Question

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

• A. $e$
• B. $e^2$
• C. $1/e$
• D. $1/e^2$

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }{\left\{\tan \left(\frac{\pi }{4}+x\right)\right\}}^{1/x}=\underset{x\to 0}{\lim }{\left\{\frac{1+\tan x}{1-\tan x}\right\}}^{1/x}$
$=\underset{x\to 0}{\lim }{\left\{1+\left(\frac{1+\tan x}{1-\tan x}\right)\right\}}^{1/x}$
$=\underset{x\to 0}{\lim }{\left\{1+\frac{2\tan x}{1-\tan x}\right\}}^{1/x}$
(Form $1^{\infty }$ )
$={e^{\underset{x\to 0}{\lim }\frac{2\tan x}{1-\tan x}.}}^{\frac{1}{x}}$
$=e^{2\underset{x\to 0}{\lim }\left(\frac{\tan x}{x}\right).\underset{x\to 0}{\lim }\frac{1}{1-\tan x}}$
$=e^{\mathrm{2.1.}\left(\frac{1}{1-0}\right)}=e^2$