Question

Evaluate the limit:
$\underset{x\to \pi }{\text{lim}}\frac{\sqrt{5+\cos x}-2}{(\pi -x{)}^2}$

Answer 1

We have,

$\underset{x\to \pi }{\text{lim}}\frac{\sqrt{5+\cos x}-2}{(\pi -x{)}^2}$

Put $x=\pi +h,h\to 0$

$=\underset{h\to 0}{\text{lim}}\frac{\sqrt{5+\cos (\pi +h)}-2}{(\pi -(\pi +h){)}^2}$

$=\underset{h\to 0}{\text{lim}}\frac{\sqrt{5-\cos h}-2}{(-h{)}^2}$ $[\because \cos (\pi +\theta )=-\cos \theta ]$

On rationalising the numerator, we get

$=\underset{h\to 0}{\text{lim}}[\frac{\sqrt{5-\cos h}-2}{h^2}\times \frac{\sqrt{5-\cos h}+2}{\sqrt{5-\cos h}+2}]$

$=\underset{h\to 0}{\text{lim}}\left[\frac{(\sqrt{5-\cos h}{)}^2-2^2}{h^2(\sqrt{5-\cos h}+2)}\right]$

$[\because (a+b\left)\right(a-b)=a^2-b^2]$

$=\underset{h\to 0}{\text{lim}}\frac{5-\cos h-4}{h^2(\sqrt{5-\cos h}+2)}$

$=\underset{h\to 0}{\text{lim}}\frac{1-\cos h}{h^2(\sqrt{5-\cos h}+2)}$

$=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^2\frac{h}{2}}{h^2(\sqrt{5-\cos h}+2)}$

$[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^2\frac{h}{2}}{4\times {\left(\frac{h}{2}\right)}^2(\sqrt{5-\cos h}+2)}$

$=\underset{h\to 0}{\text{lim}}\frac{1}{2}\times {\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2\times \frac{1}{(\sqrt{5-\cos h}+2)}$

$[\because \underset{x\to 0}{\text{lim}}\frac{\sin x}{x}=1]$

$=\frac{1}{2}\times (1{)}^2\times \frac{1}{(\sqrt{5-1}+2)}$

$=\frac{1}{(\sqrt{4}+2)\left(2\right)}$

$=\frac{1}{(2+2)\left(2\right)}$

$=\frac{1}{8}$

Therefore,

$\underset{x\to \pi }{\text{lim}}\frac{\sqrt{5+\cos x}-2}{(\pi -x{)}^2}=\frac{1}{8}$