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Q. $ \underset{x\to 0}{\mathop{\lim }}\,\,{{\left\{ \tan \left( \frac{\pi }{4}+x \right) \right\}}^{1/x}} $ is equal to

J & K CETJ & K CET 2013Limits and Derivatives

Solution:

$ \underset{x\to 0}{\mathop{\lim }}\,\,{{\left\{ \tan \left( \frac{\pi }{4}+x \right) \right\}}^{1/x}}=\underset{x\to 0}{\mathop{\lim }}\,\,{{\left\{ \frac{1+\tan x}{1-\tan x} \right\}}^{1/x}} $
$ =\underset{x\to 0}{\mathop{\lim }}\,\,{{\left\{ 1+\left( \frac{1+\tan x}{1-\tan x} \right) \right\}}^{1/x}} $
$ =\underset{x\to 0}{\mathop{\lim }}\,\,\,{{\left\{ 1+\frac{2\tan x}{1-\tan x} \right\}}^{1/x}} $
(Form $ {{1}^{\infty }} $ )
$ ={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{2\tan \,x}{1-\tan x}.}}^{\frac{1}{x}} $
$ ={{e}^{2\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\,\left( \frac{\tan x}{x} \right)\,\,.\,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1}{1-\tan x}}}\,\,\,\, $
$ ={{e}^{2.1.\left( \frac{1}{1-0} \right)}}={{e}^{2}} $