Answer:
Option C
Explanation:
We have , $f(x) =\left[ \tan \left(\frac{\pi}{4}+x\right)\right]^{1/x}$ =K
Since , f(x) is continuous at x=0, then
$f(o)=\lim_{x \rightarrow 0}f(x)$
$=\lim_{x \rightarrow 0}\left[ \tan \left(\frac{\pi}{4}+x\right)\right]^{1/x}$
$\Rightarrow K=\lim_{x \rightarrow 0}\left[\frac{1+\tan x}{1-\tan x}\right]^{1/x}$ [$1^{\infty}$ form]
$=e^{\lim_{x \rightarrow 0}\left[\frac{1+\tan x}{1-\tan x}-1\right].\frac{1}{x}}$
$=e^{\lim_{x \rightarrow 0}\left[\frac{2\tan x}{1-\tan x}\right].\frac{1}{x}}$
$=e^{2\lim_{x \rightarrow 0}\frac{\tan x}{x}.\lim_{x \rightarrow 0}\frac{1}{1-\tan x}}$
$\left[\because \lim_{x\rightarrow 0}\frac{\tan x}{x}=1\right]$
$\therefore$ $K=e^{2.1(\frac{1}{1-0})}=e^{2}$
