text{x}^{\frac{5}{7}}-\text{a}^{\frac{5}{7}}}{\text{x}-\text{a}}}{\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}}=\frac{\frac{5}{7}\text{a}^{\frac{5}{7}-1}}{\frac{2}{7}(\text{a})^{\frac{2}{7}-1}}$
$=\frac{\frac{5}{7}\text{a}^{\frac{-2}{7}}}{\frac{2}{7}\text{a}^{\frac{-5}{7}}}$
$=\frac{5}{2}\text{a}^{\frac{-2}{7}+\frac{5}{7}}$
$=\frac{5}{2}\text{a}^{\frac{3}{7}}$
View full question & answer→Question 1174 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$
Answer$\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$$=\lim\limits_{\text{x}\rightarrow2}\Bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}\big(\text{x}^2-2\text{x}-\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}(\text{x}-2)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}(\text{x}-1)-2(2\text{x}-3)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}^2-\text{x}-4\text{x}+6}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}^2-2\text{x}-3\text{x}+6}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}(\text{x}-2)-3(\text{x}-2)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{(\text{x}-2)(\text{x}-3)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}\frac{2(2\text{x}-3)}{\text{x}(\text{x}-2)(\text{x}-1)}\bigg\}$
$=\frac{-1}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\infty}\Big\{\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big\}^{\frac{3\text{x}-2}{3\text{x}+2}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}}{\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}}{\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}\big)}\times\frac{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}^2\big)-\big(1+\text{x}\big)\times\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}\big)\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{x}^2-\text{x}\big)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\big(1+\text{x}^3-1-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(\text{x}-1)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\times\big(\text{x}^2-1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(\text{x}-1)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)(\text{x})(\text{x}-1)(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)(\text{x}+1)\big)}$
$=\frac{2}{2}=1$
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Evaluate the following limit: $\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1},\lim\limits_{\text{x}\rightarrow0}\text{ f(x)}=1 $ and $\lim\limits_{\text{x}\rightarrow\infty}\text{f(x)}=1,$ then prove that $\text{f}(-2)=\text{f}(2)=1.$
Answer$\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}$ Also $\lim\limits_{\text{x}\rightarrow0}\text{ f(x)}=1\cdots{(\text{i})}$ [Given] $\Rightarrow\ \lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1$ $\Rightarrow\ \frac{\lim\limits_{\text{x}\rightarrow0}\text{ax}^2+\text{b}}{\lim\limits_{\text{x}\rightarrow0}\text{x}^2+1}=1$ $\Rightarrow\ \text{b}=1$ Also, it is given that $\lim\limits_{\text{x}\rightarrow\infty}\text{f(x)}=1$ $\therefore\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1\ \cdots(\text{ii})$ $\Rightarrow\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1$ $\Rightarrow\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}+\frac{1}{\text{x}^2}}{1+\frac{1}{\text{x}^2}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $\Rightarrow\text{ a}=1$ Thus, $\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=\frac{\text{x}^2+1}{\text{x}^2+1}=1$ [From (ii)] $\text{f}(-2)=1$ $\text{f}(2)=1$ $\text{f}(-2)=1=\text{f}(2)$ Hence, proved.
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{5}-\text{a}^5}{\text{x}-\text{a}}=405,$ find all possible value of a.
AnswerIf $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{5}-\text{a}^5}{\text{x}-\text{a}}=405\ \cdots{\text{(i)}}$ $\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{5}-\text{a}^5}{\text{x}-\text{a}}$ $=5(\text{a})^{5-1}$ $=5\text{a}^{4}$ It is given that $5\text{a}^4=405$ $\Rightarrow5\text{a}^4=405$ $\text{a}^4=\frac{405}{5}=81$ $\text{a}^4=(3)^4,\text{a}^2=9$ $\text{a}=\pm3$ $\Rightarrow\text{a}=3 \text{ and }\text{a}=-3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^3+3\text{x}^2+6\text{x}+2}{\text{x}^3+3\text{x}^2-3\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^3+3\text{x}^2+6\text{x}+2}{\text{x}^3+3\text{x}^2-3\text{x}-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}+1)\big(\text{x}^2+4\text{x}-2\big)}{(\text{x}-1)\big(\text{x}^2-4\text{x}+1\big)}$
$=\frac{(1)^24(1)-2}{(1)^24(1)+1}$
$=\frac{1+4-2}{1+4+1}$
$=\frac{3}{6}$
$=\frac{1}{2}$
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Show that $\lim\limits_{\text{x}\rightarrow0}\ \sin\frac{1}{\text{x}}$ does not exist.
Answer$\lim\limits_{\text{x}\rightarrow0^-}\sin\frac{1}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\sin\frac{1}{0-\text{h}}=-\lim\limits_{\text{h}\rightarrow0}\ \sin\frac{1}{\text{h}}$ = - (Anoscillating number which ascillates between - 1 and 1) So, $\lim\limits_{\text{x}\rightarrow0^-}\ \sin\frac{1}{\text{x}}$ does not exist. Similarly, $\lim\limits_{\text{x}\rightarrow0^+}\ \sin\frac{1}{\text{x}}$ does not exist $\lim\limits_{\text{x}\rightarrow0}\ \sin\frac{1}{\text{x}}$ does not exist.
