Question

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

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

Answer 1

Answer: (C)

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