Achievable AP Calculus AB

1. Limits

1.3. Algebraic limits

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Indeterminate forms

5 min read

Font

Discuss

Feedback

What you’ll learn

How to simplify indeterminate forms with algebraic techniques:
1. Factoring
2. Rationalizing
3. Common denominators

Sometimes, direct substitution yields an expression like $\frac{0}{0}$ , which is known as an indeterminate form. This result does not mean that the limit does not exist, but that the value is yet to be determined. In these cases, more algebraic manipulation is required to determine the limit, if it exists.

While $\frac{0}{0}$ is the most common, there are exactly 7 indeterminate forms:

Indeterminate forms:

$\frac{0}{0}$
$\frac{\pm \infty}{\pm \infty}$
$0 \times \infty$
$\infty - \infty$
$0^{0}$
$1^{\infty}$
$\infty^{0}$

This section focuses on the first two indeterminate forms, which are the most commonly encountered and can be resolved using the algebraic techniques below.

Method 1: Factoring

Many limits simplify after factoring. The main idea is to rewrite the expression so that the factor that creates the “problem area” is canceled, turning an indeterminate form into an expression that can be evaluated.

$x \to - 2 lim \frac{x ^{2} - 4}{3 x ^{3} + 24} =$

Solution

(spoiler)

Start with direct substitution:

$x \to - 2 lim \frac{x ^{2} - 4}{3 x ^{3} + 24} = \frac{( - 2 ) ^{2} - 4}{3 ( - 2 ) ^{3} + 24} = \frac{0}{0}$

To resolve the indeterminate form, factor both the numerator and denominator:

Numerator: difference of squares
Denominator: factor out the GCF, then factor the remaining sum of cubes using the formula

$(a^{3} + b^{3}) = (a + b) (a^{2} - ab + b^{2})$

After factoring,

$x \to - 2 lim \frac{( x + 2 ) ( x - 2 )}{3 ( x + 2 ) ( x ^{2} - 2 x + 4 )}$

Cancel the common factor, then directly substitute again:

$x \to - 2 lim \frac{( x - 2 ) ( x + 2 )}{3 ( x + 2 ) ( x ^{2} - 2 x + 4 )} = x \to - 2 lim \frac{x - 2}{3 ( x ^{2} - 2 x + 4 )} = \frac{- 2 - 2}{3 (( - 2 ) ^{2} - 2 ( - 2 ) + 4 )} = - \frac{1}{9}$

AP tip:

Keep the limit notation in front of the expression at every step until the value of $x$ is substituted. Dropping the limit notation too early can make it look like the limit equals the function value, which isn’t always true.

Method 2: Rationalizing with conjugates

When square roots lead to an indeterminate form after substitution, multiply by a conjugate so the square root expression simplifies using a difference of squares.

Conjugate:

A binomial expression with the same two terms but the opposite sign between them. The conjugate of $(a + b)$ is $(a - b)$ .

For example, the conjugate of

$x - 2 + 5$

$x - 2 - 5$

Multiplying them gives a difference of squares:

$(x - 2 + 5) (x - 2 - 5)$

$= ((x - 2)^{2} - 5^{2})$

$= ((x - 2) - 25)$

$= (x - 27)$

AP tip:

Since the goal is to cancel the factor causing the $0$ in the denominator, expand only the product of the conjugates. Keep the rest of the expression in factored form so you can spot and cancel the common terms.

Example

Evaluate

$x \to 4 lim \frac{x - 2}{4 - x}$

Solution

(spoiler)

Start with direct substitution:

$x \to 4 lim \frac{x - 2}{4 - x} = \frac{4 - 2}{4 - 4} = \frac{0}{0}$

To resolve the indeterminate form, multiply the numerator and denominator by the conjugate of the square root expression in the numerator. The conjugate of $(x - 2)$ is $(x + 2)$ :

$x \to 4 lim \frac{x - 2}{4 - x} \cdot \frac{x + 2}{x + 2}$

Now multiply the conjugates in the numerator:

$x \to 4 lim \frac{x - 4}{( 4 - x ) ( x + 2 )}$

Rewrite $(x - 4)$ as $- (4 - x)$ so it cancels with the denominator:

