Question 14 Marks
Evaluate the following limits: $\lim _{x \rightarrow \infty}\left(\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}$
Answer$\lim _{x \rightarrow \infty}\left(\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}$
$=\left[\lim _{x \rightarrow \infty}\left(\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right)\right]^{\lim _{x \rightarrow \infty} \frac{4 x+3}{8 x-1}}$
Consider
$\lim _{x \rightarrow \infty} \frac{8 x^2+5 x+3}{2 x^2-7 x-5}$
$=\lim _{x \rightarrow \infty}\left(\frac{\frac{8 x^2+5 x+3}{x^2}}{\left.\frac{2 x^2-7 x-5}{x^2}\right)} \cdots\left[\begin{array}{l} \text { Divide Numerator and } \\ \text { Denominatorby } x^2 \end{array}\right]\right.$
$=\frac{\lim _{x \rightarrow \infty}\left(8+\frac{5}{x}+\frac{3}{x^2}\right)}{\lim _{x \rightarrow \infty}\left(2-\frac{7}{x}-\frac{5}{x^2}\right)} \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, \mathrm{k}>0\right]$
$=\frac{8+0+0}{2-7-0}$
$=4$
Consider
$\lim _{x \rightarrow \infty} \frac{4 x+3}{8 x-1}$
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x+3}{x}}{\frac{8 x-1}{x}}$
$[$Divide Numerator and Denominator by $x ]$
$=\frac{\lim _{x \rightarrow \infty}\left(4+\frac{3}{x}\right)}{\lim _{x \rightarrow \infty}\left(8-\frac{1}{x}\right)}$
$=\frac{4+0}{8-0} \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x}=0\right]$
$=\frac{1}{2}$
$=\frac{1}{2}$
Required limit $=(4)^{\frac{1}{2}}$
$=2$
View full question & answer→Question 24 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{1}{x^{12}}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^4}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^4}{4}\right)\right]$
Answer$\lim _{x \rightarrow 0} \frac{1}{x^{12}}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^4}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^4}{4}\right)\right]$
consider
$1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^4}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^4}{4}\right)$
$=\left[1-\cos \left(\frac{x^2}{2}\right)\right]-\cos \left(\frac{x^4}{4}\right)\left[1-\cos \left(\frac{x^2}{2}\right)\right]$
$=\left[1-\cos \left(\frac{x^2}{2}\right)\right]\left[1-\cos \left(\frac{x^4}{2}\right)\right]$
$=2 \sin ^2\left(\frac{x^2}{4}\right) \times 2 \sin ^2\left(\frac{x^4}{8}\right) $
Required limit
$ =\lim _{x \rightarrow 0} \frac{4\left[\sin \left(\frac{x^2}{4}\right)\right]^2 \cdot\left[\sin \left(\frac{x^4}{8}\right)\right]^2}{x^{12}}$
$=4 \lim _{x \rightarrow 0} \frac{\left[\sin \left(\frac{x^2}{4}\right)\right]^2}{x^4} \times \frac{\left[\sin \left(\frac{x^4}{8}\right)\right]^2}{x^8}$
$=4 \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{x^2}{4}\right)}{x^2}\right]^2 \times\left[\frac{\sin \left(\frac{x^4}{8}\right)}{x^4}\right]^2$
$=4 \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{x^2}{4}\right)}{\frac{x^2}{4}}\right]^2 \times \frac{1}{16} \times \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{x^4}{8}\right)}{\frac{x^4}{8}}\right]^2 \times \frac{1}{64}$
$=4 \times(1)^2 \times \frac{1}{16} \times(1)^2 \times \frac{1}{64} \quad \cdots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right]$
$=\frac{1}{256}$
View full question & answer→Question 34 Marks
Evaluate the following limits:$\lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)$
Answer$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x}}{x}$
$=\lim _{x \rightarrow 0} \frac{\sqrt{2 \sin ^2 \frac{x}{2}}}{x}=\lim _{x \rightarrow 0} \frac{\sqrt{2}\left|\sin \frac{x}{2}\right|}{x}$
$\begin{aligned} |x| & =x & & ; x \geq 0 \\ & =-x & & ; x<0 \end{aligned}$
Left hand limit $=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{2}\left|\sin \frac{x}{2}\right|}{x}$
$ =\sqrt{2} \lim _{x \rightarrow 0} \frac{\sin \left(\frac{-x}{2}\right)}{x}$
$=\sqrt{2} \lim _{x \rightarrow 0} \frac{-\sin \frac{x}{2}}{\frac{x}{2}} \times \frac{1}{2}$
$=\sqrt{2} \times(-1) \times \frac{1}{2}$
$\cdots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right]$
$=\frac{-1}{\sqrt{2}}$
Right hand limit $=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{2}\left|\sin \frac{x}{2}\right|}{x}$
$=\sqrt{2} \lim _{x \rightarrow 0} \frac{\sin \left(\frac{x}{2}\right)}{x}$
$=\sqrt{2} \lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}} \times \frac{1}{2}$
$=\sqrt{2} \times(1) \times \frac{1}{2}$
$\ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right]$
$=\frac{1}{\sqrt{2}}$
$\therefore$ Left hand limit is not equal to the right hand limit
$\therefore \lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)$ does not exist.
