Continuity of Graph when the function because of the form 0/0 Will a graph be continuous where there exists a limit found through L'hopital's rule. For example:
$f(x)=\frac{x^2-9}{x-3}$ at $x=3$

There is no notion of "continuity" of a graph. A graph is a visual representation of a function and can itself not be continuous. The function, on the other hand, can be continuous.
Since your function equals $$f(x)=\frac{(x-3)(x+3)}{x-3},$$ it is clear that, for $x\neq 3$, the function is actually $f(x) = x+3$. For $x=3$, the function is not defined.
What you can say is that the function has a limit when $x$ approaches $3$ and that that limit equals $6$. This means that there exists a continuous function $g$ which equals $f$ wherever $f$ is defined. Of course, the function $g$ is simply $$g(x)=x+3.$$
