Question 15 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 25 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 35 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 45 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 55 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 65 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 75 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 85 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 95 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 105 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 115 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 125 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 135 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 145 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 155 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$
View full question & answer→Question 165 Marks
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$
View full question & answer→Question 175 Marks
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 185 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$
View full question & answer→Question 195 Marks
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$
View full question & answer→Question 205 Marks
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)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}^2\big)-\big(1-\text{x}^2\big)}{\text{x}\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}^2}{\text{x}\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}^2}\big)}$
$=\frac{2\times0}{\big(\sqrt{1}+\sqrt{1}\big)}$
$=\frac{2}{2}\times0$
$=0$
View full question & answer→Question 215 Marks
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)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(2-\text{x})-(2+\text{x})}{\text{x}\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-2\text{x}}{\text{x}\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\frac{-2}{\sqrt{2}+\sqrt{2}}$
$=\frac{-2}{2\sqrt{2}}$
$=\frac{-1}{\sqrt{2}}$
View full question & answer→Question 225 Marks
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$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}+4\sin3\text{x}}{4\sin2\text{x}+7\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{5+\frac{4\sin3\text{x}}{\text{x}}}{\frac{4\sin2\text{x}}{\text{x}}+7}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}5+4\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3}{4\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2+7}$
$=\frac{5+4\times1\times3}{4\times2+7}$
$=\frac{5+12}{8+7}$
$=\frac{17}{15}$
View full question & answer→Question 235 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\text{cosec}^2\text{x}-2}{\cot\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\text{cosec}^2\text{x}-2}{\cot\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\cot^2\text{x}+1-2}{\cot\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\cot^2\text{x}-1}{\cot\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{(\cot\text{x}-1)(\cot\text{x}+1)}{\cot\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}(\cot\text{x}+1)$
$=\cot\frac{\pi}{4}+1$
$=1+1$
$=2$
View full question & answer→Question 245 Marks
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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