View full question & answer→Question 44 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^x+1}{x \cdot \tan x}\right]$
Answer$ \lim _{x \rightarrow 0} \frac{\mathrm{a}^{3 x}-\mathrm{a}^{2 x}-\mathrm{a}^x+1}{x \tan x}$
$=\lim _{x \rightarrow 0} \frac{\mathrm{a}^{2 x} \cdot \mathrm{a}^x-\mathrm{a}^{2 x}-\mathrm{a}^x+1}{x \tan x}$
$=\lim _{x \rightarrow 0} \frac{\mathrm{a}^{2 x}\left(\mathrm{a}^x-1\right)-1\left(\mathrm{a}^x-1\right)}{x \tan x}$
$=\lim _{x \rightarrow 0} \frac{\left(\mathrm{a}^x-1\right) \cdot\left(\mathrm{a}^{2 x}-1\right)}{x \tan x}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(\mathrm{a}^x-1\right) \cdot\left(\mathrm{a}^{2 x}-1\right)}{x^2}}{\frac{x \tan x}{x^2}} \ldots[\because x \rightarrow 0, x \neq 0]$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{\mathrm{a}^x-1}{x}\right) \cdot\left(\frac{\mathrm{a}^{2 x}-1}{2 x}\right) \cdot 2}{\frac{\tan x}{x}}$
$=\frac{2 \lim _{x \rightarrow 0}\left(\frac{\mathrm{a}^x-1}{x}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{\mathrm{a}^{2 x}-1}{2 x}\right)}{\lim _{x \rightarrow 0} \frac{\tan x}{x}}$
$=\frac{2(\log a)(\log a)}{1}$
$\cdots\left[\because x \rightarrow 0,2 x \rightarrow 0 \text { and } \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}\right]$
$=2(\log a)^2$
View full question & answer→Question 54 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{x\left(6^x-3^x\right)}{\cos (6 x)-\cos (4 x)}\right]$
Answer$\lim _{x \rightarrow 0} \frac{x\left(6^x-3^x\right)}{\cos 6 x-\cos 4 x}$
$=\lim _{x \rightarrow 0} \frac{x \cdot 3^x\left(2^x-1\right)}{-2 \sin \left(\frac{6 x+4 x}{2}\right) \sin \left(\frac{6 x-4 x}{2}\right)}$
$=\frac{1}{-2} \lim _{x \rightarrow 0} \frac{x \cdot 3^x\left(2^x-1\right)}{\sin 5 x \cdot \sin x}$
$=\frac{1}{-2} \lim _{x \rightarrow 0} \frac{\frac{x \cdot 3^x\left(2^x-1\right)}{x^2}}{\frac{\sin 5 x \cdot \sin x}{x^2}}
...[$ Divide Numerator and Denominator by $x^2]$.
$=\frac{1}{-2} \frac{\lim _{x \rightarrow 0}\left[3^x \frac{\left(2^x-1\right)}{x}\right]}{\lim _{x \rightarrow 0}\left(\frac{\sin 5 x}{x} \cdot \frac{\sin x}{x}\right)}$
$=\frac{1}{-2} \frac{\lim _{x \rightarrow 0}\left(3^x\right) \cdot \lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)}{\lim _{x \rightarrow 0}\left(\frac{\sin x}{5 x} \times 5\right) \cdot \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)}$
$=\frac{1}{-2} \times \frac{3^0 \times \log 2}{1 \times 5 \times 1} \quad \ldots\left(\begin{array}{l} \because x \rightarrow 0,5 x \rightarrow 0 \\ \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \text { a } \\
\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{array}\right)$
$=\frac{-1}{10} \log 2$
View full question & answer→Question 64 Marks
Evaluate the following limits: $\lim _{x \rightarrow \infty}[\sqrt{x}(\sqrt{x+1}-\sqrt{x})]$
Answer$\lim _{x \rightarrow 0}[\sqrt{x}(\sqrt{x+1}-\sqrt{x})]$
$=\lim _{x \rightarrow 0} \sqrt{x}(\sqrt{x+1}-\sqrt{x}) \times \frac{(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}
...[$By rationalization$]$
$=\lim _{x \rightarrow 0} \frac{\sqrt{x}(x+1-x)}{\sqrt{x+1}+\sqrt{x}}$
$=\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}$
$=\lim _{x \rightarrow \infty} \frac{1}{\frac{\sqrt{x+1}}{\sqrt{x}}+1} \quad \cdots\left[\begin{array}{l}
\text { Divide numerator and } \\\text { denominator by } \sqrt{x}\end{array}\right]$
$=\lim _{x \rightarrow \infty} \frac{1}{\sqrt{\frac{x+1}{x}+1}}$
$=\frac{1}{\lim _{x \rightarrow \infty}\left(\sqrt{1+\frac{1}{x}}+1\right)}$
$=\frac{1}{\sqrt{1+0}+1} \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, k>0\right]$
