AHSME 1952 Problem 36

To be continuous at $x = - 1$, the value of $\frac {x^3 + 1}{x^2 - 1}$ is taken to be: $\textbf{(A)}\ - 2 \textbf{(B)}\ 0 \textbf{(C)}\ \frac {3}{2} \textbf{(D)}\ \infty \textbf{(E)}\ -\frac{3}{2}$