Question 515 Marks
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 525 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$
View full question & answer→Question 535 Marks
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 545 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$
View full question & answer→Question 555 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{\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 565 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 575 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 $\frac{1}{4\sqrt{2}}.$
View full question & answer→Question 585 Marks
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 $\text{x}-1=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos\pi(\text{y}+1)}{(-\text{y})^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos(\pi+\pi\text{y})}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos(\pi\text{y})}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2-\sin^2\frac{\pi\text{y}}{2}}{\text{y}^2}$
$=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\pi\text{y}}{2}}{\frac{\pi\text{y}}{2}}\Bigg)^2\times\frac{\pi^2}{4}$
$=2\times1\times\frac{\pi^2}{4}$
$=\frac{\pi^2}{2}$
View full question & answer→Question 595 Marks
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 605 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$
View full question & answer→Question 615 Marks
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 625 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 635 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$
View full question & answer→Question 645 Marks
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 655 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 665 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}}$$=\lim\limits_{\text{x}\rightarrow2}\frac{\big(\sqrt{3-\text{x}}-1\big)}{2-\text{x}}\times\frac{\big(\sqrt{3-\text{x}}+1\big)}{\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(3-\text{x})-1}{(2-\text{x})\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(2-\text{x})}{(2-\text{x})\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{1}{\big(\sqrt{3-\text{x}}+1\big)}$
$=\frac{1}{\sqrt{3-2}+1}=\frac{1}{1+1}$
$=\frac{1}{2}$
View full question & answer→Question 675 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{{\text{x}}-\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{{\text{x}}-\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$$=\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 685 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
$\Rightarrow\lim\limits_{({\text{x}+2)}\rightarrow{(\text{a}+2)}}\frac{(\text{y})^\frac{3}{2}-(\text{b})^\frac{3}{2}}{(\text{y})-(\text{b})},$ $\Big[\text{Using formula}\ \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}\Big]$
$=\frac{3}{2}(\text{b})^{\frac{3}{2}-1}$
$=\frac{3}{2}(\text{a}+2)^{\frac{3}{2}-1}$
$=\frac{3}{2}(\text{a}+2)^{\frac{1}{2}}$
View full question & answer→Question 695 Marks
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$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\sqrt{\text{x}}-\cos\sqrt{\text{a}}}{\text{x}-\text{a}}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{-2\sin\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)\times\sin\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}{\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)\times\Big({\sqrt{\text{x}}-\sqrt{\text{a}}}\Big)}$ $=-2\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)\times\lim\limits_{\text{x}\rightarrow{\text{a}}}\sin\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)\times\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}\times\frac12$ $=-2\sin\sqrt{\text{a}}\times1\times\frac{1}{2\sqrt{\text{a}}}\times\frac12$ $=-\frac{1}{2\sqrt{\text{a}}}\sin\sqrt{\text{a}}$
View full question & answer→Question 705 Marks
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}}}$$=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\big(\sqrt{\text{x}-2}-\sqrt{4-\text{x}}\big)}\times\frac{\big(\sqrt{\text{x}}-2+\sqrt{4-\text{x}}\big)}{\big(\sqrt{\text{x}}-2+\sqrt{4-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{(\text{x}-2)-(4-\text{x})}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{\text{x}-2-4+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{2(\text{x}-3)}$
$=\frac{1}{2}\lim\limits_{\text{x}\rightarrow3}\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)$
$=\frac{1}{2}\big(\sqrt{3-2}+\sqrt{4-\text{x}}\big)$
$=\frac{1}{2}\big(\sqrt{1}+\sqrt{1}\big)$