$=\frac{1}{1+1}$
$=\frac{1}{2}$
View full question & answer→Question 74 Marks
Evaluate the following limits:
$\lim _{x \rightarrow \infty}\left[\frac{\left(3 x^2+4\right)\left(4 x^2-6\right)\left(5 x^2+2\right)}{4 x^6+2 x^4-1}\right]$
Answer$ \lim _{x \rightarrow \infty} \frac{\left(3 x^2+4\right)\left(4 x^2-6\right)\left(5 x^2+2\right)}{4 x^6+2 x^4-1}$
$=\lim _{x \rightarrow \infty} \frac{\frac{\left(3 x^2+4\right)\left(4 x^2-6\right)\left(5 x^2+2\right)}{x^6}}{\frac{4 x^6+2 x^4-1}{x^6}} $
$ =\lim _{x \rightarrow \infty} \frac{\left(\frac{3 x^2+4}{x^2}\right)\left(\frac{4 x^2-6}{x^2}\right)\left(\frac{5 x^2+2}{x^2}\right)}{4+\frac{2}{x^2}-\frac{1}{x^6}}$
$=\frac{\lim _{x \rightarrow \infty}\left(3+\frac{4}{x^2}\right)\left(4-\frac{6}{x^2}\right)\left(5+\frac{2}{x^2}\right)}{\lim _{x \rightarrow \infty}\left(4+\frac{2}{x^2}-\frac{1}{x^6}\right)}$
$=\frac{\lim _{x \rightarrow \infty}\left(3+\frac{4}{x^2}\right) \cdot \lim _{x \rightarrow \infty}\left(4-\frac{6}{x^2}\right) \cdot \lim _{x \rightarrow \infty}\left(5+\frac{2}{x^2}\right)}{\lim _{x \rightarrow \infty} 4+\lim _{x \rightarrow \infty} \frac{2}{x^2}-\lim _{x \rightarrow \infty} \frac{1}{x^6}}$
$=\frac{(3+0)(4-0)(5+0)}{4+0-0} \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^{ k }}=0, k >0\right]$
$=\frac{3 \times 4 \times 5}{4}$
$=15 $
View full question & answer→Question 84 Marks
Evaluate the following limits:
$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+4 x+16}-\sqrt{x^2+16}\right]$
Answer$\lim _{x \rightarrow \infty} \sqrt{x^2+4 x+16}-\sqrt{x^2+16}$
$=\lim _{x \rightarrow \infty} \frac{\left(\sqrt{x^2+4 x+16}-\sqrt{x^2+16}\right)\left(\sqrt{x^2+4 x+16}+\sqrt{x^2+16}\right)}{\left(\sqrt{x^2+4 x+16}+\sqrt{x^2+16}\right)} $
...[By rationalization]
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x}{x}}{\sqrt{\frac{x^2+4 x+16}{x^2}+\sqrt{\frac{x^2+16}{x^2}}}}$
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x}{x}}{\sqrt{\frac{x^2+4 x+16}{x^2}}+\sqrt{\frac{x^2+16}{x^2}}}$
$=\lim _{x \rightarrow \infty} \frac{4}{\sqrt{1+\frac{4}{x}+\frac{16}{x^2}}+\sqrt{1+\frac{16}{x^2}}}$
$=\frac{4}{\lim _{x \rightarrow \infty} \sqrt{1+\frac{4}{x}+\frac{16}{x^2}}+\lim _{x \rightarrow \infty} \sqrt{1+\frac{16}{x^2}}}$
$=\frac{4}{\sqrt{1+0+0}+\sqrt{1+0}} \quad \cdots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, k>0\right]$
$=\frac{4}{2}$
$=2$
View full question & answer→Question 94 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{(49)^x-2(35)^x+(25)^x}{\sin x \cdot \log (1+2 x)}\right]$
Answer$\lim _{x \rightarrow 0} \frac{(49)^x-2(35)^x+(25)^x}{\sin x \cdot \log (1+2 x)}$
$=\lim _{x \rightarrow 0} \frac{\left(7^x\right)^2-2\left(7^x\right)\left(5^x\right)+\left(5^x\right)^2}{\sin x \cdot \log (1+2 x)}$
$=\lim _{x \rightarrow 0} \frac{\left(7^x-5^x\right)^2}{\sin x \cdot \log (1+2 x)}$
$=\lim _{x \rightarrow 0} \frac{\left[\left(7^x-1\right)-\left(5^x-1\right)\right]^2}{\frac{\sin x \cdot \log (1+2 x)}{x^2}} \cdots\left[\begin{array}{l} \text { Divide Numerator and } \\ \text { Denominator by } x^2 . \\ \because x \rightarrow 0, x \neq 0 \\ \therefore x^2 \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow 0}\left[\frac{\left(7^x-1\right)-\left(5^x-1\right)}{x}\right]^2}{\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{\log (1+2 x)}{x}}$
$=\frac{\lim _{x \rightarrow 0}\left[\frac{7^x-1}{x}-\frac{5^x-1}{x}\right]^2}{\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0} \frac{\log (1+2 x)}{x}}$