$x \to 4 lim \frac{- ( 4 - x )}{( 4 - x ) ( x + 2 )} = x \to 4 lim \frac{- 1}{x + 2} = \frac{- 1}{4 + 2} = - \frac{1}{4}$

Alternatively, the same problem can be solved by factoring:

(spoiler)

The denominator $4 - x$ is a difference of squares:

$(4 - x) = (2 + x) (2 - x)$

Then

$x \to 4 lim \frac{x - 2}{4 - x} = x \to 4 lim \frac{x - 2}{( 2 + x ) ( 2 - x )}$

In the numerator, rewrite $(x - 2)$ as $- (2 - x)$ , then cancel:

$x \to 4 lim \frac{- ( 2 - x )}{( 2 + x ) ( 2 - x )} = x \to 4 lim \frac{- 1}{2 + x} = - \frac{1}{4}$

Method 3: Common denominators

A fraction over a fraction can be simplified by multiplying through by a common denominator.

Evaluate

$x \to 2 lim \frac{\frac{1}{x} - \frac{1}{2}}{x - 2}$

Solution

(spoiler)

Direct substitution gives the indeterminate form $\frac{0}{0}$ . In the numerator, the fraction can be simplified by multiplying through the common denominator $2 x$ :

$x \to 2 lim \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} \cdot \frac{2 x}{2 x} = x \to 2 lim \frac{2 - x}{2 x ( x - 2 )} = x \to 2 lim \frac{- ( x - 2 )}{2 x ( x - 2 )} = x \to 2 lim \frac{- 1}{2 x} = - \frac{1}{4}$

Indeterminate forms

Expressions like $\frac{0}{0}$ , $0 \times \infty$ , $1^{\infty}$ , etc.
7 main types to recognize
Require algebraic manipulation before evaluating limits

Factoring

Factor numerator and denominator to cancel problematic terms
Use formulas:
- Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$
- Sum of cubes: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$
Always keep limit notation until substitution

Rationalizing

Multiply by conjugate to eliminate square roots
Conjugate: same terms, opposite sign (e.g., $a + b$ and $a - b$ )
Use difference of squares to simplify and cancel terms

Common denominators

Combine fractions using a common denominator
Multiply numerator and denominator by the common denominator to simplify
Cancel terms to resolve indeterminate forms

General strategies

Always try direct substitution first
If indeterminate, apply algebraic techniques before substituting again
Keep limit notation until final substitution step

Indeterminate forms

What you’ll learn

How to simplify indeterminate forms with algebraic techniques:
1. Factoring
2. Rationalizing
3. Common denominators

While $\frac{0}{0}$ is the most common, there are exactly 7 indeterminate forms:

Indeterminate forms:

$\frac{0}{0}$
$\frac{\pm \infty}{\pm \infty}$
$0 \times \infty$
$\infty - \infty$
$0^{0}$
$1^{\infty}$
$\infty^{0}$

This section focuses on the first two indeterminate forms, which are the most commonly encountered and can be resolved using the algebraic techniques below.

Method 1: Factoring

$x \to - 2 lim \frac{x ^{2} - 4}{3 x ^{3} + 24} =$

Solution

(spoiler)

Start with direct substitution:

$x \to - 2 lim \frac{x ^{2} - 4}{3 x ^{3} + 24} = \frac{( - 2 ) ^{2} - 4}{3 ( - 2 ) ^{3} + 24} = \frac{0}{0}$

To resolve the indeterminate form, factor both the numerator and denominator:

Numerator: difference of squares
Denominator: factor out the GCF, then factor the remaining sum of cubes using the formula

$(a^{3} + b^{3}) = (a + b) (a^{2} - ab + b^{2})$

After factoring,

$x \to - 2 lim \frac{( x + 2 ) ( x - 2 )}{3 ( x + 2 ) ( x ^{2} - 2 x + 4 )}$

Cancel the common factor, then directly substitute again:

$x \to - 2 lim \frac{( x - 2 ) ( x + 2 )}{3 ( x + 2 ) ( x ^{2} - 2 x + 4 )} = x \to - 2 lim \frac{x - 2}{3 ( x ^{2} - 2 x + 4 )} = \frac{- 2 - 2}{3 (( - 2 ) ^{2} - 2 ( - 2 ) + 4 )} = - \frac{1}{9}$

AP tip:

Method 2: Rationalizing with conjugates

When square roots lead to an indeterminate form after substitution, multiply by a conjugate so the square root expression simplifies using a difference of squares.

