Question:
$\lim _{x \rightarrow 0}\left(\tan \left(\frac{\pi}{4}+x\right)\right)^{1 / x}$ is equal to :
Correct Option: , 4
Solution:
$\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{1 / x}$
$\Rightarrow e^{x \rightarrow 0 x}\left[\tan \left(\frac{\pi}{4}+x\right)-1\right] \Rightarrow e^{x \rightarrow 0} \frac{1}{x}\left(\frac{1+\tan x}{1-\tan x}-1\right)$
$\Rightarrow e^{\lim _{x \rightarrow 0}\left(\frac{2 \tan x}{1-\tan x}\right) \frac{1}{x}}=e^{\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)\left(\frac{2}{1-\tan x}\right)}=e^{2}$
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