AP Calculus AB and BC

Limit of (x-3)/(x^2-9) as x Approaches 3 Is 1/6

Q: Why is x = 3 a hole while x = -3 is an asymptote?

Because the numerator vanishes at 3 and not at -3. At 3 the factor x - 3 sits on both sides of the fraction and cancels, leaving a finite value. At -3 the numerator is -6 while the denominator goes to 0, so the quotient grows without bound and no finite limit exists.

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

Yes. The form is \frac{0}{0}, and differentiating top and bottom separately gives \frac{1}{2x}, which is \frac{1}{6} at x = 3. Factoring is faster and is the skill Unit 1 is testing, but the two agree.

Q: What if the fraction were flipped, \frac{x^2-9}{x-3}?

The limit would be 6. The same cancellation leaves x + 3 rather than \frac{1}{x+3}, and that is 6 at x = 3. Reciprocal functions have reciprocal limits whenever both values are finite and nonzero, which makes a quick check on your work.

The limit of (x - 3)/(x^2 - 9) as x approaches 3 is 1/6. Direct substitution gives 0/0, so factor the denominator as (x - 3)(x + 3) and cancel the shared x - 3. What is left is 1/(x + 3), which is 1/6 at x = 3. The graph has a hole at x = 3, not an asymptote.

\lim_{x \to 3} \frac{x-3}{x^2-9} = \frac{1}{6}

Settled by factoring and cancelling.

Factoring the denominator and cancelling

The denominator is a difference of squares, $x^2 - 3^2$ , so it splits into two linear factors and one of them is the entire numerator.

\frac{x-3}{x^2-9} = \frac{x-3}{(x-3)(x+3)}

For every $x \neq 3$ the shared factor is a nonzero number over itself, so it cancels. That step is legal precisely because a limit at $3$ looks only at nearby inputs, where $x - 3 \neq 0$ .

\lim_{x \to 3} \frac{x-3}{x^2-9} = \lim_{x \to 3} \frac{1}{x+3} = \frac{1}{3+3} = \frac{1}{6}

The survivor $\frac{1}{x+3}$ is continuous at $3$ , so substitution finally works and returns $\frac{1}{6}$ . Testing $x = 3.001$ in the original fraction gives $0.16664$ , which agrees.

What direct substitution gives

At $x = 3$ the numerator is $3 - 3 = 0$ and the denominator is $9 - 9 = 0$ , so the fraction reports nothing.

\frac{3-3}{3^2-9} = \frac{0}{0}

Two polynomials vanishing at the same input share the factor $(x-3)$ , by the factor theorem, so the cancellation is guaranteed before you attempt it. On a rational function $\frac{0}{0}$ is an instruction to factor, not a verdict.

The hole at 3 and the asymptote at -3

Only one of the denominator's two zeros cancels, and the two behave nothing alike.

Input	What the graph does	Why
$x = 3$	Removable hole at height $\frac{1}{6}$	The factor $x-3$ cancels, leaving a continuous function
$x = -3$	Vertical asymptote	The factor $x+3$ has nothing to cancel against

So the same function has a finite limit at one zero of its denominator and no limit at all at the other. At $x = -3$ the simplified form $\frac{1}{x+3}$ is unbounded, positive on the right of $-3$ and negative on the left.

\lim_{x \to -3^+} \frac{x-3}{x^2-9} = \infty, \qquad \lim_{x \to -3^-} \frac{x-3}{x^2-9} = -\infty

The mistakes students make

Answering that the limit does not exist because the denominator is $0$ at $x = 3$ . A zero over a zero is the removable case, and the cancellation recovers the value.
Answering $6$ instead of $\frac{1}{6}$ by flipping the survivor. After cancelling, the numerator is $1$ and the denominator is $x + 3$ .
Cancelling the $3$ against the $9$ , or the $x$ against the $x^2$ . Only whole factors cancel, which is why $x^2 - 9$ has to be factored first.
Assuming every zero of a denominator is a vertical asymptote. Here $x = 3$ is a hole and only $x = -3$ is an asymptote.

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

Why is $x = 3$ a hole while $x = -3$ is an asymptote?

Because the numerator vanishes at $3$ and not at $-3$ . At $3$ the factor $x - 3$ sits on both sides of the fraction and cancels, leaving a finite value. At $-3$ the numerator is $-6$ while the denominator goes to $0$ , so the quotient grows without bound and no finite limit exists.

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

Yes. The form is $\frac{0}{0}$ , and differentiating top and bottom separately gives $\frac{1}{2x}$ , which is $\frac{1}{6}$ at $x = 3$ . Factoring is faster and is the skill Unit 1 is testing, but the two agree.

What if the fraction were flipped, $\frac{x^2-9}{x-3}$ ?

The limit would be $6$ . The same cancellation leaves $x + 3$ rather than $\frac{1}{x+3}$ , and that is $6$ at $x = 3$ . Reciprocal functions have reciprocal limits whenever both values are finite and nonzero, which makes a quick check on your work.