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Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\frac{{(\text{n}+2)!}+{(\text{n}+1)!}}{{(\text{n}+2)!}+{(\text{n}+1)!}}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{{(\text{n}+2)!}+{(\text{n}+1)!}}{{(\text{n}+2)!}+{(\text{n}+1)!}}$ We know that (n + 2) = (n + 2)(n + 1)! $\Rightarrow\lim\limits_{\text{n}\rightarrow\infty}\frac{(\text{n}+2)(\text{n}+1)!+(\text{n}+1)!}{(\text{n}+2)(\text{n}+1)!-(\text{n}+1)!}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{(\text{n}+1)!\big[(\text{n}+2)+1\big]}{(\text{n}+1)\big[(\text{n}+2)-1\big]}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{n}+3}{\text{n}+1}$ $\Big[\frac\infty\infty\text{ from}\Big]$ $=\lim\limits_{\text{n}\rightarrow{\infty}}\frac{1+\frac{3}{\text{n}}}{1+\frac{1}{\text{n}}}$ $=\frac{1+0}{1+0}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\frac{1}{\text{x}}}{\sin\pi(\text{x}-1)}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\frac{1}{\text{x}}}{\sin\pi(\text{x}-1)}$ As x → 1, then x - 1 → 0 let x - 1 = y $=\lim\limits_{{\text{x}-{1\rightarrow0}}}\frac{(\text{x}-1)}{\text{x}\times\sin\pi(\text{x}-1)}$ $=\lim\limits_{{\text{y}\rightarrow0}}\frac{\text{y}}{(\text{y}+1)\sin(\pi\text{y})}$ $=\lim\limits_{{\text{y}\rightarrow0}}\frac{\text{y}}{(\text{y}+1)\sin(\pi\text{y})}$ $=\lim\limits_{{\text{y}\rightarrow0}}\frac{1}{\frac{(\text{y}+1)\sin(\pi\text{y})}{\text{y}}}$ $=\frac{1}{\Big(\lim\limits_{\text{y}\rightarrow0}(\text{y}+1)\Big)\times\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\pi\text{y}}{\text{y}\times\pi}\times\pi\Big)}$ $=\frac{1}{(1)(1\times\pi)}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac{1}{\pi}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{2}+1\big)}{\text{x}^2-2}$
Answer$\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{2}+1\big)}{\text{x}^2-2}$$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\Big(\sqrt{\big(\sqrt{2}+1\big)^2}\Big)}{\text{x}^2-2}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\big(\sqrt{3+2\sqrt{2}}\big)}{\text{x}^2-2}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\big(\sqrt{3+2\sqrt{2}}\big)}{\text{x}^2-2}\times\frac{\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)}{\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{{3+2\text{x}}-{3-2\sqrt{2}}}{\big(\text{x}^2-2\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{2\big(\text{x}-\sqrt{2}\big)}{\big(\text{x}^2-2\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{2}{\big(\text{x}^2-\sqrt{2}\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\frac{2}{\big(\sqrt{2}+\sqrt{2}\big)\Big(\sqrt{3+2\sqrt{2}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\frac{2}{\big(2\sqrt{2}\big)\Big(2\sqrt{3+2\sqrt{2}}\Big)}$
$=\frac{1}{\big(2\sqrt{2}\big)\Big(\sqrt{3+2\sqrt{2}}\Big)}$
$=\frac{1}{\big(2\sqrt{2}\big)\Big(\sqrt{2}+1\Big)}$
$=\frac{\big(\sqrt{2}-1\big)}{\big(2\sqrt{2}\big)}$
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Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \cdots+\text{n}^2}{\text{n}^3}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \cdots+\text{n}^2}{\text{n}^3}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac16\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{\text{n}^3}$ $\Big[1^2+2^2+3^2+\ \cdots+\text{n}^2=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+1)\Big]$ $=\frac{1}{6}\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\text{n}^2+\text{n}\big)(2\text{n}+1)}{\text{n}^3}$ $=\frac{1}{6}\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(2\text{n}^3+\text{n}^2+2\text{n}^2+\text{n}\big)}{\text{n}^3}$ $=\frac16\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\big(2\text{n}^2+3\text{n}^2+\text{n}\big)}{\text{n}^3}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\frac16\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(2+\frac{3}{\text{n}}+\frac{1}{\text{n}^2}\Big)}{1}$ $=\frac16\frac{(2+0+0)}{1}=\frac{1}{3}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2\tan2\text{x}}{\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2\tan2\text{x}}{\tan\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\big(\frac{\text{x}^2}{2\text{x}}-\frac{\tan2\text{x}}{2\text{x}}\big)\times2\text{x}}{\frac{\tan\text{x}}{\text{x}}\times\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\big(\frac{\text{x}}{2}-\frac{\tan2\text{x}}{2\text{x}}\big)}{\frac{\tan\text{x}}{\text{x}}}$ $=2\Big(\frac{0-1}{1}\Big)$ $=-2$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\infty}\sqrt{\text{x}}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\sqrt{\text{x}}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}$$=\lim\limits_{\text{x}\rightarrow\infty}\Big(\sqrt{\text{x}^2+\text{x}}-{\text{x}}\Big)$
$=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\big(\sqrt{\text{x}^2+\text{x}}-\text{x}\big)\times\frac{\big(\sqrt{\text{x}^2+\text{x}}+\text{x}\big)}{\sqrt{\text{x}^2+\text{x}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\frac{\big(\text{x}^2+\text{x}\big)-\text{x}^2}{\sqrt{\text{x}^2+\text{x}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}}+\text{x}}\bigg)$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\lim\limits_{\text{n}\rightarrow{\infty}}\begin{pmatrix}\frac{1}{\sqrt{\frac{\text{x}^2}{\text{x}^2}}+\frac{\text{x}}{\text{x}^2}+1}\end{pmatrix}$
$=\lim\limits_{\text{n}\rightarrow{\infty}}\begin{pmatrix}\frac{1}{\sqrt{1+\frac{1}{\text{x}}}+1}\end{pmatrix}$
$=\frac{1}{1+1}$
$=\frac12$
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