Solve the Following Question.(5 Marks)

Question 15 Marks

Evaluate the following limits:
$\lim _{x \rightarrow 1}\left(\frac{x+3 x^2+5 x^3+\cdots \cdots \cdots \cdots \cdots+(2 n-1) x^n-n^2}{x-1}\right)$

Answer

$\lim _{x \rightarrow 1}\left(\frac{x+3 x^2+5 x^3+\cdots+(2 n-1) x^n-n^2}{x-1}\right)$
Consider
$1+3+5+\ldots+(2 n-1)$
$=\sum_{r=1}^n(2 r-1)$
$=2 \sum_{r=1}^n r-\sum_{r=1}^n 1$
$=2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}-\mathrm{n}$
$=n(n+1)-n$
$=n^2+n-n$
$=\mathrm{n}^2$
$\therefore \quad n^2=1+3+5+\ldots+(2 n-1) \text {. }$
$\therefore \quad \text { Required limit }$
$=\lim _{x \rightarrow 1} \frac{\left[x+3 x^2+5 x^3+\cdots+(2 n-1) x^n\right]-[1+3+5+\cdots+(2 n-1)]}{x-1}$
$=\lim _{x \rightarrow 1} \frac{(x-1)+\left(3 x^2-3\right)+\left(5 x^2-5\right)+\cdots+(2 \mathrm{n}-1) x^n-(2 \mathrm{n}-1)}{x-1}$
$=\lim _{x \rightarrow 1}\left[\frac{x-1}{x-1}+\frac{3\left(x^2-1\right)}{x-1}+\frac{5\left(x^3-1\right)}{x-1}+\cdots+\frac{(2 n-1)\left(x^n-1\right)}{x-1}\right]$
$=\lim _{x \rightarrow 1}\left(\frac{x^1-1^1}{x-1}\right)+3 \lim _{x \rightarrow 1}\left(\frac{x^2-1^2}{x-1}\right)+5 \lim _{x \rightarrow 1}\left(\frac{x^3-1^3}{x-1}\right)$
$+\cdots+(2 n-1) \lim _{x \rightarrow 1}\left(\frac{x^n-1^n}{x-1}\right)$
$=1(1)^0+3(2)(1)^1+5(3)(1)^2+\ldots+(2 n-1) n(1)^{n-1}$
$=1(1)+3(2)+5(3)+\ldots+(2 n-1) n$
$=\sum_{r=1}^n(2 \mathrm{r}-1) \mathrm{r}$
$=\sum_{r=1}^{\infty}\left(2 r^2-r\right)$
$=2 \sum_{r=1}^n r^2-\sum_{r=1}^n r$
$=2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
$=n(n+1)\left(\frac{2 n+1}{3}-\frac{1}{2}\right)$
$=n(n+1)\left(\frac{4 n+2-3}{6}\right)=\frac{n(n+1)(4 n-1)}{6}$

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Question 25 Marks

Evaluate the following limits: $\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{(\sin x-\cos x)^2}{\sqrt{2}-\sin x-\cos x}\right]$

Answer

$ \lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{(\sin x-\cos x)^2}{\sqrt{2}-\sin x-\cos x}\right]$
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sin 2 x}{\sqrt{2}-(\sin x+\cos x)}$
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sin 2 x}{\sqrt{2}-\sqrt{1+\sin 2 x}} $
Put $1+\sin 2 x=\mathrm{t}$
$ \therefore \sin 2 x=t-1$
$ \text { As } x \rightarrow \frac{\pi}{4}, t \rightarrow 1+\sin 2\left(\frac{\pi}{4}\right) $
$\therefore \mathrm{t} \rightarrow 1+\sin \frac{\pi}{2}$
$\therefore \mathrm{t} \rightarrow 1+1$
$\therefore \mathrm{t} \rightarrow 2$
$\therefore$ Required limit
$=\lim _{t \rightarrow 2} \frac{1-(t-1)}{\sqrt{2}-\sqrt{t}}$
$=\lim _{t \rightarrow 2} \frac{2-t}{2^{\frac{1}{2}}-t^{\frac{1}{2}}}$
$=\lim _{t \rightarrow 2} \frac{t-2}{t^{\frac{1}{2}}-2^{\frac{1}{2}}}$
$=\lim _{t \rightarrow 2} \frac{1}{\frac{t^{\frac{1}{2}}-2^{\frac{1}{2}}}{t-2}}\cdots\left[\begin{array}{l}
\text { Divide Numerator and Denominator by } \mathrm{t}-2 \\
\text { Ast } \rightarrow 2, \mathrm{t} \neq 2, \therefore \mathrm{t}-2 \neq 0 \end{array}\right] \\
=\frac{1}{\frac{1}{2}(2)^{\frac{-1}{2}}}$
$=2(2)^{\frac{1}{2}}$
$=2 \sqrt{2}$