$=\frac{1}{2}(1+1)=\frac22$
$=1$
View full question & answer→Question 715 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}^2+\text{cx}}-\text{x}}{}$
Answer$\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}^2+\text{cx}}-\text{x}}{}$$=\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}$
View full question & answer→Question 725 Marks
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$
View full question & answer→Question 735 Marks
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}}$
View full question & answer→Question 745 Marks
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)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(5\text{x}-4-\text{x})}{(\text{x}-1)\big(\text{x}^2+1+\text{x}\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4(\text{x}-1)}{(\text{x}-1)\big(\text{x}^2+\text{x}+1\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\frac{4}{(1+1+1)\big(\sqrt{5-4}+\sqrt{1}\big)}$
$=\frac{4}{(3)(1+1)}$
$=\frac{4}{3\times2}=\frac23$
View full question & answer→Question 755 Marks
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 765 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$
View full question & answer→Question 775 Marks
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}}$
View full question & answer→Question 785 Marks
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$
View full question & answer→Question 795 Marks
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)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h}-\text{x})}{\text{h}\big(\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big)}$
$=\frac{1}{\sqrt{\text{x}}+\sqrt{\text{x}}}$
$=\frac{1}{2\sqrt{\text{x}}}$
View full question & answer→Question 805 Marks
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$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\frac{\text{a}+\text{x}+\text{a}-\text{x}}{2}\big)\cos\big(\frac{\text{a}+\text{x}-\text{a}+\text{x}}{2}\big)-\sin\text{a}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}(\cos\text{x}-1)}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin\text{a}(1-\cos\text{x})}{\text{x}\sin\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}\frac{\text{x}}{2}}{\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}\frac{\text{x}}{2}}{\big(\cos\frac{\text{x}}{2}\big)\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{\tan\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\times\frac12$
$=-2\sin\text{a}\times1\times\frac12$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=-\sin\text{a}$
View full question & answer→Question 815 Marks
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$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^4+2^4+3^4+\ \cdots+\text{n}^4}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \cdots+\text{n}^3}{\text{n}^5}$
$=\ \lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(1+\text{n})(1+2\text{n})\big(-1+3\text{n}+3\text{n}^2\big)}{30}}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)^2}{\text{n}^5}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\frac{1}{\text{n}}+1\big)\big(\frac{1}{\text{n}}+2\big)\big(-\frac{1}{\text{n}^2}+\frac{3}{\text{n}}+3\big)}{30}\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}^5}\Bigg(\frac{\text{n}^2\big(\text{n}^2+2\text{n}+1\big)}{4}\Bigg)$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\frac{1}{\text{n}}+1\big)\big(\frac{1}{\text{n}}+2\big)\big(-\frac{1}{\text{n}^2}+\frac{3}{\text{n}}+3\big)}{30}\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(\frac{1}{\text{n}}+\frac{2}{\text{n}^2}+\frac{1}{\text{n}^3}\Big)}{4}$
$=\frac{1\times2\times3}{30}-0$
$=\frac15$
View full question & answer→Question 825 Marks
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$\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{1+\text{cosec}^3\text{x}}{\cot^2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{(1+\text{cosec}\text{x})\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{ x}\big)}{\big(\text{cosec}^2\text{x}-1\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{(\text{cosec}\text{x}+1)\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{x}\big)}{(\text{cosecx}-1)\big(\text{cosecx+1}\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{ x}\big)}{(\text{cosec }\text{x}-1)}$
$=\frac{1+\text{cosec}^2\frac{3\pi}{2}-\text{cosec}\frac{3\pi}{2}}{\text{cosec}\frac{3\pi}{2}-1}$
$=\frac{1+(-1)^2-(-1)}{(-1)-1}$ $\Big[\therefore\text{cosec}\frac{3\pi}{2}=-1\Big]$
$=\frac{1+1+1}{-2}$
$=\frac{-3}{2}$
