Mathematics - Limits Question with Solution | TestHub

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

$=e^{\underset{x\to 0}{\lim }\frac{\left(\tan \left(\frac{\pi }{4}+x\right)-1\right)}{x}} \left(\underset{x\to a}{\lim }{\left(f\left(x\right)\right)}^{g\left(x\right)}=e^{\left(\underset{x\to a}{\lim }g\left(x\right)\left(f\left(x\right)-1\right)\right)}, \underset{x\to a}{\lim }f\left(x\right)=1 \& \underset{x\to a}{\lim }g\left(x\right)=\infty \right)$

$=e^{\underset{x\to 0}{\lim }\frac{\left(\frac{1+\tan x}{1-\tan x}-1\right)}{x}}$

$=e^{\underset{x\to 0}{\lim }\frac{2\tan x}{x\left(1-\tan x\right)}}$

$=e^{\underset{x\to 0}{\lim }\left(\frac{\tan x}{x}\right)\left(\frac{2}{1-\tan x}\right)}$

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