Question

Evaluate the limit:
$\underset{x\to \pi }{lim}\frac{\sqrt{(2+\cos x)}-1}{(\pi -x{)}^2}$

Answer 1

We have:
$\underset{x\to \pi }{lim}\frac{\sqrt{(2+\cos x)}-1}{(\pi -x{)}^2}$

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

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

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

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

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

$=\underset{x\to \pi }{lim}\frac{1+\cos x}{(\pi -x{)}^2(\sqrt{(2+\cos x)}+1)}$
Put $x=\pi -h$
If $x\to \pi$, then $h\to 0$

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

$=\underset{h\to 0}{lim}\frac{1-\cos h}{h^2(\sqrt{2-\cos h}+1)}$
$[\because \cos (\pi -\theta )=-\cos \theta ]$

$=\underset{h\to 0}{lim}\frac{2{\sin }^2\frac{h}{2}}{4\times \frac{h^2}{4}}\times \frac{1}{(\sqrt{2-\cos h}+1)}$

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

$=\frac{1}{2}\times 1^2\times \frac{1}{(\sqrt{2-\cos 0}+1)}$

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

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

$=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Therefore,
$\underset{x\to \pi }{lim}\frac{\sqrt{(2+\cos x)}-1}{(\pi -x{)}^2}=\frac{1}{4}$