Question

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

Answer 1

We have,

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

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

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

$[\because \sin (\frac{\pi }{2}+\theta )=\cos \theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{1-\cos h}{h^2}$ $[\because 1-\cos \theta =2{\sin }^2\frac{\theta }{2}]$

$=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^2\frac{h}{2}}{h^2}=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^2\frac{h}{2}}{4\times \frac{h^2}{4}}$

$=\underset{h\to 0}{\text{lim}}\frac{2}{4}\times \frac{{\sin }^2\frac{h}{2}}{{\left(\frac{h}{2}\right)}^2}$

$=\frac{1}{2}\times 1^2$ $[\because \underset{x\to 0}{\text{lim}}\frac{\sin x}{x}=1]$


$=\frac{1}{2}$

Therefore,

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