Answer
not continuous at $x=-1$ or $x=1$.
Work Step by Step
1. See graph for $R(x)=\frac{x-1}{x^2-1}=\frac{x-1}{(x+1)(x-1)}=\frac{1}{x+1}, x\ne1$ with V.A. $x=-1$ and a hole at $(1, \frac{1}{2})$.
2. Although $\lim_{x\to1}R(x)=\frac{1}{2}$, the function is not continuous at $x=1$ because of the hole.
3. We have $\lim_{x\to -1^-}R(x)=-\infty$ and $\lim_{x\to -1^+}R(x)=\infty$, thus the function is not continuous at $x=-1$.
