Answer
To make the function $ f\left( x \right)=\frac{{{x}^{2}}-81}{x-9}$ continuous $ x=9$, we define $f(9)=18$.
Work Step by Step
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$
