# 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 $\boxed{\textbf{(D)}\ 2}$.

~ MATH__is__FUN

## Solution 2

The numerator of $\frac {x^2-1}{x-1}$ can be factored as $(x+1)(x-1)$. The $x-1$ terms in the numerator and denominator cancel, so the expression is equal to $x+1$ so long as $x$ does not equal $1$. Looking at the function's behavior near 1, we see that as $x$ approaches one, the expression approaches $\boxed{\textbf{(D)}\ 2}$.

Note: Alternatively, we can ignore the domain restriction and just plug in $x = 1$ into the reduced expression.

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