Question

Evaluate: ${lim}_{x\to \frac{\pi }{2}}\frac{\tan 2x}{x-\frac{\pi }{2}}$

Answer 1

${lim}_{x\to \frac{\pi }{2}}\left(\frac{\tan \left(2x\right)}{x-\frac{\pi }{2}}\right)$

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

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

$=2{lim}_{y\to 0}(\frac{\sin 2y}{2y}.\frac{1}{\cos 2y})[\because \frac{\sin x}{x}=1;\cos 0=1]=2\times 1\times 1=2$

Method 2:

${lim}_{x\to \frac{\pi }{2}}\left(\frac{\tan \left(2x\right)}{x-\frac{\pi }{2}}\right)={lim}_{x\to \frac{\pi }{2}}\left(\frac{{\sec }^2\left(2x\right)}{1}\right).2$

$={lim}_{x\to \frac{\pi }{2}}\left(2.{\sec }^2\left(2x\right)\right)=2{\sec }^2(2\times \frac{\pi }{2})=2{\sec }^2\left(\pi \right)=2{(-1)}^2=2$