Question

$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {\dfrac{\pi }{4} + x} \right)} \right)^{\dfrac{1}{x}}}$ is equal to
1. $e$
2. ${e^2}$
3. $2$
4. $1$

Answer 1

Hint: Solve the function by applying trigonometric $\tan $ function formula. Now, try to solve this further using the limit. If the answer will be infinity then use rules or formulas of limit to get the appropriate answer.

Formula Used:
Trigonometric formula –
$\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$
Limit formula –
$\mathop {\lim }\limits_{x \to 0} {\left( {1 + f\left( x \right)} \right)^n} = {e^{\mathop {\lim }\limits_{x \to 0} \left( {f(x) \times n} \right)}}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1$

Complete step by step Solution:
Given that,
$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {\dfrac{\pi }{4} + x} \right)} \right)^{\dfrac{1}{x}}}$
$ = \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan x}}{{1 - \tan \left( {\dfrac{\pi }{4}} \right)\tan x}}} \right)^{\dfrac{1}{x}}}$
$ = \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)^{\dfrac{1}{x}}}$
$ = \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{1 + \tan x + \tan x - \tan x}}{{1 - \tan x}}} \right)^{\dfrac{1}{x}}}$
$ = \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{1 - \tan x}}{{1 - \tan x}} + \dfrac{{\tan x + \tan x}}{{1 - \tan x}}} \right)^{\dfrac{1}{x}}}$
$ = \mathop {\lim }\limits_{x \to 0} {\left( {1 + \dfrac{{2\tan x}}{{1 - \tan x}}} \right)^{\dfrac{1}{x}}}$
$ = {e^{\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{2\tan x}}{{1 - \tan x}}} \right)\left( {\dfrac{1}{x}} \right)}}$
$ = {e^{2\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{1 - \tan x}}} \right)\left( {\dfrac{{\tan x}}{x}} \right)}}$
$ = {e^{2\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{1 - \tan x}}} \right)\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\tan x}}{x}} \right)}}$
$ = {e^{2(1)(1)}}$
$ = {e^2}$


Therefore, the correct option is 2.

Note: In such questions, if after solving limit you will get the power as infinity apply $\mathop {\lim }\limits_{x \to 0} {\left( {1 + f\left( x \right)} \right)^n} = {e^{\mathop {\lim }\limits_{x \to 0} \left( {f(x) \times n} \right)}}$ this formula if and only if the term is in this form. Otherwise again write the problem as $e$ to the natural log of your function.