AP Calculus AB and BC

Limit of (x^2-4)/(x^2-x-2) at x = 2 Is 4/3

Q: What is the limit of the same function as x \to -1?

It does not exist. At x = -1 the denominator (x-2)(x+1) is 0 but the numerator is (-3)(1) = -3, so the form is \frac{-3}{0}, not \frac{0}{0}. The function runs to +\infty from the right of -1 and to -\infty from the left, so x = -1 is a vertical asymptote.

Q: Does L'Hopital's rule give the same answer?

Yes. The form is \frac{0}{0}, so differentiating top and bottom gives \frac{2x}{2x-1}, which is \frac{4}{3} at x = 2. It is a legitimate check, though AP Unit 1 expects the factoring version and L'Hopital is not introduced until Unit 4.

Q: What happens to this function as x \to \infty?

The limit is 1. Both polynomials are degree 2 with leading coefficient 1, so the ratio of leading terms decides it. That is a different question from the one at x = 2, and mixing the two is the most common way students reach the wrong number here.

The limit of (x^2 - 4)/(x^2 - x - 2) as x approaches 2 is 4/3. Direct substitution gives 0/0, and both polynomials factor with an (x - 2): (x - 2)(x + 2) over (x - 2)(x + 1). Cancelling leaves (x + 2)/(x + 1), which is 4/3 at x = 2.

\lim_{x \to 2} \frac{x^2-4}{x^2-x-2} = \frac{4}{3}

Settled by factoring both polynomials.

Factoring top and bottom

The numerator is a difference of squares. The denominator needs two numbers multiplying to $-2$ and adding to $-1$ , which are $-2$ and $+1$ .

\frac{x^2-4}{x^2-x-2} = \frac{(x-2)(x+2)}{(x-2)(x+1)}

An $(x-2)$ sits on both levels, so it cancels for every $x \neq 2$ , and those are the only inputs the limit uses.

\lim_{x \to 2} \frac{(x-2)(x+2)}{(x-2)(x+1)} = \lim_{x \to 2} \frac{x+2}{x+1} = \frac{4}{3}

The surviving quotient is continuous at $x = 2$ , since its denominator is $3$ there and not $0$ , so substitution closes the problem.

What direct substitution gives

At $x = 2$ the numerator is $4 - 4 = 0$ and the denominator is $4 - 2 - 2 = 0$ .

\frac{2^2-4}{2^2-2-2} = \frac{0}{0}

Both polynomials vanish at the same input, so both carry a factor of $(x-2)$ and the fraction can be simplified. Only one copy of that factor sits in each, so they cancel exactly and the leftover is an ordinary finite number.

How the count of factors decides the answer

What survives the cancellation decides everything. If the denominator held $(x-2)^2$ against one copy on top, a single $(x-2)$ would be left underneath: the form is $\frac{\text{nonzero}}{0}$ , the function blows up, and because that leftover changes sign across $2$ it runs to $+\infty$ on one side and $-\infty$ on the other, so the two-sided limit does not exist. That is exactly what this function does at $x = -1$ . Leave $(x-2)^2$ underneath instead and the sign never flips, so both sides run to $+\infty$ . If the extra copy sat on top, the limit would be $0$ . A finite nonzero answer like $\frac{4}{3}$ says the two zeros are the same order.

The mistakes students make

Cancelling the $x^2$ terms, or the $4$ against the $2$ , straight across the fraction bar. Only complete factors cancel, never individual terms of a sum.
Factoring the denominator as $(x+2)(x-1)$ . That expands to $x^2 + x - 2$ , which has the wrong middle sign.
Answering $1$ because the leading terms match. $\frac{x^2}{x^2} \to 1$ describes the behaviour as $x \to \infty$ , not the behaviour at $x = 2$ .
Stopping at $\frac{x+2}{x+1}$ and calling that the limit. Simplifying is the middle of the work; the answer is the number that expression approaches.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of the same function as $x \to -1$ ?

It does not exist. At $x = -1$ the denominator $(x-2)(x+1)$ is $0$ but the numerator is $(-3)(1) = -3$ , so the form is $\frac{-3}{0}$ , not $\frac{0}{0}$ . The function runs to $+\infty$ from the right of $-1$ and to $-\infty$ from the left, so $x = -1$ is a vertical asymptote.

Does L'Hopital's rule give the same answer?

Yes. The form is $\frac{0}{0}$ , so differentiating top and bottom gives $\frac{2x}{2x-1}$ , which is $\frac{4}{3}$ at $x = 2$ . It is a legitimate check, though AP Unit 1 expects the factoring version and L'Hopital is not introduced until Unit 4.

What happens to this function as $x \to \infty$ ?

The limit is $1$ . Both polynomials are degree $2$ with leading coefficient $1$ , so the ratio of leading terms decides it. That is a different question from the one at $x = 2$ , and mixing the two is the most common way students reach the wrong number here.