A) \[{{e}^{2}}\]
B) e
C) \[{{e}^{-1}}\]
D) 1
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \tan \left( \frac{\pi }{4}+x \right) \right]}^{1/x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x} \right]}^{1/x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+\tan x}{1-\tan x} \right)}^{1/x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left( 1+x+\frac{{{x}^{3}}}{3}+.... \right)}^{1/x}}}{{{\left( 1-x-\frac{{{x}^{3}}}{3}-.... \right)}^{1/x}}}\] Neglecting the higher term of\[x\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{1/x}}}{{{(1-x)}^{1/x}}}\] \[=\frac{{{e}^{1}}}{{{e}^{-1}}}={{e}^{2}}\]
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