🎯⭐ INTERACTIVE LESSON

Factoring Method for Limits

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Factoring Method for Limits - Complete Interactive Lesson

Part 1: The 0/0 Trigger

🧮 Factoring Method — The $0/0$ Trigger

Part 1 of 4 — Why factoring resolves indeterminate forms

Topics in This Part

Section
The $0/0$ Trigger
🔑 Why Factoring Works
Easy First Example

🔑 Why this matters: $0/0$ is the most common indeterminate form on the AP exam. Factoring is the most common technique to resolve it.

⚠️ The $0/0$ Trigger

You computed $\lim_{x \to a} \dfrac{p(x)}{q(x)}$ by direct substitution and got . This means:

🔑 Why Factoring Is Justified

The functions $\dfrac{p(x)}{q(x)}$ and the simplified version $\dfrac{p(x)/(x-a)}{q(x)/(x-a)}$ are .

📝 Easy First Example

Find $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$ .

Concept Check 🎯

Factor and Limit 🧮

Compute by factoring:

1) $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} = ?$

Part 2: The Factoring Toolbox

🛠️ The Factoring Toolbox

Part 2 of 4 — Common patterns to recognize

Topics in This Part

Section
🔑 Difference of Squares
🔑 Quadratic Trinomials
Common Factor Extraction

🔑 Why this matters: Knowing your factoring patterns lets you spot the cancellation in seconds.

◯² Difference of Squares

$a^2 - b^2 = (a - b)(a + b)$

Part 3: Worked Examples — Workflow

📝 Worked Examples — The Factoring Workflow

Part 3 of 4 — Practice the full pipeline

Topics in This Part

Section
The 4-Step Workflow
Two-Variable Setting (denominator factors too)
When the Limit Equals 0 vs. a Nonzero Number

🔑 Why this matters: AP problems rarely give you a "just factor the numerator" setup. Both top and bottom may need factoring.

🔄 The 4-Step Workflow

For $\lim_{x \to a} \dfrac{p(x)}{q(x)}$ :

Part 4: Beyond Quadratics

📐 Beyond Quadratics — Higher-Degree Factoring

Part 4 of 4 — Cubes, grouping, and synthetic division

Topics in This Part

Section
🔑 Sum and Difference of Cubes
Factoring by Grouping
Synthetic Division to Find $(x - a)$

🔑 Why this matters: AP problems sometimes hide $(x - a)$ inside a cubic or higher. Knowing extra patterns saves the day.

$p(a) = 0$ → so $(x - a)$ is a factor of $p(x)$ .
$q(a) = 0$ → so $(x - a)$ is a factor of $q(x)$ .

Both top and bottom share the factor $(x - a)$ . We can cancel that common factor — for $x \ne a$ , the simplified expression equals the original.

💡 Big idea: the $(x - a)$ in numerator and denominator are causing the $0/0$ . Cancel them, and the surviving expression is what the limit "really wants to be."

q(x)/(x−a)p(x)/(x−a)

identical for every $x \ne a$

Limits ignore the value at $x = a$ — they only care about behavior near $a$ . So the two functions have the same limit as $x \to a$ , even though one is $0/0$ at $a$ and the other is well-defined.

$\lim_{x \to a} \frac{p(x)}{q(x)} = \lim_{x \to a} \frac{p(x)/(x-a)}{q(x)/(x-a)}$

Step 1: Try direct substitution: $\dfrac{4 - 4}{2 - 2} = \dfrac{0}{0}$ . Indeterminate.

Step 2: Factor: $x^2 - 4 = (x - 2)(x + 2)$ .

Step 3: Cancel: $\dfrac{(x-2)(x+2)}{x - 2} = x + 2$ for $x \ne 2$ .

Step 4: Take the limit of the simplified expression: $\lim_{x \to 2} (x + 2) = 4$ .

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

2) $\lim_{x \to -2} \dfrac{x^2 - 4}{x + 2} = ?$

Pops up constantly in $0/0$ problems with $x^2$ in the numerator.

$x^2 - 9 = (x - 3)(x + 3)$
$x^2 - 16 = (x - 4)(x + 4)$
$4x^2 - 25 = (2x - 5)(2x + 5)$
$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x-1)(x+1)(x^2 + 1)$

📐 Quadratic Trinomials

$ax^2 + bx + c = a(x - r_1)(x - r_2)$

where $r_1, r_2$ are the roots. Use the quadratic formula or AC-method.

Worked example: $\lim_{x \to 2} \dfrac{x^2 - 5x + 6}{x - 2}$ .

Direct sub: $\dfrac{4 - 10 + 6}{0} = 0/0$ . Indeterminate.
Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$ .

💡 If the denominator is $(x - 2)$ , the numerator factors must contain $(x - 2)$ — otherwise $0/0$ wouldn't happen.

🔧 Common Factor Extraction (GCF)

Sometimes you can pull out an obvious common factor first.

Example: $\lim_{x \to 0} \dfrac{x^3 - 4x}{x}$ .

Direct sub: $0/0$ .
Factor numerator: $x^3 - 4x = x(x^2 - 4)$ .

