What does 0/0 mean in a limit?

\frac{0}{0} is an indeterminate form: it means the limit might exist, but you cannot find it by direct substitution alone. Limits by Factoring (and other techniques like L'Hôpital's Rule) are designed to resolve indeterminate forms and find the actual limit value.

Can I use Limits by Factoring if the common factor is (x-a)²?

Yes. If both the numerator and denominator have (x-a)^2 (or higher powers) as a factor, you cancel the highest power you can from both. For example, \frac{(x-2)^2(x+1)}{(x-2)^2(x-3)} = \frac{x+1}{x-3} after canceling (x-2)^2. Always factor completely to see all the common factors.

Why must the original function be undefined at x=a for Limits by Factoring to work?

It does not have to be undefined at x=a. Limits by Factoring works whenever the numerator and denominator have a common factor of (x-a). However, the method is most useful when direct substitution gives \frac{0}{0}, which happens precisely when both the original numerator and denominator equal zero at x=a.

What if the numerator becomes a constant after canceling?

No problem. If you cancel (x-a) and are left with just a constant c in the numerator and a function h(x) in the denominator, the limit is \frac{c}{h(a)} (provided h(a) \neq 0). For example, \lim_{x \to 0} \frac{x}{x^2+x} = \lim_{x \to 0} \frac{1}{x+1} = 1.

Limits by Factoring

Limits by Factoring removes common factors from numerator and denominator to resolve 0/0 indeterminate forms and evaluate limits algebraically.

\lim_{x \to a} \frac{(x-a)g(x)}{(x-a)h(x)} = \frac{g(a)}{h(a)}

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What each symbol means

What Limits by Factoring takes

Symbol	Meaning
$g$	The polynomial or function that remains in the numerator after factoring out $(x-a)$ ; it must be continuous and well-defined at $x=a$ .
$h$	The polynomial or function that remains in the denominator after factoring out $(x-a)$ ; it must be continuous at $x=a$ and non-zero so that the final limit is defined.
$x$	The independent variable that is approaching the point $a$ ; as $x$ moves closer to $a$ , the limit describes the value the function approaches.
$a$	The point being approached; both the original numerator and denominator evaluate to zero at $x=a$ in a typical Limits by Factoring problem.

When to use it

Use this method when direct substitution gives the indeterminate form 0/0 in a rational expression.

Level

Usually taught in: Calculus I

Worked examples

1. Evaluate a limit by factoring a difference of squares

Problem

Find

\lim_{x \to 4} \frac{x^2 - 16}{x - 4}

$\lim_{x \to 4} \frac{x^2 - 16}{x - 4} = \lim_{x \to 4} \frac{(x-4)(x+4)}{x-4}$
Factor the numerator as a difference of squares: $x^2 - 16 = (x-4)(x+4)$ .
$= \lim_{x \to 4} (x+4)$
Cancel the common factor $(x-4)$ from numerator and denominator.
$= 4 + 4 = 8$
Substitute $x=4$ into the simplified expression to find the limit.

Answer: $8$

The original expression is undefined at $x=4$ because substitution gives $\frac{0}{0}$ . By factoring the numerator, we identify the common factor $(x-4)$ in both numerator and denominator. After canceling, the expression simplifies to $x+4$ , and we can evaluate the limit by substitution: $\lim_{x \to 4} (x+4) = 8$ .

2. Factor a quadratic numerator to find a limit at a negative point

Problem

Find

\lim_{x \to -3} \frac{x^2 + 5x + 6}{x + 3}

$\lim_{x \to -3} \frac{x^2 + 5x + 6}{x + 3} = \lim_{x \to -3} \frac{(x+3)(x+2)}{x+3}$
Factor the quadratic numerator by finding two numbers that multiply to $6$ and add to $5$ : these are $3$ and $2$ , giving $(x+3)(x+2)$ .
$= \lim_{x \to -3} (x+2)$
Cancel the common factor $(x+3)$ ; for any $x \neq -3$ , the expression equals $x+2$ .
$= -3 + 2 = -1$
Substitute $x=-3$ into the simplified form $x+2$ to evaluate the limit.

Answer: $-1$

The method works the same for negative approach points. The numerator is a quadratic that factors nicely over the integers. Once both numerator and denominator are in factored form, the common $(x+3)$ becomes visible and can be canceled. This cancellation reveals that the limit is simply $-3+2=-1$ .

3. Interpret a profit limit at a school bake sale

Problem

At a school bake sale, the profit from selling

x

boxes of cookies (where

x > 0

and

x \neq 9

) is given by

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

dollars. What profit does the model predict as sales approach

9

boxes?

$\lim_{x \to 9} \frac{x^2 - 9x}{x - 9} = \lim_{x \to 9} \frac{x(x-9)}{x-9}$
Factor the numerator by extracting $x$ as a common factor: $x^2 - 9x = x(x-9)$ .
$= \lim_{x \to 9} x$
Cancel the common factor $(x-9)$ . For $x \neq 9$ , the profit function simplifies to just $x$ .
$= 9$
Substitute $x=9$ to find that the profit approaches $\$9$ as sales near $9$ boxes.