$=\frac{\left[\lim _{x \rightarrow 0} \frac{7^x-1}{x}-\lim _{x \rightarrow 0} \frac{5^x-1}{x}\right]^2}{\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x} \times 2}$
$=\frac{(\log 7-\log 5)^2}{1 \times 1 \times 2}\ldots\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1, \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a, \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
$=\frac{1}{2}\left(\log \frac{7}{5}\right)^2$
View full question & answer→Question 104 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left(\frac{8^{\sin x}-2^{\tan x}}{e^{2 x}-1}\right)$
Answer$\lim _{x \rightarrow 0} \frac{8^{\sin x}-2^{\tan x}}{e^{2 x}-1}$
$=\lim _{x \rightarrow 0} \frac{\left(8^{\operatorname{tin} x}-1\right)-\left(2^{\sin x}-1\right)}{e^{2 x}-1}$
$\left(8^{\sin x}-1\right)-\left(2^{\tan x}-1\right)$
$=\lim _{x \rightarrow 0} \frac{x}{\frac{\mathrm{e}^{2 x}-1}{x} }\ldots\left[\begin{array}{l} \text { Divide Numerator and } \\ \text { Denominator by } x . \\
\because x \rightarrow 0, \quad x \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{8^{\sin x}-1}{x}-\frac{2^{\sin x}-1}{x}\right)}{\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{8^{\sin x}-1}{\sin x} \cdot \frac{\sin x}{x}-\frac{2^{\tan x}-1}{\tan x} \cdot \frac{\tan x}{x}\right)}{\lim _{x \rightarrow 0} \frac{\mathrm{e}^{2 x}-1}{x}}$
$=\frac{\left(\lim _{x \rightarrow 0} \frac{8^{\sin x}-1}{\sin x}\right)\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)-\left(\lim _{x \rightarrow 0} \frac{2^{\operatorname{man} x}-1}{\tan x}\right)\left(\lim _{x \rightarrow 0} \frac{\tan x}{x}\right)}{\left(\lim _{x \rightarrow 0} \frac{\mathrm{e}^{2 x}-1}{2 x}\right) \times 2}$
$=\frac{(\log 8)(1)-(\log 2)(1)}{(1) \times 2} \ldots\left[\begin{array}{l} \because x \rightarrow 0,2 x \rightarrow 0, \\
\sin x \rightarrow 0, \tan x \rightarrow 0 \\ \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log a \end{array}\right]$
$=\frac{\log \frac{8}{2}}{2}$
$=\frac{\log 4}{2}$
$=\frac{\log (2)^2}{2}$
$=\frac{2 \log 2}{2}$
$=\log 2$
View full question & answer→Question 114 Marks
Evaluate the following limits:
$\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{\cos 3 x+3 \cos x}{(2 x-\pi)^3}\right]$
Answer$ \lim _{x \rightarrow \frac{\pi}{2}} \frac{3 \cos x+\cos 3 x}{(2 x-\pi)^3}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{3 \cos x+4 \cos ^3 x-3 \cos x}{(2 x-\pi)^3}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{4 \cos ^3 x}{(2 x-\pi)^3}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{4 \cos ^3 x}{8\left(x-\frac{\pi}{2}\right)^3} $
Put $x-\frac{\pi}{2}=\mathrm{h}$,
$ \therefore \quad x=\frac{\pi}{2}+\mathrm{h}$
$\text { As } x \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{2}} \frac{4 \cos ^3 x}{8\left(x-\frac{\pi}{2}\right)^3}$
$=\lim _{h \rightarrow 0} \frac{4 \cos ^3\left(\frac{\pi}{2}+h\right)}{8 h^3}$
$=\lim _{h \rightarrow 0} \frac{4(-\sin h)^3}{8 h^3} \quad \ldots\left[\because \cos \left(\frac{\pi}{2}+\theta\right)=-\sin \theta\right]$
$=-\frac{1}{2} \cdot \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^3$
$=-\frac{1}{2} \cdot\left(\lim _{h \rightarrow 0} \frac{\sin h}{h}\right)^3$
$=\frac{-1}{2} \cdot(1)^3 \quad \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=\frac{-1}{2}$
View full question & answer→Question 124 Marks
Evaluate the following limits:
$\lim _{x \rightarrow a}\left[\frac{\sin (\sqrt{x})-\sin (\sqrt{a})}{x-a}\right]$
Answer$\lim _{x \rightarrow \mathrm{a}}\left[\frac{\sin (\sqrt{x})-\sin (\sqrt{\mathrm{a}})}{x-\mathrm{a}}\right]$
Put $\sqrt{x}=y, \sqrt{\mathrm{a}}=\mathrm{b}$ and $y=\mathrm{b}+\mathrm{h}$.
