Answer
No, it is not possible to define $ f\left( x \right)=\frac{1}{x-9}$ at $ x=9$ so that the function becomes continuous at $9$.
Work Step by Step
Consider the function $ f\left( x \right)=\frac{1}{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 $\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{1}{x-9}$
Take some points near to the left of $9$
Suppose the points are $8.9,8.99\text{ and }8.999$
Take some points near to the right of $9$
Suppose the points are $9.001,9.01\text{ and }9.1$
Now evaluate the values of the function at the above chosen points.
As x nears $9$ from the left or right the value of the function does not near to some unique number.
Thus, $\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{1}{x-9}$ does not exist.
So, the function does not satisfy the second condition of being continuous.
Then, defining the function $ f\left( x \right)=\frac{1}{x-9}$ at the point $9$ will not make it continuous at $9$.
Thus, it’s not possible to define $ f\left( x \right)=\frac{1}{x-9}$ at $ x=9$ so that the function becomes continuous at $9$.
The discontinuity of the function $ f\left( x \right)=\frac{{{x}^{2}}-81}{x-9}$ at $ x=9$ can be removed by defining the function at $9$ and keeping the value $ f\left( 9 \right)$ equal to $\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{{{x}^{2}}-81}{x-9}$.
The discontinuity of the function $ f\left( x \right)=\frac{1}{x-9}$ at $ x=9$ cannot be removed as $\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{1}{x-9}$ does not exist.