Answer: $9 \text{ dollars}$

The original profit formula $P(x) = \frac{x^2 - 9x}{x - 9}$ is undefined at $x=9$ because both numerator and denominator equal zero. The limit addresses this by asking what value the profit approaches as $x$ gets close to $9$ . Factoring reveals that for all $x \neq 9$ , the profit is exactly $x$ dollars, so the predicted profit at approximately $9$ boxes is $\$9$ . This illustrates how limits can predict a function's behavior even where the formula itself breaks down.

Common mistakes

Where Limits by Factoring usually goes wrong

The mistake	Why it is wrong	The fix
When I substitute $x=1$ into $\frac{x^2-1}{x-1}$ and get $\frac{0}{0}$ , I conclude that the limit does not exist.	$\frac{0}{0}$ is an indeterminate form, which signals that the limit might exist but requires further work—it does not mean the limit is undefined or nonexistent.	Treat $\frac{0}{0}$ as a signal to factor the numerator and denominator, cancel common factors, and re-evaluate the limit using the simplified form.
Looking at $\frac{x^3 - 8}{x - 2}$ , I decide there is no common factor because I do not recognize that the numerator factors.	The expression $x^3 - 8$ is a difference of cubes ( $x^3 - 2^3$ ) and factors as $(x-2)(x^2+2x+4)$ , but you must know or recognize this special factorization pattern.	Learn factorization patterns for difference of squares ( $a^2 - b^2$ ), difference of cubes ( $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ ), trinomials, and other standard forms.
After canceling the $(x-3)$ factor in $\frac{(x-3)(x+1)}{x-3}$ to get $x+1$ , I substitute $x=3$ back into the original fraction $\frac{(x-3)(x+1)}{x-3}$ .	The original fraction is undefined at $x=3$ ; you must use the simplified form $(x+1)$ to evaluate the limit, not the original undefined expression.	After factoring and canceling, always substitute into the simplified form, not the original. The simplified form is valid for all $x$ near $a$ , even though the original is undefined at $x=a$ .

Tips and when to use something else

Always factor both numerator and denominator completely before canceling. Look for difference of squares ( $a^2 - b^2$ ), trinomials, and special patterns like difference or sum of cubes.
When direct substitution gives $\frac{0}{0}$ , Limits by Factoring is often the quickest method, provided the numerator and denominator factor nicely.
If factoring does not reveal a common $(x-a)$ factor, or if you encounter $\frac{0}{0}$ again after factoring, switch to L'Hôpital's Rule or another technique.
Remember that the cancellation is valid for $x \neq a$ , so the function may remain undefined at $x=a$ (a removable discontinuity) even though the limit exists.

Related formulas

Indeterminate FormsIndeterminate forms are expressions like 0/0 and ∞/∞ that arise when evaluating limits, signaling that direct substitution doesn't work.L'Hopital's RuleL'Hopital's Rule efficiently solves limits that reduce to 0/0 or ∞/∞ forms by taking derivatives of the numerator and denominator separately.Removable DiscontinuityA removable discontinuity is a hole in a graph where the limit exists but the function value differs; it can be fixed by redefining that point.Definition of a LimitDefinition of a Limit describes what value a function approaches as its input approaches a specific point—essential for all of calculus.

Frequently asked questions

What does 0/0 mean in a limit?: $\frac{0}{0}$ is an indeterminate form: it means the limit might exist, but you cannot find it by direct substitution alone. Limits by Factoring (and other techniques like L'Hôpital's Rule) are designed to resolve indeterminate forms and find the actual limit value.
Can I use Limits by Factoring if the common factor is (x-a)²?: Yes. If both the numerator and denominator have $(x-a)^2$ (or higher powers) as a factor, you cancel the highest power you can from both. For example, $\frac{(x-2)^2(x+1)}{(x-2)^2(x-3)} = \frac{x+1}{x-3}$ after canceling $(x-2)^2$ . Always factor completely to see all the common factors.
Why must the original function be undefined at x=a for Limits by Factoring to work?: It does not have to be undefined at $x=a$ . Limits by Factoring works whenever the numerator and denominator have a common factor of $(x-a)$ . However, the method is most useful when direct substitution gives $\frac{0}{0}$ , which happens precisely when both the original numerator and denominator equal zero at $x=a$ .
What if the numerator becomes a constant after canceling?: No problem. If you cancel $(x-a)$ and are left with just a constant $c$ in the numerator and a function $h(x)$ in the denominator, the limit is $\frac{c}{h(a)}$ (provided $h(a) \neq 0$ ). For example, $\lim_{x \to 0} \frac{x}{x^2+x} = \lim_{x \to 0} \frac{1}{x+1} = 1$ .

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Reviewed 2026-09-18