Question

$li{m}_{x\to \pi }\frac{\sqrt{2+\cos x-1}}{{(\pi -x)}^2}$

Answer 1

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

we know that $\cos x=2{\cos }^2\frac{x}{2}-1$

$\sqrt{1+\cos x}=\sqrt{2}\cos \frac{x}{2}$

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

Apply L - hospital Rule two times

$\underset{x\to \pi }{lim}-\frac{\frac{\sqrt{2}}{2}}{-4\sqrt{2}}$
$=0[\because \cos \frac{\pi }{2}=0]$