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Question 35 Marks

Evaluate the following limits: $\lim _{x \rightarrow a}\left[\frac{x \cos a-a \cos x}{x-a}\right]$

Answer

$ \lim _{x \rightarrow \mathrm{a}} \frac{x \cos \mathrm{a}-\mathrm{a} \cos x}{x-\mathrm{a}}$
$=\lim _{x \rightarrow \mathrm{a}} \frac{x \cos \mathrm{a}-\mathrm{a} \cos \mathrm{a}-\mathrm{a} \cos x+\mathrm{a} \cos \mathrm{a}}{x-\mathrm{a}}$
$=\lim _{x \rightarrow \mathrm{a}} \frac{(x-\mathrm{a}) \cos \mathrm{a}-\mathrm{a}(\cos x-\cos \mathrm{a})}{x-\mathrm{a}}$
$=\lim _{x \rightarrow \mathrm{a}}\left[\frac{(x-\mathrm{a}) \cos \mathrm{a}}{x-\mathrm{a}}-\frac{\mathrm{a}(\cos x-\cos \mathrm{a})}{x-\mathrm{a}}\right]$
$=\lim _{x \rightarrow \mathrm{a}}(\cos \mathrm{a})-\mathrm{a} \lim _{x \rightarrow \mathrm{a}}\left(\frac{\cos x-\cos \mathrm{a}}{x-\mathrm{a}}\right)\ldots .[\because x \rightarrow \mathrm{a}, x \neq \mathrm{a}, \therefore x-\mathrm{a} \neq 0] $
Consider, $\lim _{x \rightarrow a} \frac{\cos x-\cos a}{x-a}$
$ =\lim _{x \rightarrow 0} \frac{-2 \sin \left(\frac{x+a}{2}\right) \cdot \sin \left(\frac{x-a}{2}\right)}{x-a}$
$=-\lim _{x \rightarrow a} \sin \left(\frac{x+a}{2}\right) \cdot \frac{\sin \left(\frac{x-a}{2}\right)}{\frac{x-a}{2}}$
$=-\lim _{x \rightarrow a} \sin \left(\frac{x+\mathrm{a}}{2}\right) \cdot \lim _{x \rightarrow \mathrm{a}} \frac{\sin \left(\frac{x-\mathrm{a}}{2}\right)}{\frac{x-\mathrm{a}}{2}}$
$=-1 \times \sin \left(\frac{\mathrm{a}+\mathrm{a}}{2}\right) \times 1\ldots\left[\begin{array}{l}\because x \rightarrow a,(x-a) \rightarrow 0 \\ \therefore \frac{x-a}{2} \rightarrow 0 \text { and } \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\end{array}\right]$
$=-\sin a $
Required limit
$ =\cos a-a(-\sin a)$
$=\cos a+a \sin a $

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Question 45 Marks

Evaluate the following limits: $\lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^2+5}-\sqrt{x^2-3}}{\sqrt{x^2+3}-\sqrt{x^2+1}}\right]$

