Solve this

Given:

$f(x)=\left\{\begin{array}{l}\frac{\sin 3 x}{x}, \text { when } x \neq 0 \\ 1, \quad \text { when } x=0\end{array}\right.$

We observe

$(\mathrm{LHL}$ at $x=0)=\lim _{\mathrm{x} \rightarrow 0} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$

$=\lim _{h \rightarrow 0} \frac{\sin (-3 h)}{-h}=\lim _{h \rightarrow 0} \frac{-\sin (3 h)}{-h}=\lim _{h \rightarrow 0} \frac{3 \sin (3 h)}{3 h}=3 \lim _{h \rightarrow 0} \frac{\sin (3 h)}{3 h}=3 \cdot 1=3$

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)$

$=\lim _{h \rightarrow 0} \frac{\sin 3 h}{h}=\lim _{h \rightarrow 0} \frac{3 \sin 3 h}{3 h}=3 \lim _{h \rightarrow 0} \frac{\sin (3 h)}{3 h}=3 \cdot 1=3$

Given:

$f(0)=1$

It is known that for a function $f(x)$ to be continuous at $x=a$,

$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$

But here,

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$

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