Question

If the function $f ( x )=\begin{cases}{\left[\tan \left(\frac{\pi}{4}+ x \right)\right] \frac{1}{ x },} & \text { for } x \neq0 \\ K , & x =0\end{cases}$
is continuous at $x =0$, then $K =$ ?

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

Answer 1

Answer: (C)

Given : $f ( x )$ is continuous at $x =0$
$\therefore f(0)=\displaystyle\lim _{x \rightarrow 0} f(x)$
$=\displaystyle\lim _{x \rightarrow 0}\left[\tan \left(\frac{\pi}{4}+x\right)\right] \frac{1}{x}$
$=\displaystyle\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{x}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\left[(1+\tan x)^{\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}{\left[(1-\tan x)^{-\frac{1}{\tan x}}\right]^{\frac{-\tan x}{x}}}$
$=\frac{ e ^{1}}{ e ^{-1}}= e ^{1} \cdot e ^{1}= e ^{2}$