**Question:**

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

**Solution:**

(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 $x=0$.