Classifying the discontinuity at x = 3

Factor the numerator: $x^2 - 9 = (x-3)(x+3)$, so for $x \neq 3$, $f(x) = \frac{(x-3)(x+3)}{x-3} = x+3.$ This simplification is valid everywhere except at $x=3$ itself, since we divided by $x-3$.

Since $f(x) = x+3$ for all $x$ near 3 (but not at 3), the limit is $\lim_{x\to 3} f(x) = 3+3 = 6$, which exists. However, $f(3)$ is undefined because the original expression gives $0/0$ there. So the limit exists but $f(3)$ is not defined — condition 1 of continuity fails, and this is a removable discontinuity (a hole at $(3,6)$).