_ x → 0 ( ((π)/(4) + x)) ^ 1 / x is equal to

$\lim_{x \to 0}{\left(\text{ tan }\left(\dfrac{\pi}{4} + x\right)\right)}^{1 / x}{\left(1\right)}^{\infty}$form

$= {e}^{\lim_{x \to 0}\dfrac{\left(\tan \left(\dfrac{\pi}{4} + x\right) - 1\right)}{x}}\left(\lim_{x \to a}{\left(f\left(x\right)\right)}^{g\left(x\right)} = {e}^{\left(\lim_{x \to a}g\left(x\right)\left(f\left(x\right) - 1\right)\right)}, \lim_{x \to a}f\left(x\right) = 1 \text{ and } \lim_{x \to a}g\left(x\right) = \infty\right)$

$= {e}^{\lim_{x \to 0}\dfrac{\left(\dfrac{1 + \tan x}{1 - \tan x} - 1\right)}{x}}$

$= {e}^{\lim_{x \to 0}\dfrac{2\tan x}{x\left(1 - \tan x\right)}}$

$= {e}^{\lim_{x \to 0}\left(\dfrac{\tan x}{x}\right)\left(\dfrac{2}{1 - \tan x}\right)}$

$= {e}^{\left(1\right)\left(\dfrac{2}{1 - 0}\right)} = {e}^{2}$