As $x \rightarrow \mathrm{a}, y \rightarrow \mathrm{b}$ and $\mathrm{h} \rightarrow 0$.
$ \lim _{x \rightarrow a} \frac{\sin \sqrt{x}-\sin \sqrt{a}}{x-a}$
$=\lim _{y \rightarrow b} \frac{\sin y-\sin b}{y^2-b^2}$
$=\lim _{y \rightarrow b} \frac{\sin y-\sin b}{(y-b)(y+b)}$
$=\lim _{h \rightarrow 0} \frac{\sin (b+h)-\sin b}{h(b+h+b)} $
$=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{b+h+b}{2}\right) \sin \left(\frac{b+h-b}{2}\right)}{h(2 b+h)}$
$=\lim _{h \rightarrow 0} \frac{2 \cos \left(b+\frac{h}{2}\right) \cdot \sin \left(\frac{h}{2}\right)}{h(2 b+h)}$
$=\lim _{h \rightarrow 0} \frac{\cos \left[b+\frac{h}{2}\right]}{2 b+h} \cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$=\lim _{h \rightarrow 0} \frac{\cos \left[b+\frac{h}{2}\right]}{2 b+h} \cdot\left[\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}\right]$
$=\frac{\cos (b+0)}{2 b+0} \cdot 1$
$=\frac{\cos b}{2 b}$
$=\frac{\cos \sqrt{a}}{2 \sqrt{a}} $
View full question & answer→Question 134 Marks
Evaluate the following limits: $\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{\cos x-\sqrt{3} \sin x}{\pi-6 x}\right]$
Answer$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{\cos x-\sqrt{3} \sin x}{\pi-6 x}\right]$
Put $\frac{\pi}{6}-x=\mathrm{h}$
$\therefore x=\frac{\pi}{6}-\mathrm{h}$
As $x \rightarrow \frac{\pi}{6}, \mathrm{~h} \rightarrow 0$
$\therefore \lim _{x \rightarrow \frac{\pi}{6}} \frac{\cos x-\sqrt{3} \sin x}{\pi-6 x}$
$=\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}-h\right)-\sqrt{3} \sin \left(\frac{\pi}{6}-h\right)}{\pi-6\left(\frac{\pi}{6}-h\right)}$
$=\lim _{h \rightarrow 0} \frac{\left[
\left(\cos \frac{\pi}{6} \cdot \cos h+\sin \frac{\pi}{6} \cdot \sin h\right) -\sqrt{3}\left(\sin \frac{\pi}{6} \cdot \cos h-\cos \frac{\pi}{6} \cdot \sin h\right)
\right]}{\pi-\pi+6 h}$
$=\lim _{h \rightarrow 0} \frac{\left(\frac{\sqrt{3}}{2} \cos h+\frac{1}{2} \sin h\right)-\sqrt{3}\left(\frac{1}{2} \cos h-\frac{\sqrt{3}}{2} \sin h\right)}{6 h}$
$=\lim _{h \rightarrow 0} \frac{\frac{\sqrt{3}}{2} \cos h+\frac{1}{2} \sin h-\frac{\sqrt{3}}{2} \cos h+\frac{3}{2} \sin h}{6 h}$
$=\lim _{h \rightarrow 0} \frac{2 \sin h}{6 h}$
$=\frac{1}{3} \lim _{h \rightarrow 0} \frac{\sin h}{h}$
$=\frac{1}{3}(1)\cdots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=\frac{1}{3}$
View full question & answer→Question 144 Marks
Evaluate the following limits: $\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt[5]{x}-\sqrt[5]{a}}$
Answer$\lim _{x \rightarrow \mathrm{a}} \frac{\sin x-\sin \mathrm{a}}{\sqrt[5]{x}-\sqrt[5]{\mathrm{a}}}$
$=\lim _{x \rightarrow \mathrm{a}} \frac{2 \cos \left(\frac{x+\mathrm{a}}{2}\right) \cdot \sin \left(\frac{x-\mathrm{a}}{2}\right)}{x^{\frac{1}{5}}-\mathrm{a}^{\frac{1}{5}}}$
$=\lim _{x \rightarrow a} \frac{2 \cos \left(\frac{x+a}{2}\right) \cdot \frac{\sin \left(\frac{x-a}{2}\right)}{x-a}}{\frac{x^{\frac{1}{5}}-a^{\frac{1}{5}}}{x-a}}\cdots\left[\begin{array}{l}
\text { Divide numerator and denominator by } x-\mathrm{a} . \\ \because x \rightarrow \mathrm{a}, x \neq \mathrm{a}, \therefore x-\mathrm{a} \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow \mathrm{a}} \cos \left(\frac{x+\mathrm{a}}{2}\right) \cdot \lim _{x \rightarrow \mathrm{a}}\left[\frac{\sin \left(\frac{x-\mathrm{a}}{2}\right)}{\frac{x-\mathrm{a}}{2}}\right]}{\lim _{x \rightarrow \mathrm{a}} \frac{x^{\frac{1}{5}}-\mathrm{a}^{\frac{1}{5}}}{x-\mathrm{a}}}$
$=\frac{\cos \left(\frac{a+a}{2}\right) \times 1}{\frac{1}{5} \cdot a^{\frac{-4}{5}}}$
