01 / 21
lim

Determining Limits Using
Algebraic Manipulation

Learn how to evaluate limits that produce indeterminate forms by applying factoring, conjugates, and other algebraic techniques to simplify and solve.

AP Calculus AB

What Is a Limit?

A limit describes the value a function \( f(x) \) approaches as \( x \) gets arbitrarily close to some number \( c \). We write:

$$ \lim_{x \to c} f(x) = L $$

This means that as \( x \to c \), the output \( f(x) \) gets closer and closer to \( L \), even if \( f(c) \) itself is undefined.

The Indeterminate Form Problem

Direct substitution is the first strategy: plug \( c \) into \( f(x) \). But sometimes you get:

$$ \frac{0}{0} $$

This is called an indeterminate form: it does not tell us the limit's value. It signals that the numerator and denominator share a common factor that cancels at \( x = c \).

Visualizing the Removable Discontinuity

x y 1 2 3 4 1 2 3 4 5 6 hole at (2, 4) The limit is 4, even though f(2) is undefined

The Core Principle

If \( f(x) = g(x) \) for all \( x \neq c \), then:

$$ \lim_{x \to c} f(x) = \lim_{x \to c} g(x) $$

This is why algebraic manipulation works: cancelling a common factor creates a simpler function \( g(x) \) that equals \( f(x) \) everywhere except at the point of interest. Since limits only care about values near \( c \), not at \( c \), the two limits are identical.

Three Key Techniques

Factoring
Factor numerator and denominator, then cancel the common factor that causes \( \frac{0}{0} \).
Conjugate Multiplication
Multiply by the conjugate to eliminate square roots and reveal a cancellable factor.
Simplifying Complex Fractions
Combine fractions in numerator or denominator into a single fraction, then simplify.

Factoring vs. Conjugates

Factoring Conjugate
When to use Polynomials in numerator and/or denominator Square roots in numerator or denominator
Action Factor, cancel \( (x - c) \) Multiply by conjugate, simplify, cancel
Example trigger \( \frac{x^2 - 9}{x - 3} \) \( \frac{\sqrt{x+1} - 2}{x - 3} \)
Result Polynomial after cancel Rationalized expression after cancel

How to Apply the Technique

1
Try direct substitution. Plug \( c \) into \( f(x) \). If you get a real number, that is the limit.
2
Identify the indeterminate form. If you get \( \frac{0}{0} \), algebraic manipulation is needed.
3
Choose your technique: factor (polynomials), conjugate (radicals), or combine fractions (complex fractions).
4
Simplify and cancel the factor that created the \( \frac{0}{0} \) form.
5
Substitute again. Plug \( c \) into the simplified expression to find the limit.

Example 1: Factoring

Evaluate: \( \displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3} \)

1
Direct substitution: \( \frac{9 - 9}{3 - 3} = \frac{0}{0} \). Indeterminate.
2
Factor the numerator: \( x^2 - 9 = (x - 3)(x + 3) \).
3
Cancel: \( \frac{(x-3)(x+3)}{x-3} = x + 3 \) for \( x \neq 3 \).
4
Substitute: \( 3 + 3 = \) \( 6 \).

Example 2: Conjugate Method

Evaluate: \( \displaystyle\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \)

1
Direct substitution: \( \frac{\sqrt{4} - 2}{0} = \frac{0}{0} \). Indeterminate.
2
Multiply by conjugate: \( \frac{\sqrt{x+4}-2}{x} \cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} \).
3
Simplify numerator: \( (x+4) - 4 = x \). Expression becomes \( \frac{x}{x(\sqrt{x+4}+2)} \).
4
Cancel and substitute: \( \frac{1}{\sqrt{0+4}+2} = \frac{1}{4} \).

Example 3: Complex Fractions

Evaluate: \( \displaystyle\lim_{x \to 0} \frac{\frac{1}{x+3} - \frac{1}{3}}{x} \)

1
Direct substitution: \( \frac{\frac{1}{3} - \frac{1}{3}}{0} = \frac{0}{0} \). Indeterminate.
2
Combine the numerator: \( \frac{1}{x+3} - \frac{1}{3} = \frac{3 - (x+3)}{3(x+3)} = \frac{-x}{3(x+3)} \).
3
Divide by \( x \): \( \frac{-x}{3(x+3)} \cdot \frac{1}{x} = \frac{-1}{3(x+3)} \).
4
Substitute: \( \frac{-1}{3(0+3)} = \frac{-1}{9} \).

Graphing the Conjugate Example

x y 0 5 0.25 hole at (0, 1/4) The limit equals 1/4 as x approaches 0

Combining with Trig Identities

Sometimes you need both a trig identity and algebraic manipulation. The key identity:

$$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $$

Example: \( \displaystyle\lim_{x \to 0} \frac{\sin(3x)}{x} \)

1
Rewrite: Multiply and divide by 3: \( \frac{\sin(3x)}{x} = 3 \cdot \frac{\sin(3x)}{3x} \).
2
Apply the identity: As \( x \to 0 \), \( 3x \to 0 \), so \( \frac{\sin(3x)}{3x} \to 1 \).
3
Result: \( 3 \cdot 1 = \) \( 3 \).

Higher-Degree Factoring

Evaluate: \( \displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x - 1} \)

1
Direct substitution: \( \frac{1 - 1}{1 - 1} = \frac{0}{0} \). Indeterminate.
2
Factor using the difference of cubes: \( x^3 - 1 = (x-1)(x^2 + x + 1) \).
3
Cancel: \( \frac{(x-1)(x^2+x+1)}{x-1} = x^2 + x + 1 \).
4
Substitute: \( 1 + 1 + 1 = \) \( 3 \).

Common Mistakes to Avoid

✕ Claiming \( \frac{0}{0} = 0 \)
\( \frac{0}{0} \) is indeterminate, not zero. You must simplify first before substituting.
✕ Cancelling Before Factoring
You cannot cancel individual terms: \( \frac{x^2 - 4}{x} \neq x - 4 \). Factor first, then cancel common factors.
✕ Forgetting the Domain Restriction
After cancelling \( (x - c) \), the simplified function equals the original only for \( x \neq c \). The limit still exists, but \( f(c) \) may not.

Summary of Techniques

Technique Trigger Key Step Example
Factoring Polynomial / polynomial Factor, cancel \( (x-c) \) \( \frac{x^2-9}{x-3} \)
Conjugate Square root in expression Multiply by conjugate \( \frac{\sqrt{x+4}-2}{x} \)
Combine Fractions Nested fractions Common denominator \( \frac{\frac{1}{x+3}-\frac{1}{3}}{x} \)
Trig + Algebra Trig functions at 0 Rewrite to match \( \frac{\sin u}{u} \) \( \frac{\sin 3x}{x} \)

Question 1 of 5

Evaluate: \( \displaystyle\lim_{x \to 5} \frac{x^2 - 25}{x - 5} \)

Question 2 of 5

Evaluate: \( \displaystyle\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x} \)

Question 3 of 5

Evaluate: \( \displaystyle\lim_{x \to 0} \frac{\frac{1}{x+2} - \frac{1}{2}}{x} \)

Question 4 of 5

Evaluate: \( \displaystyle\lim_{x \to 2} \frac{x^3 - 8}{x - 2} \)

Question 5 of 5

Which algebraic technique is most appropriate for evaluating \( \displaystyle\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} \)?