Evaluate the following limits:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2}-\sqrt{1+\sin x}}{\sqrt{2} \cos ^{2} x}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2}-\sqrt{1+\sin x}}{\sqrt{2} \cos x \cos x}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2}-\sqrt{1+\sin x}}{\sqrt{2} \cos x \cos x} \times \frac{\sqrt{2}+\sqrt{1+\sin x}}{\sqrt{2}+\sqrt{1+\sin x}}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\sqrt{2}+\sqrt{1+\sin x}(\sqrt{2} \cos x \cos x)}$
Let,
$\mathrm{y}=\mathrm{x}-\frac{\pi}{2}$
$=\lim _{y \rightarrow 0} \frac{1-\cos y}{\sqrt{2}+\sqrt{1+\cos y}(\sqrt{2} \sin y \sin y)}$
$=\frac{1}{2 \sqrt{2}} \times \frac{1}{2 \sqrt{2}}$
$=\frac{1}{8}$
$\therefore \lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2}-\sqrt{1+\sin x}}{\sqrt{2} \cos x \cos x}=\frac{1}{8}$