Answer
$f(x)=\dfrac{2x+5}{x^2-4}$ is continuous for all real numbers except for $-2$ and $2$
Work Step by Step
Given: $f(x)=\dfrac{2x+5}{x^2-4}$
Re-write as: $\dfrac{2x+5}{x^2-4}=\dfrac{2x+5}{(x+2)(x-2)}$
When the denominator is $0$, then the fraction $\dfrac{2x+5}{x^2-4}=\dfrac{2x+5}{(x+2)(x-2)}$ is undefined. By the zero product rule, we have: $x+2\ne0$ and $x-2\ne0$. So, $x\ne2$ and $x\ne-2$,
The fraction function is continuous everywhere, except from where it is undefined. So, $f(x)=\dfrac{2x+5}{x^2-4}$ is continuous for all real numbers except for $-2$ and $2$.