Answer

$\lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^2+5}-\sqrt{x^2-3}}{\sqrt{x^2+3}-\sqrt{x^2+1}}\right]$
$\begin{aligned} & =\lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^2+5}-\sqrt{x^2-3}}{\sqrt{x^2+3}-\sqrt{x^2+1}} \times \frac{\sqrt{x^2+5}+\sqrt{x^2-3}}{\sqrt{x^2+3}+\sqrt{x^2+1}}\right. \\ & \left.\times \frac{\sqrt{x^2+3}+\sqrt{x^2+1}}{\sqrt{x^2+5}+\sqrt{x^2-3}}\right]\end{aligned}$
$\ldots\left[\begin{array}{l}\text { By taking conjugates of both, the } \\ \text { numerator as well as the denominator. }\end{array}\right]$
$\begin{aligned} =\lim _{x \rightarrow \infty}\left[\frac{\left(x^2+5\right)-\left(x^2-3\right)}{\left(x^2+3\right)-\left(x^2+1\right)} \times \frac{\sqrt{x^2+3}+\sqrt{x^2+1}}{\sqrt{x^2+5}+\sqrt{x^2-3}}\right] \end{aligned}$
$ =\lim _{x \rightarrow \infty} \frac{8\left(\sqrt{x^2+3}+\sqrt{x^2+1}\right)}{2\left(\sqrt{x^2+5}+\sqrt{x^2-3}\right)} $
$ =4 \lim _{x \rightarrow \infty}\left(\frac{\frac{\sqrt{x^2+3}+\sqrt{x^2+1}}{x}}{\frac{\sqrt{x^2+5}+\sqrt{x^2-3}}{x}}\right)$
$\left[\begin{array}{l}\text { Divide numerator and } \\ \cdots \\ \text { denominator by } x\end{array}\right]$
$\begin{aligned} & =4 \lim _{x \rightarrow \infty}\left(\frac{\frac{\sqrt{x^2+3}}{x}+\frac{\sqrt{x^2+1}}{x}}{\frac{\sqrt{x^2+5}}{x}+\frac{\sqrt{x^2-3}}{x}}\right) \end{aligned} $
$ =\frac{4 \lim _{x \rightarrow \infty}\left(\sqrt{\frac{x^2+3}{x^2}}+\sqrt{\frac{x^2+1}{x^2}}\right)}{\lim _{x \rightarrow \infty}\left(\sqrt{\frac{x^2+5}{x^2}}+\sqrt{\frac{x^2-3}{x^2}}\right)} $
$ =\frac{4\left(\lim _{x \rightarrow \infty} \sqrt{1+\frac{3}{x^2}}+\lim _{x \rightarrow \infty} \sqrt{1+\frac{1}{x^2}}\right)}{\lim _{x \rightarrow \infty} \sqrt{1+\frac{5}{x^2}}+\lim _{x \rightarrow \infty} \sqrt{1-\frac{3}{x^2}}}$
$\begin{aligned} & =\frac{4(\sqrt{1+0}+\sqrt{1+0})}{\sqrt{1+0}+\sqrt{1-0}} \quad \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, k>0\right] \end{aligned} $
$ =\frac{4(1+1)}{1+1} $
$ =4$

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Question 55 Marks

Evaluate the following limits:
$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^2}\right]$

Answer

$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^2}$
$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{2-\sqrt{3} \cos x-\sin x}{36\left(x-\frac{\pi}{6}\right)^2} $
Put $x-\frac{\pi}{6}=\mathrm{h}$,
$x=\frac{\pi}{6}+\mathrm{h}$
As $x \rightarrow \frac{\pi}{6}, \mathrm{~h} \rightarrow 0$
$ \therefore \lim _{x \rightarrow \frac{\pi}{6}} \frac{2-\sqrt{3} \cos x-\sin x}{36\left(x-\frac{\pi}{6}\right)^2}$
$=\lim _{h \rightarrow 0} \frac{2-\sqrt{3} \cos \left(\frac{\pi}{6}+h\right)-\sin \left(\frac{\pi}{6}+h\right)}{36 h^2}$
$ =\lim _{h \rightarrow 0} \frac{2-\sqrt{3}\left(\cos \frac{\pi}{6} \cdot \cos h-\sin \frac{\pi}{6} \cdot \sinh \right)-\left(\sin \frac{\pi}{6} \cdot \cos h+\cos \frac{\pi}{6} \cdot \sinh \right)}{36 h^2}$
$=\lim _{h \rightarrow 0} \frac{2-\sqrt{3}\left(\frac{\sqrt{3}}{2} \cdot \cos h-\frac{1}{2} \sin h\right)-\left(\frac{1}{2} \cos h+\frac{\sqrt{3}}{2} \cdot \sin h\right)}{36 h^2} $
$\frac{\left[2-\frac{3}{2} \cdot \cosh +\frac{\sqrt{3}}{2} \cdot \sin h-\frac{1}{2} \cosh -\frac{\sqrt{3}}{2} \sin h\right]}{36 h^2}$
$=\lim _{h \rightarrow 0} \frac{2-2 \cos h}{36 h^2}$
$=\lim _{h \rightarrow 0} \frac{2(1-\cos h)}{36 h^2}$
$=\frac{1}{18} \lim _{h \rightarrow 0} \frac{1-\cos h}{h^2} \times \frac{1+\cos h}{1+\cos h}$
$=\frac{1}{18} \lim _{\mathbf{h} \rightarrow 0} \frac{1-\cos ^2 h}{h^2(1+\cos h)}$
$=\frac{1}{18} \lim _{h \rightarrow 0} \frac{\sin ^2 h}{h^2(1+\cos h)}$
$=\frac{1}{18}\left(\lim _{h \rightarrow 0} \frac{\sin h}{h}\right)^2 \lim _{h \rightarrow 0} \frac{1}{(1+\cos h)}$
$=\frac{1}{18}(1)^2 \frac{1}{(1+\cos 0)}$
$\ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=\frac{1}{18(1+1)}$
$=\frac{1}{36}$