Conjugate:

A binomial expression with the same two terms but the opposite sign between them. The conjugate of $(a + b)$ is $(a - b)$ .

For example, the conjugate of

$x - 2 + 5$

$x - 2 - 5$

Multiplying them gives a difference of squares:

$(x - 2 + 5) (x - 2 - 5)$

$= ((x - 2)^{2} - 5^{2})$

$= ((x - 2) - 25)$

$= (x - 27)$

AP tip:

Example

Evaluate

$x \to 4 lim \frac{x - 2}{4 - x}$

Solution

(spoiler)

Start with direct substitution:

$x \to 4 lim \frac{x - 2}{4 - x} = \frac{4 - 2}{4 - 4} = \frac{0}{0}$

To resolve the indeterminate form, multiply the numerator and denominator by the conjugate of the square root expression in the numerator. The conjugate of $(x - 2)$ is $(x + 2)$ :

$x \to 4 lim \frac{x - 2}{4 - x} \cdot \frac{x + 2}{x + 2}$

Now multiply the conjugates in the numerator:

$x \to 4 lim \frac{x - 4}{( 4 - x ) ( x + 2 )}$

Rewrite $(x - 4)$ as $- (4 - x)$ so it cancels with the denominator:

$x \to 4 lim \frac{- ( 4 - x )}{( 4 - x ) ( x + 2 )} = x \to 4 lim \frac{- 1}{x + 2} = \frac{- 1}{4 + 2} = - \frac{1}{4}$

Alternatively, the same problem can be solved by factoring:

(spoiler)

The denominator $4 - x$ is a difference of squares:

$(4 - x) = (2 + x) (2 - x)$

Then

$x \to 4 lim \frac{x - 2}{4 - x} = x \to 4 lim \frac{x - 2}{( 2 + x ) ( 2 - x )}$

In the numerator, rewrite $(x - 2)$ as $- (2 - x)$ , then cancel:

$x \to 4 lim \frac{- ( 2 - x )}{( 2 + x ) ( 2 - x )} = x \to 4 lim \frac{- 1}{2 + x} = - \frac{1}{4}$

Method 3: Common denominators

A fraction over a fraction can be simplified by multiplying through by a common denominator.

Evaluate

$x \to 2 lim \frac{\frac{1}{x} - \frac{1}{2}}{x - 2}$

Solution

(spoiler)

Direct substitution gives the indeterminate form $\frac{0}{0}$ . In the numerator, the fraction can be simplified by multiplying through the common denominator $2 x$ :

Key points

Indeterminate forms

Expressions like $\frac{0}{0}$ , $0 \times \infty$ , $1^{\infty}$ , etc.
7 main types to recognize
Require algebraic manipulation before evaluating limits

Factoring

Factor numerator and denominator to cancel problematic terms
Use formulas:
- Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$
- Sum of cubes: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$
Always keep limit notation until substitution

Rationalizing

Multiply by conjugate to eliminate square roots
Conjugate: same terms, opposite sign (e.g., $a + b$ and $a - b$ )
Use difference of squares to simplify and cancel terms

Common denominators

Combine fractions using a common denominator
Multiply numerator and denominator by the common denominator to simplify
Cancel terms to resolve indeterminate forms

General strategies

Always try direct substitution first
If indeterminate, apply algebraic techniques before substituting again
Keep limit notation until final substitution step

What you’ll learn

How to simplify indeterminate forms with algebraic techniques:
1. Factoring
2. Rationalizing
3. Common denominators

While $\frac{0}{0}$ is the most common, there are exactly 7 indeterminate forms:

Indeterminate forms:

$\frac{0}{0}$
$\frac{\pm \infty}{\pm \infty}$
$0 \times \infty$
$\infty - \infty$
$0^{0}$
$1^{\infty}$
$\infty^{0}$

This section focuses on the first two indeterminate forms, which are the most commonly encountered and can be resolved using the algebraic techniques below.

