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:

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

Solution

Factoring the numerator using the sum of cubes identity and denominator using the difference of squares identity gives \[\dfrac{(x+1)(x^{2}-x+1)}{(x+1)(x-1)}\] Cancelling out a factor of $x+1$ from the numerator and denominator gives \[\dfrac{(x^{2}-x+1)}{(x-1)}\] Plugging in $x= -1$ gives $\dfrac{3}{-2}$ or $\fbox{E}$.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 35
Followed by
Problem 37
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