$\displaystyle \lim _{x \rightarrow 0} e^{\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)^\frac{1}{x} $
since e is continuous
$\displaystyle e^{\lim _{x \rightarrow 0}\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)^\frac{1}{x} $
$\displaystyle \lim _{x \rightarrow 0}\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)^\frac{1}{x}$
$\displaystyle \lim _{x \rightarrow 0}\left(\frac{1}{x}\right)\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)$
$\displaystyle \lim _{x \rightarrow 0}\left(\frac{\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)}{x}\right)$
0/0
use lopital rule