Question

If the function $f\left(x\right)={\left[\tan (\frac{\pi }{4}+x)\right]}^{\frac{1}{x}}$ of $x\ne 0$
$=K$ for $x=0$
is continuous at $x=0$ then $K=$?

Answer 1

$\tan (\frac{\pi }{4}+x)=\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}.\tan x}$

$=\frac{1+\tan x}{1-\tan x}$

$=1+\frac{1+\tan x}{1-\tan x}-1$

$=1+\frac{1+\tan x-1+\tan x}{1-\tan x}$

$=1+\frac{2\tan x}{1-\tan x}$

$\therefore \underset{x\to 0}{lim}{[1+\frac{2\tan x}{1-\tan }]}^{\frac{1}{x}}={[1+\frac{2\tan x}{1-\tan x}]}^{\frac{1-\tan x}{2\tan x}\times \frac{2\tan x}{x(1+\tan x)}}$

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

$=e^2$

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

$=\frac{2\times 1}{1-0}=2$.