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Question 65 Marks

Evaluate the following limits:
$\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\sqrt{2}-\cos x-\sin x}{(4 x-\pi)^2}\right]$

Answer

$ \lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\sqrt{2}-\cos x-\sin x}{(4 x-\pi)^2}\right]$
$=\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\sqrt{2}-(\cos x+\sin x)}{\left[4\left(x-\frac{\pi}{4}\right)\right]^2}\right] $
Put $x-\frac{\pi}{4}=\mathrm{h}$
$ \therefore \quad x =\frac{\pi}{4}+\mathrm{h}$
$ \text { As } x \rightarrow \frac{\pi}{4}, \mathrm{~h} \rightarrow 0 $
As $x \rightarrow \frac{\pi}{4}, \mathrm{~h} \rightarrow 0$
Also,
$ \cos x+\sin x =\cos x+\cos \left(\frac{\pi}{2}-x\right)$
$ =\cos \left(\frac{\pi}{4}+h\right)+\cos \left[\frac{\pi}{2}-\left(\frac{\pi}{4}+h\right)\right]$
$ =\cos \left(\frac{\pi}{4}+h\right)+\cos \left(\frac{\pi}{4}-h\right)$
$ =2 \cos \frac{\pi}{4} \cdot \cosh $
(By defactorisation)
$ =2\left(\frac{1}{\sqrt{2}}\right) \cosh$
$=\sqrt{2} \cosh $
$\therefore \quad$ Required limit
$ =\lim _{h \rightarrow 0} \frac{\sqrt{2}-\sqrt{2} \cosh }{(4 h)^2}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{2}(1-\cos h)}{16 h^2}$
$=\frac{\sqrt{2}}{16} \lim _{h \rightarrow 0}\left(\frac{1-\cos h}{h^2} \times \frac{1+\cos h}{1+\cos h}\right)$
$=\frac{1}{8 \sqrt{2}} \lim _{h \rightarrow 0} \frac{\sin ^2 h}{h^2(1+\cosh )}$
$=\frac{1}{8 \sqrt{2}} \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2 \times \frac{1}{1+\cosh }$
$=\frac{1}{8 \sqrt{2}} \times \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2 \times \lim _{h \rightarrow 0} \frac{1}{1+\cosh }$
$=\frac{1}{8 \sqrt{2}} \times(1)^2 \times \frac{1}{1+1} \quad \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=\frac{1}{8 \sqrt{2}} \times \frac{1}{2} = \frac{1}{16 \sqrt{2}} $

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Question 75 Marks

Evaluate the following limits:
$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin x-1}{\pi-6 x}\right]$

