The function $f(x)=\left\{\begin{array}{rr}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ 0 & , x=0\end{array}\right.$
(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
(b) is not continuous at x = 0
Given: $f(x)=\left\{\begin{array}{l}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}, x \neq 0 \\ 0, x=0\end{array}\right.$
We have
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)$
If $e^{\frac{1}{x}}=t$, then
$x \rightarrow 0, \quad t \rightarrow \infty$
$\lim _{x \rightarrow 0} f(x)=\lim _{t \rightarrow \infty}\left(\frac{t-1}{t+1}\right)=\lim _{t \rightarrow \infty}\left(\frac{1-\frac{1}{t}}{1+\frac{1}{t}}\right)=\frac{1-0}{1+0}=1$
Also, $f(0)=0$
$\therefore \lim _{x \rightarrow 0} f(x) \neq f(0)$
Hence, $f(x)$ is discontinuous at $x=0$.