Evaluate $\underset{x\to 0}{\lim }{\left(\tan \left(\frac{\pi }{4}+x\right)\right)}^{\frac{1}{x}}.$.

 Given: $\underset{x\to 0}{\lim }{\left(\tan \left(\frac{\pi }{4}+x\right)\right)}^{\frac{1}{x}}.$

We have, 

$\Rightarrow \underset{x\to 0}{\lim }{\left(\tan \left(\frac{\pi }{4}+x\right)\right)}^{\frac{1}{x}}=\underset{x\to 0}{\lim }{\left(\frac{\tan \left(\frac{\pi }{4}\right)+\tan x}{1-\tan \left(\frac{\pi }{4}\right)·\tan x}\right)}^{\frac{1}{x}}(\because this is 1^{\infty } model)$

$\Rightarrow \underset{x\to 0}{\lim }{\left(\tan \left(\frac{\pi }{4}+x\right)\right)}^{\frac{1}{x}}=e^{li{m}_{x\to 0} \left(\frac{1+\tan x}{1-\tan x}-1\right)·\left(\frac{1}{x}\right)}$

$=e^{li{m}_{x\to 0}\left(\frac{2\tan x}{x}\right)·\left(\frac{1}{1-\tan x}\right)}$

Since $li{m}_{x\to 0}\left(\frac{\tan x}{x}\right)=1,$

$\Rightarrow \underset{x\to 0}{\lim }{\left(\tan \left(\frac{\pi }{4}+x\right)\right)}^{\frac{1}{x}}=e^2.$

Value of given limit is $e^2.$

Therefore, the correct option is $\left(2\right).$