Answer

$ \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin x-1}{\pi-6 x}$
$\text { Put } \frac{\pi}{6}-\mathrm{h}=x$
$\therefore \quad \mathrm{h}=\frac{\pi}{6}-x$
$\text { As } x \rightarrow \frac{\pi}{6}, \mathrm{~h} \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin x-1}{\pi-6 x}$
$=\lim _{h \rightarrow 0} \frac{2 \sin \left(\frac{\pi}{6}-h\right)-1}{\pi-6\left(\frac{\pi}{6}-h\right)}$
$=\lim _{h \rightarrow 0} \frac{2\left[\sin \left(\frac{\pi}{6}-h\right)-\frac{1}{2}\right]}{6 h}$
$=\frac{1}{3} \lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{6}-h\right)-\sin \frac{\pi}{6}}{h}$
$=\frac{1}{3} \lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{\frac{\pi}{6}-h+\frac{\pi}{6}}{2}\right) \cdot \sin \left(\frac{\frac{\pi}{6}-h-\frac{\pi}{6}}{2}\right)}{h}$
$=\frac{1}{3} \lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{\pi}{6}-\frac{h}{2}\right) \cdot \sin \left(\frac{-h}{2}\right)}{h}$
$=\frac{-1}{3} \lim _{h \rightarrow 0} \cos \left(\frac{\pi}{6}-\frac{h}{2}\right) \cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$=\frac{-1}{3} \lim _{h \rightarrow 0} \cos \left(\frac{\pi}{6}-\frac{h}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$=\frac{-1}{3} \cdot \cos \left(\frac{\pi}{6}-0\right) \cdot(1) \quad \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right]$
$=\frac{-1}{3} \times \frac{\sqrt{3}}{2}=\frac{-1}{2 \sqrt{3}}$

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Question 85 Marks

Evaluate the following limits:
$\lim _{x \rightarrow \pi}\left[\frac{\sqrt{5+\cos x-2}}{(\pi-x)^2}\right]$

Answer

$\lim _{x \rightarrow x}\left[\frac{\sqrt{5+\cos x}-2}{(\pi-x)^2}\right]$
Put $\pi-x=h$
$ \therefore x=\pi-\mathrm{h}$
$ \text { As } x \rightarrow \pi, \mathrm{h} \rightarrow 0$
$\therefore \lim _{x \rightarrow x}\left[\frac{\sqrt{5+\cos x}-2}{(\pi-x)^2}\right] $
$ =\lim _{h \rightarrow 0}\left[\frac{\sqrt{5+\cos (\pi-h)}-2}{(h)^2}\right]$
$=\lim _{h \rightarrow 0}\left[\frac{\sqrt{5-\cos h}-2}{h^2}\right]$
$=\lim _{h \rightarrow 0}\left[\frac{\sqrt{5-\cosh }-2}{h^2} \times \frac{\sqrt{5-\cosh }+2}{\sqrt{5-\cos h+2}}\right] $
...[By rationalization]
$ =\lim _{h \rightarrow 0} \frac{5-\cos h-4}{h^2(\sqrt{5-\cos h}+2)}$
$=\lim _{h \rightarrow 0} \frac{1-\cosh ^{}}{h^2(\sqrt{5-\cos h}+2)}$
$=\lim _{h \rightarrow 0} \frac{1-\cosh }{h^2(\sqrt{5-\cosh }+2)} \times \frac{1+\cosh }{1+\cosh }$
$=\lim _{h \rightarrow 0} \frac{\sin ^2 h}{h^2(\sqrt{5-\cos h}+2)(1+\cos h)}$
$=\lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2 \times \frac{1}{(\sqrt{5-\cos h}+2)(1+\cos h)}$
$=\lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2 \times \lim _{h \rightarrow 0} \frac{1}{(\sqrt{5-\cos h}+2)(1+\cos h)}$
$=(1)^2 \times \frac{1}{(\sqrt{5-1}+2)(1+1)} \quad \cdots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=\frac{1}{(\sqrt{4}+2)(2)}=\frac{1}{(2+2)(2)}$
$=\frac{1}{8}$
$ $

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Question 95 Marks

Evaluate the following limits:
$\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{\operatorname{cosec} x-1}{\left(\frac{\pi}{2}-x\right)^2}\right]$

