Evaluate the following limits:

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

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

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

Let,

$y=x-\pi$

$=\lim _{y \rightarrow 0} \frac{1-\cos y}{x^{2} \times \sqrt{2-\cos y}+1}$

$=\frac{1}{4}$

$\therefore \lim _{x \rightarrow \pi} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}=\frac{1}{4}$