Question

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

• A. $e$
• B. $e^2$
• C. $2$
• D. $1$

Answer 1

The correct option is B $e^2$

$\underset{x\to 0}{\text{lim}}{\left(\tan (\frac{\pi }{4}+x)\right)}^{\frac{1}{x}}$
This is of the form $1^{\infty }$
We know that for $\underset{x\to 0}{lim}f(x{)}^{g\left(x\right)},$ where $f\left(x\right)\to 1,g\left(x\right)\to \infty ,$ the limit is $L=e^{g\left(x\right)\left[f\right(x)-1]}$

Thus $L=e^{\underset{x\to 0}{\text{lim}}\frac{\tan (\frac{\pi }{4}+x)-1}{x}}$
$\Rightarrow L=e^{\underset{x\to 0}{\text{lim}}\frac{\frac{1+\tan x}{1-\tan x}-1}{x}}$
$\Rightarrow L=e^{\underset{x\to 0}{lim}2\left(\frac{\tan x}{x}\right)\cdot \left(\frac{1}{1-\tan x}\right)}$
$\Rightarrow L=e^{+2}$