Answer
all real numbers apart from $-3$ and $3$.
Work Step by Step
We know that the fraction $\frac{x^2-4}{x^2-9}=\frac{2x+5}{(x+3)(x-3)}$ is undefined if its denominator is $0$. By the zero product rule $x+3\ne0$ and $x-3\ne0$, hence $x\ne3$ and $x\ne-3$, hence it is continuous everywhere, apart from where it is undefined, therefore it is continuous for all real numbers apart from $-3$ and $3$.
