Precalculus (6th Edition) Blitzer

The complete statement is, “A function f is continuous at a when three conditions are satisfied. (a) f is defined at a, so that $f\left( a \right)$ is a real number. (b) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists. (c) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ $=$ $f\left( a \right)$ ”
For a function to be continuous at a point a, the function must satisfy the following three conditions: (a) f is defined at a. (b) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists. (c) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ For example, consider a function $f\left( x \right)=2x$ To check whether the function is continuous at the point $a=3$ or not, Find the value of $f\left( x \right)$ at $a=3$, $f\left( 3 \right)=2\left( 3 \right)=6$ The function is defined at the point $a=3$. Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,2x$, \begin{align} & \,\underset{x\to 3}{\mathop{\lim }}\,2x=2\underset{x\to 3}{\mathop{\lim }}\,x \\ & =2\left( 3 \right) \\ & =6 \end{align} Thus, $\,\underset{x\to 3}{\mathop{\lim }}\,2x=6$ From the above two steps, $\,\underset{x\to 3}{\mathop{\lim }}\,2x=6=f\left( 3 \right)$. Thus, the function satisfies all the conditions of being continuous. Hence, the function $f\left( x \right)=2x$ is continuous at $3$.