Precalculus (6th Edition) Blitzer

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 $a=\pi$.
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 $\pi$.