Question

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

Answer 1

We have,

$\underset{x\to \pi /2}{\text{lim}}\frac{\sqrt{2}-\sqrt{1+\sin x}}{{\cos }^{2}x}$

On rationalising the numerator, we get

$=\underset{x\to \pi /2}{\text{lim}}\frac{\sqrt{2}-\sqrt{1+\sin x}}{{\cos }^{2}x}\times \frac{\sqrt{2}+\sqrt{1+\sin x}}{\sqrt{2}+\sqrt{1+\sin x}}$

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

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

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

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

$[\because {\sin }^{2}x+{\cos }^{2}x=1]$

$=\underset{x\to \pi /2}{\text{lim}}\frac{1-\sin x}{(1-\sin x)(1+\sin x)(\sqrt{2}+\sqrt{1+\sin x})}$

$=\underset{x\to \pi /2}{\text{lim}}\frac{1}{(1+\sin x)(\sqrt{2}+\sqrt{1+\sin x})}$

$=\frac{1}{(1+\sin \frac{\pi }{2})(\sqrt{2}+\sqrt{1+\sin \frac{\pi }{2}})}$

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

$=\frac{1}{2\left(2\sqrt{2}\right)}$

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

Therefore,

$\underset{x\to \pi /2}{\text{lim}}\frac{\sqrt{2}-\sqrt{1+\sin x}}{{\cos }^{2}x}=\frac{1}{4\sqrt{2}}$