🎯⭐ INTERACTIVE LESSON

Rationalizing to Evaluate Limits

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Rationalizing to Evaluate Limits - Complete Interactive Lesson

Part 1: When to Use the Conjugate

🔁 Rationalizing — When to Reach for the Conjugate

Part 1 of 4 — The signal: a $\sqrt{}$ in a $0/0$ limit

Topics in This Part

Section
🔑 The Trigger: $\sqrt{}$ + $0/0$
What Is a Conjugate?
Why Multiplying by 1 Helps

🔑 Why this matters: Factoring fails when a square root creates the $0/0$ . Conjugates clear the radical and unlock cancellation.

⚠️ The Trigger

Use rationalizing when:

Direct substitution gives $0/0$ , AND
The expression contains a square root (or sometimes higher root) that's causing the zero.

Classic shape: $\lim_{x \to a} \frac{\sqrt{f(x)} - c}{x - a} \quad \text{or} \quad \lim_{x \to a} \frac{x - a}{\sqrt{f(x)} - c}$

🔄 What Is a Conjugate?

The conjugate of $a + b$ is $a - b$ , and vice versa. The key identity:

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

✖️ Multiplying by 1 (in disguise)

The trick: multiply by $\dfrac{\text{conjugate}}{\text{conjugate}}$ . This equals 1, so the value of the expression is unchanged — but the form is transformed.

$\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{x - 4}{(x - 4) (\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}$

Conjugate Concept 🎯

Build the Conjugate 🧮

Write down what each multiplication produces (use the difference-of-squares identity):

1) $(\sqrt{x} - 5)(\sqrt{x} + 5) = ?$ (answer in form like )

Part 2: Multiplying by the Conjugate

🛠️ The Rationalizing Workflow (Numerator Has the Root)

Part 2 of 4 — Step-by-step

Topics in This Part

Section
🔑 The 5-Step Workflow
Worked Example #1
Common Setup

🔑 Why this matters: Most AP rationalizing problems have the radical in the numerator. Master this case first.

🔄 The 5-Step Workflow

For $\lim_{x \to a} \dfrac{\sqrt{\,\text{stuff}\,} - c}{(x - a)}$ :

Part 3: Worked Examples

📝 More Worked Examples

Part 3 of 4 — Variations on the rationalizing theme

Topics in This Part

Section
Shifted Argument: $\sqrt{x + h} - \sqrt{x}$ Style

Part 4: Conjugates in the Denominator

🔃 Conjugates in the Denominator (and Mixed Cases)

Part 4 of 4 — Less common but still tested

Topics in This Part

Section
Conjugate in the Denominator
Both Sides Have a Radical
Pitfalls

🔑 Why this matters: When the radical is on the bottom, you still rationalize — but be careful which factor you actually need to expand.

⬇️ Radical in the Denominator

Find $\lim_{x \to 0} \dfrac{x}{\sqrt{x + 1} - 1}$ .

💡 If you only have polynomials and you get $0/0$ , factor instead. Rationalize when a $\sqrt{}$ is in the way.

This eliminates radicals because $(\sqrt{x})^2 = x$ .

Expression	Conjugate
$\sqrt{x} - 2$	$\sqrt{x} + 2$
$\sqrt{x + 4} - 3$	$\sqrt{x + 4} + 3$
$1 + \sqrt{x}$	$1 - \sqrt{x}$

\frac{x - 2}{x - 4} \cdot \frac{x + 2}{x + 2} = \frac{x - 4}{( x - 4 ) ( x + 2 )} = \frac{1}{x + 2}

Now you can substitute $x = 4$ : $\dfrac{1}{2 + 2} = \dfrac{1}{4}$ .

💡 The $(x - 4)$ that was hidden inside the radical comes out and cancels with the denominator.

2) $(\sqrt{x + 7} - 3)(\sqrt{x + 7} + 3) = ?$ (simplify completely; e.g. x-2)

Substitute to confirm $0/0$ .
Multiply by 1: $\dfrac{\text{conjugate}}{\text{conjugate}}$ where the conjugate flips the sign of the $\sqrt{}$ piece.
Simplify the numerator using $(a-b)(a+b) = a^2 - b^2$ — the radical disappears.
Cancel the common factor $(x - a)$ .
Substitute $x = a$ into the simplified expression.

