1950 AHSME Problems/Problem 40

Problem

The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$

Solution

Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$, using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$. Thus, the answer is $\boxed{\textbf{(D)}\ 2}$.

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See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 41
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