Question
Evaluate $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}$

Answer

Here $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}\left[ {\frac{0}{0}{\text{from}}} \right]$
Put $x = \frac{\pi }{2} + y$ as $x \to \frac{\pi }{2},\;y \to 0$
$\therefore \;\mathop {\lim }\limits_{y \to 0} \frac{{\tan 2\left( {\frac{\pi }{2} + y} \right)}}{{\frac{\pi }{2} + y - \frac{\pi }{2}}} = \mathop {\lim }\limits_{y \to 0} \frac{{\tan (\pi + 2y)}}{y}$
$ = \mathop {\lim }\limits_{y \to 0} \frac{{\tan 2y}}{y} = \mathop {\lim }\limits_{y \to 0} \frac{{\tan 2y}}{{2y}} \times 2 = 1 \times 2 = 2$

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