Method 1: Factoring

$x \to - 2 lim \frac{x ^{2} - 4}{3 x ^{3} + 24} =$

Solution

(spoiler)

Start with direct substitution:

$x \to - 2 lim \frac{x ^{2} - 4}{3 x ^{3} + 24} = \frac{( - 2 ) ^{2} - 4}{3 ( - 2 ) ^{3} + 24} = \frac{0}{0}$

To resolve the indeterminate form, factor both the numerator and denominator:

Numerator: difference of squares
Denominator: factor out the GCF, then factor the remaining sum of cubes using the formula

$(a^{3} + b^{3}) = (a + b) (a^{2} - ab + b^{2})$

After factoring,

$x \to - 2 lim \frac{( x + 2 ) ( x - 2 )}{3 ( x + 2 ) ( x ^{2} - 2 x + 4 )}$

Cancel the common factor, then directly substitute again:

$x \to - 2 lim \frac{( x - 2 ) ( x + 2 )}{3 ( x + 2 ) ( x ^{2} - 2 x + 4 )} = x \to - 2 lim \frac{x - 2}{3 ( x ^{2} - 2 x + 4 )} = \frac{- 2 - 2}{3 (( - 2 ) ^{2} - 2 ( - 2 ) + 4 )} = - \frac{1}{9}$

AP tip:

Method 2: Rationalizing with conjugates

When square roots lead to an indeterminate form after substitution, multiply by a conjugate so the square root expression simplifies using a difference of squares.

Conjugate:

A binomial expression with the same two terms but the opposite sign between them. The conjugate of $(a + b)$ is $(a - b)$ .

For example, the conjugate of

$x - 2 + 5$

$x - 2 - 5$

Multiplying them gives a difference of squares:

$(x - 2 + 5) (x - 2 - 5)$

$= ((x - 2)^{2} - 5^{2})$

$= ((x - 2) - 25)$

$= (x - 27)$

AP tip:

Example

Evaluate

$x \to 4 lim \frac{x - 2}{4 - x}$

Solution

(spoiler)

Start with direct substitution:

$x \to 4 lim \frac{x - 2}{4 - x} = \frac{4 - 2}{4 - 4} = \frac{0}{0}$

To resolve the indeterminate form, multiply the numerator and denominator by the conjugate of the square root expression in the numerator. The conjugate of $(x - 2)$ is $(x + 2)$ :

$x \to 4 lim \frac{x - 2}{4 - x} \cdot \frac{x + 2}{x + 2}$

Now multiply the conjugates in the numerator:

$x \to 4 lim \frac{x - 4}{( 4 - x ) ( x + 2 )}$

Rewrite $(x - 4)$ as $- (4 - x)$ so it cancels with the denominator:

$x \to 4 lim \frac{- ( 4 - x )}{( 4 - x ) ( x + 2 )} = x \to 4 lim \frac{- 1}{x + 2} = \frac{- 1}{4 + 2} = - \frac{1}{4}$

Alternatively, the same problem can be solved by factoring:

(spoiler)

The denominator $4 - x$ is a difference of squares:

$(4 - x) = (2 + x) (2 - x)$

Then

$x \to 4 lim \frac{x - 2}{4 - x} = x \to 4 lim \frac{x - 2}{( 2 + x ) ( 2 - x )}$

In the numerator, rewrite $(x - 2)$ as $- (2 - x)$ , then cancel:

$x \to 4 lim \frac{- ( 2 - x )}{( 2 + x ) ( 2 - x )} = x \to 4 lim \frac{- 1}{2 + x} = - \frac{1}{4}$

Method 3: Common denominators

A fraction over a fraction can be simplified by multiplying through by a common denominator.

Evaluate

$x \to 2 lim \frac{\frac{1}{x} - \frac{1}{2}}{x - 2}$

Solution

(spoiler)

Direct substitution gives the indeterminate form $\frac{0}{0}$ . In the numerator, the fraction can be simplified by multiplying through the common denominator $2 x$ :