Pattern Spot 🎯

Apply 🧮

1) $\lim_{x \to 1} \dfrac{x^2 - 1}{x - 1} = ?$

2) $\lim_{x \to 3} \dfrac{x^2 - 6x + 9}{x - 3} = ?$

3) $\lim_{x \to -1} \dfrac{x^2 + 3x + 2}{x + 1} = ?$

Try direct substitution. If you get a clean number → done.
If you get $0/0$ , fully factor both $p(x)$ and $q(x)$ .
Cancel the common $(x - a)$ factor (or factors).
Substitute $x = a$ into the simplified expression.

💡 If after step 3 you still get $0/0$ , repeat: cancel another $(x - a)$ factor.

🔁 Both Sides Factor

Worked example: $\lim_{x \to 3} \dfrac{x^2 - 9}{x^2 - 5x + 6}$ .

Direct sub: $\dfrac{0}{0}$ . ✅ Indeterminate.
Numerator: $x^2 - 9 = (x - 3)(x + 3)$ .

$\boxed{\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 5x + 6} = 6}$

🎯 Limit = 0 vs. Limit = Nonzero

After cancellation and substitution:

Result	Meaning
Clean number $L$ (including 0)	That's the limit ✅
Still $0/0$	Factor again — there's another $(x - a)$ in both
Nonzero / 0	Cancellation didn't fully resolve it; you have a vertical asymptote — limit DNE

Repeated-factor example: $\lim_{x \to 2} \dfrac{(x-2)^3}{(x-2)^2} = \lim_{x \to 2} (x - 2) = 0$ .

Apply the Workflow 🎯

Workflow Practice 🧮

1) $\lim_{x \to 4} \dfrac{x^2 - 16}{x^2 - 3x - 4} = ?$ (decimal or fraction)

2) $\lim_{x \to -3} \dfrac{x^2 + 5x + 6}{x + 3} = ?$

🧊 Sum and Difference of Cubes

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

Example: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ .

Worked limit: $\lim_{x \to 2} \dfrac{x^3 - 8}{x - 2} = \lim_{x \to 2} (x^2 + 2x + 4) = 4 + 4 + 4 = 12$ .

💡 The leftover quadratic $x^2 + ax + a^2$ does NOT factor further over the reals (discriminant negative). Just substitute.

🔗 Factoring by Grouping

For four-term polynomials:

Example: $x^3 + x^2 - 4x - 4$ .

Group: $(x^3 + x^2) + (-4x - 4) = x^2(x + 1) - 4(x + 1) = (x + 1)(x^2 - 4) = (x + 1)(x - 2)(x + 2)$ .

Worked limit: $\lim_{x \to -1} \dfrac{x^3 + x^2 - 4x - 4}{x + 1}$ .

$0/0$ ✅
After cancelling $(x + 1)$ : $(x - 2)(x + 2) = x^2 - 4 \to (-1)^2 - 4 = -3$ .

➗ Synthetic Division to Extract $(x - a)$

When a polynomial $p(x)$ doesn't factor obviously, use synthetic division by $(x - a)$ .

Example: $\lim_{x \to 1} \dfrac{x^3 - 6x^2 + 11x - 6}{x - 1}$ .

Synthetic-divide $1 \;|\; 1 \; -6 \; 11 \; -6$ :

step	result
Bring down 1	$1$
$1 \cdot 1 = 1$ , $-6 + 1 = -5$

So $p(x) = (x - 1)(x^2 - 5x + 6)$ . Cancel $($ , take limit: .

💡 Synthetic division IS factoring; it just gives you the quotient mechanically.

Higher-Degree Practice 🎯

Cancel:

\dfrac{(x-2)(x-3)}{x-2} \to x - 3

Substitute:

2 - 3 = -1

. Answer: $-1$ .

Cancel:

\dfrac{x(x^2 - 4)}{x} = x^2 - 4

for

x \ne 0

Substitute:

0 - 4 = -4

. Answer: $-4$ .

Denominator:

x^2 - 5x + 6 = (x - 3)(x - 2)

Cancel:

\dfrac{(x-3)(x+3)}{(x-3)(x-2)} \to \dfrac{x + 3}{x - 2}

Substitute:

\dfrac{6}{1} = 6

Factoring Method for Limits

Factoring Method for Limits - Complete Interactive Lesson

Part 1: The 0/0 Trigger

🧮 Factoring Method — The 0/00/00/0 Trigger

Topics in This Part

⚠️ The 0/00/00/0 Trigger

🔑 Why Factoring Is Justified

📝 Easy First Example

Part 2: The Factoring Toolbox

🛠️ The Factoring Toolbox

Topics in This Part

◯² Difference of Squares

Part 3: Worked Examples — Workflow

📝 Worked Examples — The Factoring Workflow

Topics in This Part

🔄 The 4-Step Workflow

Part 4: Beyond Quadratics

📐 Beyond Quadratics — Higher-Degree Factoring

Topics in This Part

📐 Quadratic Trinomials

🔧 Common Factor Extraction (GCF)

🔁 Both Sides Factor

🎯 Limit = 0 vs. Limit = Nonzero

🧊 Sum and Difference of Cubes

🔗 Factoring by Grouping

➗ Synthetic Division to Extract (x−a)(x - a)(x−a)

🧮 Factoring Method — The $0/0$ Trigger

⚠️ The $0/0$ Trigger

➗ Synthetic Division to Extract $(x - a)$