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

1 Answer

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$
answered Apr 5, 2014 by thanvigandhi_1
 

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