Factoring Method for Limits

🧮 Factoring Method — The $0/0$ Trigger

Part 1 of 4 — Why factoring resolves indeterminate forms

Topics in This Part

🔑 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 $0/0$. This means:

• $p(a) = 0$ → 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."

🔑 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 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.

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

📝 Easy First Example

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

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: $\lim_{x \to 2} (x + 2) = 4$.

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

Concept Check 🎯

Factor and Limit 🧮

Compute by factoring:

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

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

🛠️ The Factoring Toolbox

Part 2 of 4 — Common patterns to recognize

Topics in This Part

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

◯${}^{2}$ Difference of Squares

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

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

Examples:

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

📐 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: $x^2 - 5x + 6 = (x - 2)(x - 3)$.
• Cancel: $\dfrac{(x-2)(x-3)}{x-2} \to x - 3$.
• Substitute: $2 - 3 = -1$. Answer: $-1$.

💡 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: $x^3 - 4x = x(x^2 - 4)$.
• Cancel: $\dfrac{x(x^2 - 4)}{x} = x^2 - 4$ for $x \ne 0$.
• Substitute: $0 - 4 = -4$. Answer: $-4$.

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

📝 Worked Examples — The Factoring Workflow

Part 3 of 4 — Practice the full pipeline

Topics in This Part

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

1. Try direct substitution. If you get a clean number → done.
2. If you get $0/0$, fully factor both $p(x)$ and $q(x)$.
3. Cancel the common $(x - a)$ factor (or factors).
4. 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: $x^2 - 9 = (x - 3)(x + 3)$.
• 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: $\dfrac{6}{1} = 6$.

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

🎯 Limit = 0 vs. Limit = Nonzero

After cancellation and substitution:

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

📐 Beyond Quadratics — Higher-Degree Factoring

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

Topics in This Part

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

🧊 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 $(x + 1)$: $(x - 2)(x + 2) = x^2 - 4 \to (-1)^2 - 4 = -3$. Answer: $-3$.

➗ 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 $1 \;|\; 1 \; -6 \; 11 \; -6$:

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

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

Higher-Degree Practice 🎯

Exit Check ✅