# Evaluate the given limit $\lim\limits_{x \to \large\frac{\pi}{2}} \large\frac{\tan \: 2x}{ x - \Large\frac{\pi}{2}}$

At $x = \large\frac{\pi}{2}$, the value of the given function takes the form $\large\frac{0}{0}.$
Now, put $x - \large\frac{\pi}{2}$$= y so that x \to \large\frac{\pi}{2}$$, \: y \to 0$
$\therefore \lim\limits_{x \to \large\frac{\pi}{2}} \large\frac{\tan \: 2x}{ x - \Large\frac{\pi}{2}}$$= \lim\limits_{y \to 0} \large\frac{ \tan \: 2 \bigg( y+ \Large\frac{\pi}{2} \bigg) }{y} = \lim\limits_{ y \to 0} \large\frac{ \tan ( \pi + 2y)}{y} = \lim\limits_{ y \to 0} \large\frac{ \tan \: 2y}{y}$$ \quad \quad [ \tan ( \pi+2y) = \tan \: 2y]$
$= \lim\limits_{ y \to 0} \large\frac{ \sin \: 2y}{y \: \cos\: 2y}$
$= \lim\limits_{ y \to 0} \bigg( \Large\frac{ \sin \: 2y}{2y}$$\times \Large\frac{2}{ \cos\: 2y} \bigg) = \bigg( \lim\limits_{ 2y \to 0} \Large\frac{ \sin \: 2y}{2y} \bigg)$$ \times \lim\limits_{y \to 0} \bigg( \Large\frac{2}{ \cos\: 2y} \bigg)$$\quad \quad [ y \to 0 \Rightarrow 2y \to 0] = 1 \times \large\frac{2}{ \cos \: 0} \quad \quad \bigg[ \lim\limits_{x \to 0} \large\frac{ \sin \: x}{x}$$ = 1 \bigg]$
$= 1 \times \large\frac{2}{1}$
$= 2$