Answer
not continuous at $x=-1$ or $x=1$.
Work Step by Step
1. See graph for $R(x)=\frac{x^2+x}{x^2-1}=\frac{x(x+1)}{(x+1)(x-1)}=\frac{x}{x-1}, x\ne-1$ with V.A. $x=1$, H.A. $y=1$ and a hole at $(-1, \frac{1}{2})$.
2. Although $\lim_{x\to-1}R(x)=\frac{1}{2}$, 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 $x=1$.
