Question

The function $f\left(x\right)=\left\{\begin{array}{ll}\frac{e^{1/x}-1}{e^{1/x}+1},& x\ne 0\\ 0,& x=0\end{array}\right.$
(a) is continuous at x = 0
(b) is not continuous at x = 0
(c) is not continuous at x = 0, but can be made continuous at x = 0
(d) none of these

Answer 1

(b) is not continuous at x = 0

Given: $f\left(x\right)=\left\{\begin{array}{l}\frac{e^{{\displaystyle \frac{1}{x}}}-1}{e^{{\displaystyle \frac{1}{x}}}+1},x\ne 0\\ 0,x=0\end{array}\right.$

We have
$\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{e^{{\displaystyle \frac{1}{x}}}-1}{e^{\frac{1}{x}}+1}\right)$

If $e^{\frac{1}{x}}=t$, then
$x\mathit{\to }0,t\to \infty$

$\underset{x\to 0}{\lim }f\left(x\right)=\underset{t\to \infty }{\lim }\left(\frac{t-1}{t+1}\right)=\underset{t\to \infty }{\lim }\left(\frac{1-{\displaystyle \frac{1}{t}}}{1+{\displaystyle \frac{1}{t}}}\right)=\frac{1-{\displaystyle 0}}{1+{\displaystyle 0}}=1$
Also, $f\left(0\right)=0$

$\underset{x\to 0}{\therefore \lim }f\left(x\right)\ne f\left(0\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.