1952 AHSME Problems/Problem 36

Problem

To be continuous at $x = - 1$ , the value of $\frac {x^3 + 1}{x^2 - 1}$ is taken to be:

$\textbf{(A)}\ - 2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ \infty \qquad \textbf{(E)}\ -\frac{3}{2}$

$\fbox{}$

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 35	Followed by Problem 37
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