Evaluate the following limits: _ x [ 5+ x-2 ( -x)^2 ] - Vidyadip

Question

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

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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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