View full question & answer→Question 835 Marks
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:
$=\lim\limits_{\text{x}\rightarrow{\infty}}\Big(\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}\Big)\begin{pmatrix}\frac{\sqrt{1+\frac{\text{a}^2}{\text{x}^2}}+\sqrt{1+\frac{\text{a}^2}{\text{x}^2}}}{\sqrt{1+\frac{1}{\text{a}^2}}+\sqrt{1+\frac{\text{b}^2}{\text{a}^2}}}\end{pmatrix}$
$\text{As x}\rightarrow\infty,\frac{1}{\text{x}},\frac{1}{\text{x}^2}\rightarrow0$
$=\Big(\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}\Big)\Big(\frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}\Big)$
$=\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}$
View full question & answer→Question 845 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(\cos3\text{x}-\cos\text{x})}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(\cos3\text{x}-\cos\text{x})}{\text{x}^3}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}\Big(-2\sin\big(\frac{3\text{x}+\text{x}}{2}\big)\sin\big(\frac{3\text{x}-\text{x}}{2}\big)\Big)}{\text{x}^3}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}(-2\sin2\text{x}\sin\text{x})}{\text{x}^3}$
$=\frac{-2\lim\limits_{\text{x} \rightarrow0}\sin2\text{x}\times\lim\limits_{\text{x} \rightarrow0}\sin2\text{x}\times\lim\limits_{\text{x} \rightarrow0}\sin\text{x}}{\text{x}^3}$
$=-2\Big(\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\Big)\times\Big(2\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\Big)\times\Big(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)$
$=-2(1\times2)\times(2)\times(1)$
$=-8$
View full question & answer→Question 855 Marks
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
$=\lim\limits_{\text{y}\rightarrow\infty}\Big(\sqrt{4\text{y}^2+7\text{y}}-2\text{y}\Big)$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{\Big(\sqrt{4\text{y}^2+7\text{y}}-2\text{y}\Big)\Big(\sqrt{4\text{y}^2+7\text{y}}+2\text{y}\Big)}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{\Big(4\text{y}^2+7\text{y}+4\text{y}^2\big)}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{(7\text{y})}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{7}{\sqrt{4+\frac{7}{\text{y}}}+2}$
$=\frac{7}{2+2}=\frac{7}{4}$
View full question & answer→Question 865 Marks
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$
View full question & answer→Question 875 Marks
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$
View full question & answer→Question 885 Marks
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}$
View full question & answer→Question 895 Marks
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}}$
$\Rightarrow \text{Let} 1 + \text{x} = \text{y}, \text{as x} → 0, \text{y} → 1$
$=\frac{\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^6-1^6}{\text{y}-1}}{\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^2-1}{\text{y}-1}}$
$=\frac{6(1)^{6-1}}{2(1)^{2-1}}$ $\Big[\text{Using formula} \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{a}-\text{a}}=\text{na}^{\text{n}-1}\Big]$
$=\frac62$
$=3$
View full question & answer→Question 905 Marks
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 $\frac{\pi}{8}-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cot4\big(\frac\pi8-\text{y}\big)-\cos4\big(\frac\pi8-\text{y}\big)}{(8)^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cot4\big(\frac\pi2-4\text{y}\big)-\cos4\big(\frac\pi2-4\text{y}\big)}{(8)^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan4\text{y}-\sin4\text{y}}{(8)^2\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{\sin4\text{y}}{\cos4\text{y}}-\sin4\text{y}}{8^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}-\sin4\text{y}\cos4\text{y}}{\cos4\text{y}\times\text{y}^3\times8^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}\big(2\sin^22\text{y}\big)}{\cos4\text{y}\times\text{y}^3\times8^3}$
$=\frac{2}{8^3}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}}{\text{y}}\times\frac{\sin^22\text{y}}{\text{y}^2}\times\frac{1}{\cos4\text{y}}$
$=\frac{2}{8^3}\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}}{4\text{y}}\times4\Big)\times\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\cos4\text{y}}$
$=\frac{2}{8^2}(1\times4)\times(1)\times4\times\frac11$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\cos\theta=1\Big]$
$=\frac{2\times4\times4}{8\times8\times8}$
$\frac{1}{16}$
