Question

If the function $f\left(x\right)=\{\begin{array}{cc}{\left[\tan (\frac{\pi }{4}+x)\right]}^{\frac{1}{x}},& \text{for}x\ne 0\\ K,& x=0\end{array}$

is continuous at $x=0$, then $K=$?

• A. $e$
• B. $e^{-1}$
• C. $e^2$
• D. $e^{-2}$

Answer 1

The correct option is C $e^2$

Given : $f\left(x\right)$ is continuous at $x=0$

$\therefore f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}{\left[\tan (\frac{\pi }{4}+x)\right]}^{\frac{1}{x}}$
$=\underset{x\to 0}{lim}{\left(\frac{1+\tan x}{1-\tan x}\right)}^{\frac{1}{x}}$
$=\underset{x\to 0}{lim}\frac{{\left[\right(1+\tan x{)}^{\frac{1}{\tan x}}]}^{\frac{\tan x}{x}}}{{\left[\right(1-\tan x{)}^{-\frac{1}{\tan x}}]}^{\frac{-\tan x}{x}}}$
taking limits
$=\frac{e^1}{e^{-1}}=e^1.e^1=e^2$