Solution:
The function provided is $f(x)= \begin{cases}\frac{|x|}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}$
Also the $\mathrm{f}(\mathrm{x})=|\mathrm{x}| / \mathrm{x}$ can be defined as,
$\frac{x}{x}=1 \text { if } x>0 \text { and } \frac{-x}{x}=-1 \text { if } x<0$
$\Rightarrow f(x)=1 \text { if } x>0, f(x)=-1 \text { if } x<0 \text { and } f(x)=0 \text { if } x=0$
Now we know that, the domain of $f(x)$ is $\mathrm{R}$ as $f(x)$ is defined for $x>0, x<0$ and $x=0$. It is constant function and continuous for all $x>0$, $f(x)=1$.
It is a constant function and continuous for all $x<0, f(x)=-1$ .
As a result $f(x)$ is continuous on $\mathrm{R}-\{0\}$.
Now,
L.H.L.= $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-1)=-1$
R.H.L.= $\lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0^{-}}(1)=1$
Since, $\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x)$
As a result, $\lim _{x \rightarrow 0} f(x)$ does not exist and $f(x)$ is discontinuous at only $x=0$.