View full question & answer→Question 915 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos5\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos5\text{x}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(-2\sin\big(\frac{3\text{x}+5\text{x}}{2}\big)\sin\big(\frac{3\text{x}-5\text{x}}{2}\big)\Big)}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{-2\sin4\text{x}\sin(-\text{x})}{\text{x}^2}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin4\text{x}\sin\text{x}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin4\text{x}}{\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$
$=2\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin4\text{x}}{4\text{x}}\times4\Big)\times\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=8$
View full question & answer→Question 925 Marks
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}}}$$\Rightarrow\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\frac{\text{x}^{\frac{5}{7}}-\text{a}^{\frac{5}{7}}}{\text{x}-\text{a}}}{\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}}$ [Dividing numrator and denominator by x - a]
$=\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{5}{7}}-\text{a}^{\frac{5}{7}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}}$
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
Here, $\text{n}=\frac57$ is numerator and applying $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\text{m}-\text{a}^\text{m}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator, where $\text{m}=\frac27$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\frac{\text{x}^{\frac{5}{7}}-\text{a}^{\frac{5}{7}}}{\text{x}-\text{a}}}{\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}}=\frac{\frac{5}{7}\text{a}^{\frac{5}{7}-1}}{\frac{2}{7}(\text{a})^{\frac{2}{7}-1}}$
$=\frac{\frac{5}{7}\text{a}^{\frac{-2}{7}}}{\frac{2}{7}\text{a}^{\frac{-5}{7}}}$
$=\frac{5}{2}\text{a}^{\frac{-2}{7}+\frac{5}{7}}$
$=\frac{5}{2}\text{a}^{\frac{3}{7}}$
View full question & answer→Question 935 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$
Answer$\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$$=\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\}$
$= \lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}(\text{x}-2)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}(\text{x}-1)-2(2\text{x}-3)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}^2-\text{x}-4\text{x}+6}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}^2-2\text{x}-3\text{x}+6}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}(\text{x}-2)-3(\text{x}-2)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{(\text{x}-2)(\text{x}-3)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}\frac{2(2\text{x}-3)}{\text{x}(\text{x}-2)(\text{x}-1)}\bigg\}$
$=\frac{-1}{2}$
View full question & answer→Question 945 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\infty}\Big\{\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big\}^{\frac{3\text{x}-2}{3\text{x}+2}}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\Big\{\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big\}^{\frac{3\text{x}-2}{3\text{x}+2}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow\infty}\Big\{\Big({\frac{3\text{x}-2}{3\text{x}+2}}\Big)\text{In}\Big(\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big)\Big\}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow\infty}\Bigg\{\Bigg(\frac{3-\frac{2}{\text{x}}}{3+\frac{2}{\text{x}}}\Bigg)\Big(\text{In}\Big(\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big)\Big)\Bigg\}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow\infty}\Bigg\{\Bigg(\frac{3-\frac{2}{\text{x}}}{3+\frac{2}{\text{x}}}\Bigg)\begin{pmatrix}\text{In}\begin{pmatrix}\frac{1+\frac{2}{\text{x}}+\frac{3}{\text{x}^2}}{2+\frac{1}{\text{x}}+\frac{5}{\text{x}^2}}\end{pmatrix}\end{pmatrix}\Bigg\}}$
$=\text{e}^{1.\text{In}\big(\frac{1}{2}\big)}$
$=\frac{1}{2}$
View full question & answer→Question 955 Marks
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$
View full question & answer→Question 965 Marks
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.
View full question & answer→Question 975 Marks
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$
View full question & answer→Question 985 Marks
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)}$
$=\frac{(1)^24(1)-2}{(1)^24(1)+1}$
$=\frac{1+4-2}{1+4+1}$
$=\frac{3}{6}$
$=\frac{1}{2}$
View full question & answer→Question 995 Marks
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.
View full question & answer→Question 1005 Marks
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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