Answer
$(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
Work Step by Step
We are given the function:
$f(x)=\dfrac{1}{x^2-4}$
$f(x)$ is a rational function, therefore according to Theorem 2.10 b), the function is continuous for all $x$ for which the denominator is not zero.
We check the zeros of the denominator:
$x^2-4=0$
$(x-2)(x+2)=0$
$x-2=0\Rightarrow x=2$
$x+2=0\Rightarrow x=-2$
The interval on which the function is continuous is:
$(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
