JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 18)
$$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$$ is equal to :
2
1
$$e$$
$$e$$2
Explanation
$$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$$
This is 1$$\infty $$ form.
= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {\tan \left( {{\pi \over 4} + x} \right) - 1} \right] \times {1 \over x}}}$$
= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{1 + \tan x} \over {1 - \tan x}} - 1} \right] \times {1 \over x}}}$$
= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{2\tan x} \over {x\left( {1 - \tan x} \right)}}} \right]}}$$
= $${e^{2\mathop {\lim }\limits_{x \to 0} \left[ {{{\tan x} \over x} \times {1 \over {\left( {1 - \tan x} \right)}}} \right]}}$$
= $${e^{2\mathop {\lim }\limits_{x \to 0} \left[ {1 \times {1 \over {\left( {1 - 0} \right)}}} \right]}}$$
= e2
This is 1$$\infty $$ form.
= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {\tan \left( {{\pi \over 4} + x} \right) - 1} \right] \times {1 \over x}}}$$
= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{1 + \tan x} \over {1 - \tan x}} - 1} \right] \times {1 \over x}}}$$
= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{2\tan x} \over {x\left( {1 - \tan x} \right)}}} \right]}}$$
= $${e^{2\mathop {\lim }\limits_{x \to 0} \left[ {{{\tan x} \over x} \times {1 \over {\left( {1 - \tan x} \right)}}} \right]}}$$
= $${e^{2\mathop {\lim }\limits_{x \to 0} \left[ {1 \times {1 \over {\left( {1 - 0} \right)}}} \right]}}$$
= e2
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