Question

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

Answer 1

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

Applying L' Hospitals Rule

$L={lim}_{x\to \frac{\pi }{2}}\frac{2(1-\sin x)(-\cos x)}{-1}={lim}_{x\to \frac{\pi }{2}}\frac{2(-\cos x+\sin x\cos x)}{-1}=\frac{2(-\cos \frac{\pi }{2}+\sin \frac{\pi }{2}\cos \frac{\pi }{2})}{-1}=\frac{0}{-1}=0$