Question

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

Answer 1

We have,

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

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

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

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

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

$=\underset{h\to 0}{\text{lim}}\frac{-\sin 2h}{-\sin h}$

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

$=\underset{h\to 0}{\text{lim}}2\cos h$

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

Therefore,

$\underset{x\to \frac{\pi }{2}}{\text{lim}}\frac{\sin 2x}{\cos x}=2$