Answer
$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$
Work Step by Step
We are given the function:
$s(x)=\dfrac{x^2-4x+3}{x^2-1}$
$s(x)$ 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:
$x^2-1=0$
$(x-1)(x+1)=0$
$x-1=0\Rightarrow x=1$
$x+1=0\Rightarrow x=-1$
The interval on which the function is continuous is:
$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$
