Answer
We can define $g(3)=6$ that extends $g(x)$ to be continuous at $x=3$.
Work Step by Step
$$g(x)=\frac{x^2-9}{x-3}$$
Currently, $g(x)$ is not defined at $x=3$, so $g(x)$ is not continuous at $x=3$ as well.
For $g(x)$ to be continuous at $x=3$, we need to extend $g(x)$ to include a value of $g(3)$ so that $g(3)=\lim_{x\to3}g(x)$
So we need to calculate $\lim_{x\to3}g(x)$:
$$\lim_{x\to3}g(x)=\lim_{x\to3}\frac{x^2-9}{x-3}=\lim_{x\to3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\to3}(x+3)$$
$$\lim_{x\to3}g(x)=3+3=6$$
So we can define $g(3)=6$ that extends $g(x)$ to be continuous at $x=3$.
