Question

Evaluate:${lim}_{x\to 0}{\left[\tan (\frac{\pi }{4}+x)\right]}^{\frac{1}{\sin 2x}}$

Answer 1

${lim}_{x\to 0}{\left[\tan (\frac{\pi }{4}+x)\right]}^{\frac{1}{\sin 2x}}$

$={lim}_{x\to 0}{\left[\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}\right]}^{\frac{1}{\sin 2x}}$

$=e^{{lim}_{x\to 0}(\frac{1+\tan x}{1-\tan x}-1)\frac{1}{\sin 2x}}$

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

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

$=e^{{lim}_{x\to 0}\left(\frac{2\frac{\sin x}{\cos x}}{1-\frac{\sin x}{\cos x}}\right)\frac{1}{2\sin x\cos x}}$

$=e^{{lim}_{x\to 0}\frac{1}{\cos x(\cos x-\sin x)}}$

$=e^{\frac{1}{\cos 0(\cos 0-\sin 0)}}$

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

$=e^1=e$