$\ldots\left[\begin{array}{l} \because x \rightarrow a, x-a \rightarrow 0 \\ \therefore \frac{x-a}{2} \rightarrow 0 ; \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 \\ \text { and } \lim _{x \rightarrow 0} \frac{x^n-a^n}{x-a}=n a^{n-1}
\end{array}\right]$
$=5 a^{\frac{4}{5}} \cdot \cos a$
View full question & answer→Question 154 Marks
Evaluate the following limits: $\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}\right]$
Answer$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}\right]$
Consider, $2 \sin ^2 x+\sin x-1$
$=2 \sin ^2 x+2 \sin x-\sin x-1$
$=2 \sin x(\sin x+1)-1(\sin x+1)$
$=(\sin x+1)(2 \sin x-1)$
Now,
$2 \sin ^2 x-3 \sin x+1 =2 \sin ^2 x-2 \sin x-\sin x+1$
$ =2 \sin x(\sin x-1)-1(\sin x-1)$
$ =(\sin x-1)(2 \sin x-1)$
$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}\right]$
$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{(\sin x+1)(2 \sin x-1)}{(\sin x-1)(2 \sin x-1)}$
$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin x+1}{\sin x-1} \quad \cdots\left[\begin{array}{l}
\because x \rightarrow \frac{\pi}{6}, \sin x \rightarrow \frac{1}{2} \\
\therefore \sin x \neq \frac{1}{2} \therefore 2 \sin x \neq 1 \\
\therefore 2 \sin x-1 \neq 0
\end{array}\right]$
$=\frac{\lim _{x \rightarrow \frac{\pi}{6}}(\sin x+1)}{\lim _{x \rightarrow \frac{\pi}{6}}(\sin x-1)}$
$=\frac{\sin \frac{\pi}{6}+1}{\sin \frac{\pi}{6}-1}$
$=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}$
$=\frac{\frac{3}{2}}{\frac{-1}{2}}$
$=-3$
View full question & answer→Question 164 Marks
Evaluate the following limits: $\lim _{x \rightarrow \pi}\left[\frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin ^2 x}\right]$
Answer$\lim _{x \rightarrow \pi} \frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin ^2 x}$
$=\lim _{x \rightarrow \pi} \frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin ^2 x} \times \frac{\sqrt{1-\cos x}+\sqrt{2}}{\sqrt{1-\cos x}+\sqrt{2}} \text {...[By rationalization] }$
$=\lim _{x \rightarrow \pi} \frac{1-\cos x-2}{\sin ^2 x} \times \frac{1}{\sqrt{1-\cos x}+\sqrt{2}}$
$=\lim _{x \rightarrow \pi} \frac{-\cos x-1}{1-\cos ^2 x} \times \frac{1}{\sqrt{1-\cos x}+\sqrt{2}}$
$=\lim _{x \rightarrow \pi} \frac{-(1+\cos x)}{(1-\cos x)(1+\cos x)} \times \frac{1}{\sqrt{1-\cos x}+\sqrt{2}}$
$=\lim _{x \rightarrow x} \frac{-1}{(1-\cos x)[\sqrt{1-\cos x}+\sqrt{2}]}$
$\cdots\left[\begin{array}{l} \because x \rightarrow \pi, \cos x \neq \cos \pi \\ \therefore \cos x \neq-1, \therefore 1+\cos x \neq 0\end{array}\right]$
$=\frac{-1}{\lim _{x \rightarrow \pi}(1-\cos x) \cdot \lim _{x \rightarrow \pi}(\sqrt{1-\cos x}+\sqrt{2})}$
$=\frac{-1}{(1-\cos \pi)[\sqrt{1-\cos \pi}+\sqrt{2}]}$
$=\frac{-1}{[1-(-1)][\sqrt{1-(-1)}+\sqrt{2}]}$
$=\frac{-1}{(2)(2 \sqrt{2})}$
$=\frac{-1}{4 \sqrt{2}}$
View full question & answer→Question 174 Marks
Evaluate the following limits: $\lim _{z \rightarrow 4}\left[\frac{3-\sqrt{5+z}}{1-\sqrt{5-z}}\right]$
Answer$\lim _{z \rightarrow 4}\left[\frac{3-\sqrt{5+z}}{1-\sqrt{5-z}}\right]$
$=\lim _{x \rightarrow 4}\left[\frac{3-\sqrt{5+z}}{1-\sqrt{5-z}} \times \frac{3+\sqrt{5+z}}{3+\sqrt{5+z}} \times \frac{1+\sqrt{5-z}}{1+\sqrt{5-z}}\right]$
$\cdots\left[\begin{array}{l}\text { By taking conjugates of both, the } \\ \text { numerator as well as the denominator. }\end{array}\right]$
$=\lim _{z \rightarrow 4}\left[\frac{9-(5+z)}{1-(5-z)} \times \frac{1+\sqrt{5-z}}{3+\sqrt{5+z}}\right]$
$=\lim _{z \rightarrow 4}\left[\frac{4-z}{-4+z} \times \frac{1+\sqrt{5-z}}{3+\sqrt{5+z}}\right]$
