Evaluate the following limits:

Question:

Evaluate the following limits:

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

 

Solution:

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

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

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

Let,

$\mathrm{y}=\mathrm{x}-\frac{\pi}{2}$

$=\lim _{y \rightarrow 0} \frac{1-\cos y}{\sqrt{2}+\sqrt{1+\cos y}(\sqrt{2} \sin y \sin y)}$

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

$=\frac{1}{8}$

$\therefore \lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2}-\sqrt{1+\sin x}}{\sqrt{2} \cos x \cos x}=\frac{1}{8}$

 

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