Answer
The number for which the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\sin x}{x-\pi }\text{if }x\ne \pi \\
& \text{1 if }x=\pi
\end{align} \right.$ is discontinuous is $ a=\pi $.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\sin x}{x-\pi }\text{if }x\ne \pi \\
& \text{1 if }x=\pi
\end{align} \right.$.
Check the discontinuity of the function at $ a=\pi $.
Find the value of $ f\left( x \right)$ at $ a=\pi $,
From the definition of the function,
$ f\left( \pi \right)=1$
The function is defined at the point $ a=\pi $.
Now find the value of $\,\underset{x\to \pi }{\mathop{\lim }}\,f\left( x \right)$,
Take some points near to the left of $\pi $.
Suppose the points are $\pi -0.1,\pi -0.01\text{ and }\pi -0.001$
Take some points near to the right of $\pi $
Suppose the points are $\pi +0.001,\pi +0.01\text{ and }\pi +0.1$
Now evaluate the values of the function at the above chosen points.
As x nears $\pi $ from the left or right the value of the function nears $-1$.
Thus $\,\underset{x\to \pi }{\mathop{\lim }}\,f\left( x \right)=-1$
From the above steps, $\,\,\underset{x\to \pi }{\mathop{\lim }}\,f\left( x \right)\ne f\left( \pi \right)$
Thus, the function does not satisfy the third property of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\sin x}{x-\pi }\text{if }x\ne \pi \\
& \text{1 if }x=\pi
\end{align} \right.$ is discontinuous at $\pi $.