$=\lim _{z \rightarrow 4}\left[\frac{-(z-4)}{z-4} \times \frac{1+\sqrt{5-z}}{3+\sqrt{5+z}}\right]$
$=\lim _{z \rightarrow 4}\left[\frac{-(1+\sqrt{5-z})}{3+\sqrt{5+z}}\right] \cdots\left[\because z \rightarrow 4, z \neq 4,\ \therefore z-4 \neq 0\right]$
$=\frac{-\lim _{z \rightarrow 4}(1+\sqrt{5-z})}{\lim _{z \rightarrow 4}(3+\sqrt{5+z})}$
$=\frac{-(1+\sqrt{5-4})}{3+\sqrt{5+4}}$
$=\frac{-(1+1)}{3+3}$
$=\frac{-2}{6}$
$=-\frac{1}{3}$
View full question & answer→Question 184 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 2}\left[\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\right]$
Answer$ \lim _{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}$
$=\lim _{x \rightarrow 2}\left[\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} \times \frac{\sqrt{1+\sqrt{2+x}}+\sqrt{3}}{\sqrt{1+\sqrt{2+x}}+\sqrt{3}}\right] $
...[By rationalization]
$ =\lim _{x \rightarrow 2} \frac{1+\sqrt{2+x}-3}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})}$
$=\lim _{x \rightarrow 2} \frac{\sqrt{2+x}-2}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})} \times \frac{\sqrt{2+x}+2}{\sqrt{2+x}+2} $
...[By rationalization]
$ =\lim _{x \rightarrow 2} \frac{2+x-4}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)}$
$=\lim _{x \rightarrow 2} \frac{x-2}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)}$
$=\lim _{x \rightarrow 2} \frac{1}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)}$
$=\frac{1}{(\sqrt{1+\sqrt{2+2}}+\sqrt{3})(\sqrt{2+2}+2)}$
$\left.=\frac{1}{(\sqrt{1+2}+\sqrt{3})(2+2)} \quad \therefore x-2 \neq 0\right]$
$=\frac{1}{(2 \sqrt{3})(4)}$
$=\frac{1}{8 \sqrt{3}} $
View full question & answer→Question 194 Marks
Evaluate the following limits: $\lim _{x \rightarrow a}\left[\frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}\right]$
Answer$\lim _{x \rightarrow a}\left[\frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}\right]$
$=\lim _{x \rightarrow a}\left[\frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} \times \frac{\sqrt{a+2 x}+\sqrt{3 x}}{\sqrt{a+2 x}+\sqrt{3 x}} \times \frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{3 a+x}+2 \sqrt{x}}\right]\cdots\left[\begin{array}{l} \text { By taking conjugates of both, the } \\ \text { numerator as well as the denominator. }
\end{array}\right]$
$=\lim _{x \rightarrow a}\left[\frac{(a+2 x)-3 x}{(3 a+x)-4 x} \times \frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right]$
$=\lim _{x \rightarrow a}\left[\frac{a-x}{3 a-3 x} \times \frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right]$
$=\lim _{x \rightarrow a}\left[\frac{-(x-a)}{-3(x-a)} \times \frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{\sqrt{3 a+x}+2 \sqrt{x}}{3(\sqrt{a+2 x}+\sqrt{3 x})}\right] \ldots[\because x \rightarrow a, x \neq a$
$=\frac{\sqrt{3 a+a}+2 \sqrt{a}}{3(\sqrt{a+2 a}+\sqrt{3 a})}$
$=\frac{\sqrt{4 a}+2 \sqrt{a}}{3(\sqrt{3 a}+\sqrt{3 a})}$
$=\frac{2 \sqrt{a}+2 \sqrt{a}}{3(2 \sqrt{3 a})}$
$=\frac{4 \sqrt{a}}{6 \sqrt{3} \sqrt{a}}$
$=\frac{2}{3 \sqrt{3}}$
View full question & answer→Question 204 Marks
Evaluate the following limits: $\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right]$
Answer$\lim _{x \rightarrow 2}\left(\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right)$
$=\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}} \times \frac{\sqrt{2+x}+\sqrt{6-x}}{\sqrt{2+x}+\sqrt{6-x}} \times \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}}\right]{\left[\begin{array}{l} \text { By taking conjugates of both, the } \\ \text { numerator as well as the denominator. }
