A) (a) e
B) (b) \[\frac{1}{e}\]
C) (c) e2
D) (d) e3
Correct Answer: C
Solution :
[c] \[\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \tan \frac{\pi }{4}+x \right]}^{\frac{1}{x}}}=\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}.\tan x} \right]}^{\frac{1}{x}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \frac{1+\tan x}{1-\tan x} \right]}^{\frac{1}{x}}}=\underset{x\to 0}{\mathop{\lim }}\,{{e}^{\frac{1}{x}\log }}^{\left[ \frac{1+\tan x}{1-\tan x} \right]}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{e}^{\frac{1}{x}\left\{ log\left( 1+tanx \right)-log\left( 1-tanx \right) \right\}}}\] \[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,{{e}^{\frac{1}{x}}}\left\{ \tan x-\frac{{{(\tan x)}^{2}}}{2}+....\left( -\tan x-\frac{{{\left( \tan x \right)}^{2}}}{2}-.....\infty \right) \right\}}}\] \[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,2.\frac{1}{x}\left\{ \tan x+\frac{{{(\tan x)}^{3}}}{3}+....\infty \right\}}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,2\left\{ \frac{\tan x}{x}+\frac{1}{x}.\frac{{{(\tan x)}^{3}}}{3}+....\infty \right\}}}\] Hence the option [c] is correct.
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