Answer

Put $\frac{\pi}{2}-x=\mathrm{h}$,
$\therefore \quad x=\frac{\pi}{2}-\mathrm{h}$
As $x \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0$
$\begin{aligned}
\therefore \quad & \lim _{x \rightarrow \frac{\pi}{2}} \frac{\operatorname{cosec} x-1}{\left(\frac{\pi}{2}-x\right)^2}=\lim _{h \rightarrow 0} \frac{\operatorname{cosec}\left(\frac{\pi}{2}-h\right)-1}{(h)^2} \\
& =\lim _{h \rightarrow 0} \frac{\operatorname{sech}-1}{h^2} \quad \cdots\left[\because \operatorname{cosec}\left(\frac{\pi}{2}-\theta\right)=\sec \theta\right] \\
& =\lim _{h \rightarrow 0} \frac{\frac{1}{\cosh }-1}{h^2} \\
& =\lim _{h \rightarrow 0} \frac{1-\cosh }{h^2 \cdot \cosh } \cdot \\
& =\lim _{h \rightarrow 0} \frac{2 \sin ^2 \frac{h}{2}}{h^2} \cdot \frac{1}{\cosh } \\
& =2 \lim _{h \rightarrow 0}\left(\frac{\frac{h}{2}}{\frac{\sin ^2}{2}}\right)^2 \cdot \frac{1}{4} \cdot \frac{1}{\cosh }
\end{aligned}
$
$ =\frac{1}{2} \lim _{\mathrm{h} \rightarrow 0}\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^2 \lim _{\mathrm{h} \rightarrow 0} \frac{1}{\cos \mathrm{h}}$
$=\frac{1}{2}(1)^2 \times \frac{1}{\cos 0} \quad \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{\mathrm{p} \theta}=1\right]$
$=\frac{1}{2} \times 1$
$=\frac{1}{2} \quad $

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Question 105 Marks

Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{\cos (a x)-\cos (b x)}{\cos (c x)-1}\right]$

Answer

$\cdots\left[\begin{array}{l} \text { Divide numerator and denominator by } x^2 \\ \because x \rightarrow 0, x \neq 0, \therefore x^2 \neq 0 \end{array}\right] \\
\frac{\lim _{x \rightarrow 0}\left[\frac{\sin ^2\left(\frac{a x}{2}\right)}{x^2}-\frac{\sin ^2\left(\frac{b x}{2}\right)}{x^2}\right]}{\lim _{x \rightarrow 0} \frac{\sin ^2\left(\frac{c x}{2}\right)}{x^2}}$
$=\frac{\lim _{x \rightarrow 0}\left[\frac{\sin \frac{a x}{2}}{x}\right]^2-\lim _{x \rightarrow 0}\left[\frac{\sin \frac{b x}{2}}{x}\right]^2}{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{c x}{2}}{x}\right)^2}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{a x}{2}}{\frac{a x}{2}}\right)^2 \cdot\left(\frac{a}{2}\right)^2-\lim _{x \rightarrow 0}\left(\frac{\sin \frac{b x}{2}}{\frac{b x}{2}}\right)^2 \cdot\left(\frac{b}{2}\right)^2}{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{c x}{2}}{\frac{c x}{2}}\right)^2 \cdot\left(\frac{c}{2}\right)^2}$
$=\frac{(1)^2 \cdot \frac{a^2}{4}-(1)^2 \cdot \frac{b^2}{4}}{(1)^2 \cdot \frac{c^2}{4}}$
$=\frac{\frac{a^2}{4}-\frac{b^2}{4}}{\frac{c^2}{4}}$
$=\frac{a^2-b^2}{c^2}$

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Question 115 Marks

Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{1-\cos (n x)}{1-\cos (m x)}\right]$

Answer

$\lim _{x \rightarrow 0}\left[\frac{1-\cos (n x)}{1-\cos (m x)}\right]=\lim _{x \rightarrow 0}\left[\frac{2 \sin ^2\left(\frac{n x}{2}\right)}{2 \sin ^2\left(\frac{m x}{2}\right)}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{\left(\sin \frac{n x}{2}\right)^2}{\left.\frac{x^2}{m x}\right)^2}\right]\cdots\left[\begin{array}{l}
\text { Divide numerator and denominator by } x^2 \\
\because x \rightarrow 0, x \neq 0 \therefore x^2 \neq 0
\end{array}\right]$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{\mathrm{n} x}{2}}{x}\right)^2}{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{\mathrm{m} x}{2}}{x}\right)^2}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{\mathrm{n} x}{2}}{\frac{\mathrm{nx}}{2}}\right)^2 \times\left(\frac{\mathrm{n}}{2}\right)^2}{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{\mathrm{m} x}{2}}{\frac{\mathrm{mx}}{2}}\right)^2 \times\left(\frac{\mathrm{m}}{2}\right)^2}$
$=\frac{(1)^2 \times \frac{n^2}{4}}{(1)^2 \times \frac{m^2}{4}} \cdots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right]$
$=\frac{\mathrm{n}^2}{\mathrm{~m}^2}$