\end{array}\right] } \\
= \lim _{x \rightarrow 2}\left[\frac{(2+x)-(6-x)}{(x-2)} \times \frac{\sqrt{x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{6-x}}\right]$
$= \lim _{x \rightarrow 2}\left[\frac{-4+2 x}{x-2} \times \frac{\sqrt{x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{6-x}}\right]$
$= \lim _{x \rightarrow 2}\left[\frac{2(\sqrt{x}+\sqrt{2})}{x-2} \times \frac{\sqrt{x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{6-x}}\right]$
$\left[\because x \rightarrow 2, x \neq 2,\cdots x-2 \neq 0\right]$
$=\frac{\lim _{x \rightarrow 2} 2(\sqrt{x}+\sqrt{2})}{\lim _{x \rightarrow 2}(\sqrt{2+x}+\sqrt{6-x})}$
$=\frac{2(\sqrt{2}+\sqrt{2})}{\sqrt{2+2}+\sqrt{6-2}}$
$=\frac{2(2 \sqrt{2})}{2+2}$
$=\frac{4 \sqrt{2}}{4}$
$=\sqrt{2}$
View full question & answer→Question 214 Marks
Evaluate the following limits: $\lim _{x \rightarrow 1}\left[\frac{x^4-3 x^2+2}{x^3-5 x^2+3 x+1}\right]$
View full question & answer→Question 224 Marks
Evaluate the following limits: $\lim _{x \rightarrow 2}\left[\frac{x^3-7 x+6}{x^3-7 x^2+16 x-12}\right]$
View full question & answer→Question 234 Marks
In the following examples, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.$\lim _{x \rightarrow 1}\left(x^2+x+1\right)=3$
AnswerWe have to find some $\delta>0$ such that
$\lim _{x \rightarrow 1}\left(x^2+x+1\right)=3$
Here $a=1, l=3$ and $f(x)=x^2+x+1$
Consider $\in>0$ and $|f(x)-\|| \in$
$ \therefore\left|\mathrm{x}^2+\mathrm{x}+1-3\right|<\in$
$\therefore\left|\mathrm{x}^2+\mathrm{x}-2\right|<\in$
$\therefore|(\mathrm{x}+2)(\mathrm{x}-1)|<\in \ldots . .(\mathrm{i}) $
We have to get rid of the factor $|x+2|$
$ \text { As }|x-1|<\delta$
$-\delta\therefore 1-\delta$ Since $\delta$ can be assumed as very small, let us choose $\delta<1$ $ \therefore 0<\mathrm{x}<2$
$\therefore 2<\mathrm{x}+2<4$
$\therefore|\mathrm{x}+2|<4$
$\therefore|(\mathrm{x}+2)(\mathrm{x}-1)|<4|\mathrm{x}-1| $
From (i) and (ii), we get
$ 4|x-1|<\in$
$\therefore|x-1|<\frac{\in}{4} $
If $\delta=\frac{\in}{4}$,
$|x-1|<\delta \Rightarrow x^2+x-2<\in$
$\therefore$ We choose $\delta=\min \left\{\frac{\in}{4}, 1\right\}$ then
$|x-1|<\delta \Rightarrow|f(x)-3|<\in$
View full question & answer→Question 244 Marks
In the following examples, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.$\lim _{x \rightarrow 2}\left(x^2-1\right)=3$
AnswerWe have to find some $\delta>0$ such that
$\lim _{x \rightarrow 2}\left(x^2-1\right)=3$
Here, $a=2, I=3$ and $f(x)=x^2-1$
Consider $\in>0$ and $|\mathrm{f}(\mathrm{x})-\||<\in$
$ \therefore\left|\left(x^2-1\right)-3\right|<\in$
$\therefore\left|x^2-4\right|<\in$
$\therefore|(x+2)(x-2)|<\in \ldots . .(i) $
We have to get rid of the factor $|x+2|$
As $|x-2|<\delta$
$ -\delta<\mathrm{x}-2<\delta$
$\therefore 2-\delta<\mathrm{x}<2+\delta $ Since $\delta$ can be assumed as very small, let us choose $\delta<1$ $ \therefore 1<\mathrm{x}<3$
$\therefore 3<\mathrm{x}+2<5 \ldots . . \text { (Adding }$
$\therefore|\mathrm{x}+2|<5$
$\therefore|(\mathrm{x}+2)(\mathrm{x}-2)|<5|\mathrm{x}-2| . $
$\text { (Adding } 2 \text { throughout) }$
From (i) and (ii), we get
$ 5|x-2|<\in$
$\therefore x-2<\frac{\in}{5}$
$\text { If } \delta=\frac{\in}{5},|x-2|<\delta \Rightarrow\left|x^2-4\right|<\in$
$\therefore \text { We } $
$\therefore$ We choose $\delta=\min \left\{\frac{\in}{5}, 1\right\}$ then
$|x-2|<\delta \Rightarrow|f(x)-3|<\in$
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