Question

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan 2x}{x-\frac{\pi }{2}}$

Answer 1

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan 2x}{x-\pi /2}$

Let $y=x-\frac{\pi }{2}$ as $x\to \frac{\pi }{2}$

So $y=\frac{\pi }{2}-\frac{\pi }{2}$

So $y\to 0$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan 2x}{x-\pi /2}$

$={lim}_{y\to 0}\left(\frac{\tan 2(\frac{\pi }{2}+y)}{y}\right)$

$={lim}_{y\to 0}\left(\frac{\tan (\pi +2y)}{y}\right)$

$={lim}_{y\to 0}\left(\frac{\tan 2y}{y}\right)$

$={lim}_{y\to 0}(\frac{1}{y}.\frac{\sin 2y}{\cos 2y})$

$={lim}_{y\to 0}(\frac{\sin 2y}{y}.\frac{1}{\cos 2y})$

$={lim}_{y\to 0}\frac{\sin 2y}{y}\times {lim}_{y\to 0}\frac{1}{\cos 2y}$

Multiply & divided by $2y$

$={lim}_{y\to 0}(\frac{\sin 2y}{y}\times \frac{2y}{2y}).{lim}_{y\to 0}\frac{1}{\cos 2y}$

$={lim}_{y\to 0}(\frac{\sin 2y}{2y}\times 2){lim}_{y\to 0}\frac{1}{\cos 2y}$

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

$=2{lim}_{y\to 0}\frac{\sin 2y}{2y}.{lim}_{y\to 0}\frac{1}{\cos 2y}$

$=\mathrm{2.1.}{lim}_{y\to 0}\frac{1}{\cos 2y}$

$=2\frac{1}{\cos 2\left(0\right)}$

$=\frac{2}{\cos 0}[\because \cos 0=1]$

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