#### Answer

There is no number for which the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\sin 3x}{x}\text{if }x\ne 0 \\
& 3\text{ if }x=0
\end{align} \right.$ is discontinuous.

#### Work Step by Step

Consider the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\sin 3x}{x}\text{if }x\ne 0 \\
& 3\text{ if }x=0
\end{align} \right.$,
Check the discontinuity of the function at $ x=0$
Find the value of $ f\left( x \right)$ at $ x=0$,
From the definition of the function,
$ f\left( 0 \right)=3$
The function is defined at the point $ x=0$.
Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$,
Take some points near to the left of $0$
Suppose the points are $-0.03,-0.02\text{ and }-0.01$
Take some points near to the right of $0$
Suppose the points are $0.01,0.02\text{ and }0.03$
Now evaluate the values of the function at the above chosen points.
As x nears $0$ from the left or right the value of the function nears $3$
Thus $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=3$
From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=3=f\left( 0 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\sin 3x}{x}\text{if }x\ne 0 \\
& 3\text{ if }x=0
\end{align} \right.$ is continuous at $0$.
Thus, there is no number for which the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\sin 3x}{x}\text{if }x\ne 0 \\
& 3\text{ if }x=0
\end{align} \right.$ is discontinuous.