A) 1
B) \[-1\]
C) \[{{e}^{2}}\]
D) e
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,{{\left\{ \tan \left( \frac{\pi }{4}+x \right) \right\}}^{1/x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x} \right]}^{1/x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \frac{1+\tan x}{1-\tan x} \right]}^{1/x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{[{{(1+\tan x)}^{1/\tan x}}]}^{\frac{\tan x}{x}}}}{{{[{{(1-\tan x)}^{1/\tan x}}]}^{\frac{\tan x}{x}}}}\] \[=\frac{{{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}}}}{{{e}^{-\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}}}}\] \[\left[ \because \underset{x\to 0}{\mathop{\lim }}\,{{\{1+f(x)\}}^{1/f(x)}}=e \right]\] \[=\frac{{{e}^{-1}}}{{{e}^{-1}}}\] \[\left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}=1 \right]\] \[={{e}^{2}}\]
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