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Question 125 Marks

Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right]$

Answer

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$
$=\lim _{x \rightarrow 0} \frac{\left(\sqrt{1+x^2}-\sqrt{1+x}\right)}{\left(\sqrt{1+x^3}-\sqrt{1+x}\right)} \times \frac{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}\times \frac{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}$
$...\left[\text { By taking conjugates of both } \text { numerator as well as the denomerator }\right.$
$\left.1+x^2-(1+x)\right]\left(\sqrt{1+x^3}+\sqrt{1+x}\right)$
$\left.1+x^3-(1+x)\right]\left(\sqrt{1+x^2}+\sqrt{1+x}\right)$
$=\lim _{x \rightarrow 0} \frac{\left(x^2-x\right)\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\left(x^3-x\right)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}$
$=\lim _{x \rightarrow 0} \frac{x(x-1)\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{x\left(x^2-1\right)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}$
$=\lim _{x \rightarrow 0} \frac{x(x-1)\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{x(x-1)(x+1)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}$
$=\lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}+\sqrt{1+x}}{(x+1)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)} \ldots[\because x \rightarrow 0, x \neq 0]$
$=\frac{\lim _{x \rightarrow 0}\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\lim _{x \rightarrow 0}(x+1) \cdot \lim _{x \rightarrow 0}\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}$
$=\frac{\sqrt{1+0^3}+\sqrt{1+0}}{(0+1)\left(\sqrt{1+0^2}+\sqrt{1+0}\right)}$
$=\frac{1+1}{1(1+1)}=1$

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Question 135 Marks

Evaluate the following limits: $\lim _{x \rightarrow 0}\left(\frac{\sqrt{x^2+9}-\sqrt{2 x^2+9}}{\sqrt{3 x^2+4}-\sqrt{2 x^2+4}}\right)$

Answer

$\lim _{x \rightarrow 0}\left[\frac{\sqrt{x^2+9}-\sqrt{2 x^2+9}}{\sqrt{3 x^2+4}-\sqrt{2 x^2+4}}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{\sqrt{x^2+9}-\sqrt{2 x^2+9}}{\sqrt{3 x^2+4}-\sqrt{2 x^2+4}} \times \frac{\sqrt{x^2+9}+\sqrt{2 x^2+9}}{\sqrt{x^2+9}+\sqrt{2 x^2+9}}\right.$
$\sqrt{\left.\times \frac{\sqrt{3 x^2+4}+\sqrt{2 x^2+4}}{\sqrt{3 x^2+4}+\sqrt{2 x^2+4}}\right]}$
$\quad \ldots\left[\begin{array}{l} \text { By taking conjugates of both, the } \\ \text { numerator as well as the denominator. } \end{array}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{\left(x^2+9\right)-\left(2 x^2+9\right)}{\left(3 x^2+4\right)-\left(2 x^2+4\right)} \times \frac{\sqrt{3 x^2+4}+\sqrt{2 x^2+4}}{\sqrt{x^2+9}+\sqrt{2 x^2+9}}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{-x^2\left(\sqrt{3 x^2+4}+\sqrt{2 x^2+4}\right)}{x^2\left(\sqrt{x^2+9}+\sqrt{2 x^2+9}\right)}\right]$
$=-\lim _{x \rightarrow 0}\left[\frac{\sqrt{3 x^2+4}+\sqrt{2 x^2+4}}{\sqrt{x^2+9}+\sqrt{2 x^2+9}}\right]$
$\left[\begin{array}{l}\because x \rightarrow 0, x \neq 0 \\ \cdots x^2 \neq 0\end{array}\right]$
$=\frac{-\lim _{x \rightarrow 0}\left(\sqrt{3 x^2+4}+\sqrt{2 x^2+4}\right)}{\lim _{x \rightarrow 0}\left(\sqrt{x^2+9}+\sqrt{2 x^2+9}\right)}$
$=\frac{-(\sqrt{0+4}+\sqrt{0+4})}{\sqrt{0+9}+\sqrt{0+9}}$
$=\frac{-(2+2)}{3+3}$
$=\frac{-2}{3}$

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