Precalculus (6th Edition) Blitzer

To make the function $f\left( x \right)=\frac{{{x}^{2}}-81}{x-9}$ continuous $x=9$, we define $f(9)=18$.
Consider the function $f\left( x \right)=\frac{{{x}^{2}}-81}{x-9}$, 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) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ Find the value of $\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{{{x}^{2}}-81}{x-9}$, \begin{align} & \underset{x\to 9}{\mathop{\lim }}\,\frac{{{x}^{2}}-81}{x-9}=\underset{x\to 9}{\mathop{\lim }}\,\frac{\left( x-9 \right)\left( x+9 \right)}{x-9} \\ & =\underset{x\to 9}{\mathop{\lim }}\,x+9 \\ & =9+9 \\ & =18 \end{align} For the function $f\left( x \right)=\frac{{{x}^{2}}-81}{x-9}$ to be continuous at $x=9$, the function must be defined at $9$, and the value of $\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{{{x}^{2}}-81}{x-9}$ must be equal to $f\left( 9 \right)$, that is $f\left( 9 \right)=\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{{{x}^{2}}-81}{x-9}=18$ Thus, $f\left( 9 \right)=18$