Question

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

Answer 1

We have,

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

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

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

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

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

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

$[\because \sin 2\theta =2\sin \theta \cos \theta ,1-\cos 2\theta =2{\sin }^2\theta ]$

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

$=2\left(\cos 0\right)$

$=2\left(1\right)=2$