Precalculus (6th Edition) Blitzer

The number for which the function f\left( x \right)=\left\{ \begin{align} & \frac{\cos x}{x-\frac{\pi }{2}}\text{if }x\ne \frac{\pi }{2} \\ & \text{1 if }x=\frac{\pi }{2} \end{align} \right. is discontinuous is $a=\frac{\pi }{2}$.
Consider the function f\left( x \right)=\left\{ \begin{align} & \frac{\cos x}{x-\frac{\pi }{2}}\text{if }x\ne \frac{\pi }{2} \\ & \text{1 if }x=\frac{\pi }{2} \end{align} \right., Check the discontinuity of the function at $a=\frac{\pi }{2}$ . Find the value of $f\left( x \right)$ at $a=\frac{\pi }{2}$, From the definition of the function, $f\left( \frac{\pi }{2} \right)=1$ The function is defined at the point $a=\frac{\pi }{2}$. Now find the value of $\,\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)$, Take some points near to the left of $\frac{\pi }{2}$. Suppose the points are $\frac{\pi }{2}-0.1,\frac{\pi }{2}-0.01\text{ and }\frac{\pi }{2}-0.001$ Take some points near to the right of $\frac{\pi }{2}$. Suppose the points are $\frac{\pi }{2}+0.001,\frac{\pi }{2}+0.01\text{ and }\frac{\pi }{2}+0.1$ Now evaluate the values of the function at the above chosen points. As x nears $\frac{\pi }{2}$ from the left or right the value of the function nears $-1$. Thus $\,\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)=-1$. From the above steps, $\,\,\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)\ne f\left( \frac{\pi }{2} \right)$. Thus, the function does not satisfy the third property of being continuous. Hence, the function f\left( x \right)=\left\{ \begin{align} & \frac{\cos x}{x-\frac{\pi }{2}}\text{if }x\ne \frac{\pi }{2} \\ & \text{1 if }x=\frac{\pi }{2} \end{align} \right. is discontinuous at $\frac{\pi }{2}$.