Question

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

Answer 1

We have,

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

On rationalising the numerator, we get

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

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

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

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

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

Put $x=\frac{\pi }{2}+h,h\to 0$

$=\underset{h\to 0}{\text{lim}}\frac{1-\sin (\frac{\pi }{2}+h)}{{(\frac{\pi }{2}-(\frac{\pi }{2}+h))}^2(\sqrt{2-\sin (\frac{\pi }{2}+h)}+1)}$

$=\underset{h\to 0}{\text{lim}}\frac{1-\cos h}{(-h{)}^2(\sqrt{2-\cos h}+1)}$

$=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^2\frac{h}{2}}{h^2(\sqrt{2-\cos h}+1)}$

$\{\because 1-\cos \theta =2{\sin }^2\frac{\theta }{2}\}$

$=\underset{h\to 0}{\text{lim}}\frac{\sin \frac{h}{2}}{2\times \frac{h}{2}}\times \frac{\sin \frac{h}{2}}{2\times \frac{h}{2}}\times \frac{2}{(\sqrt{2-\cos h}+1)}$

$=\frac{1}{2}\times \frac{1}{2}\times \frac{2}{(\sqrt{2-\cos 0}+1)}$

$[\because \underset{x\to 0}{\text{lim}}\frac{\sin x}{x}=1]$

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

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

Therefore,

$\underset{x\to \frac{\pi }{2}}{\text{lim}}\frac{\sqrt{2-\sin x}-1}{{(\frac{\pi }{2}-x)}^2}=\frac{1}{4}$