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Question

Mathematics Question on limits and derivatives

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

A

ee

B

e2{{e}^{2}}

C

1/e1/e

D

1/e21/{{e}^{2}}

Answer

e2{{e}^{2}}

Explanation

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∞{{1}^{\infty }} )
=elim⁡x→0   2tan⁡ x1−tan⁡x.1x={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{2\tan \,x}{1-\tan x}.}}^{\frac{1}{x}}
=e2lim⁡x→0     (tan⁡xx)  .  lim⁡x→0   11−tan⁡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}}}\,\,\,\,
=e2.1.(11−0)=e2={{e}^{2.1.\left( \frac{1}{1-0} \right)}}={{e}^{2}}