Precalculus (6th Edition) Blitzer

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.
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.