📝 Worked Example #1

Find $\lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9}$ .

Step 1. Substitute: $\dfrac{0}{0}$ ✅ Step 2. Multiply by $\dfrac{\sqrt{x} + 3}{\sqrt{x} + 3}$ . Numerator becomes . Cancel : Substitute : .

$\boxed{\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} = \frac{1}{6}}$

🎯 The General Pattern

For $\lim_{x \to a} \dfrac{\sqrt{x} - \sqrt{a}}{x - a} = \dfrac{1}{2\sqrt{a}}$ .

This actually IS the derivative of $\sqrt{x}$ at $x = a$ — you'll meet this idea soon.

💡 If you start seeing the same answer pattern $\dfrac{1}{2\sqrt{a}}$ , that's why.

Workflow Check 🎯

Apply the Workflow 🧮

1) $\lim_{x \to 25} \dfrac{\sqrt{x} - 5}{x - 25} = ?$ (decimal or fraction)

2) $\lim_{x \to 1} \dfrac{\sqrt{x} - 1}{x - 1} = ?$ (decimal or fraction)

🔑 Why this matters: Real AP problems vary the form. Recognizing structure is half the battle.

📐 Shifted Argument

Find $\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x}$ .

Substitute: $\dfrac{0}{0}$ ✅
Multiply by $\dfrac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}$ :

🟰 Two Radicals

Find $\lim_{x \to 3} \dfrac{\sqrt{x + 1} - \sqrt{4}}{x - 3} = \lim_{x \to 3} \dfrac{\sqrt{x + 1} - 2}{x - 3}$ .

Substitute: $0/0$ ✅
Conjugate $\sqrt{x + 1} + 2$ :

🔍 Hidden Roots

Sometimes the radical is hiding inside an expression.

Example. $\lim_{x \to 0} \dfrac{\sqrt{1 + x} - 1}{x}$ .

$0/0$ ✅
Conjugate $\sqrt{1 + x} + 1$ :

💡 The pattern $\dfrac{\sqrt{1 + x} - 1}{x} \to \dfrac{1}{2}$ is famous — it's the derivative of at .

Pattern Recognition 🎯

Mixed Practice 🧮

1) $\lim_{x \to 0} \dfrac{\sqrt{x + 16} - 4}{x} = ?$ (fraction)

2) $\lim_{x \to 7} \dfrac{\sqrt{x + 2} - 3}{x - 7} = ?$ (fraction)

Substitute: $0/0$ ✅
Multiply by $\dfrac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1}$ : $\dfrac{x(\sqrt{x + 1} + 1)}{(x + 1) - 1} = \dfrac{x(\sqrt{x + 1} + 1)}{x} = \sqrt{x + 1} + 1$
Substitute $x = 0$ : $1 + 1 = 2$ .

💡 When the radical is on the bottom, the expanded numerator is messier — but it's the denominator that simplifies and lets you cancel the offending $x$ .

⬆️⬇️ Both Sides Have Roots

Sometimes you need to rationalize both top and bottom.

Example. $\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{\sqrt{x + 1} - 1}$ .

$0/0$ ✅
Multiply by $\dfrac{(\sqrt{x + 4} + 2)(\sqrt{x + 1} + 1)}{(\sqrt{x + 4} + 2)(\sqrt{x + 1} + 1)}$ :

⚠️ Pitfalls

Pitfall	Fix
Distributing the conjugate prematurely. Leave the un-rationalized side as a single product so it cancels cleanly.	Only multiply out the side you're rationalizing.
Forgetting "in disguise". You must multiply by $\dfrac{\text{conj}}{\text{conj}}$ , not just by the conjugate.	Always pair top and bottom — that's what makes it $= 1$ .
Sign errors in the conjugate.	The conjugate flips the sign between the two terms only, not on each term.

Apply the Idea 🎯

\dfrac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)}

\dfrac{x - 9}{(x - 9)(\sqrt{x} + 3)}

\dfrac{1}{\sqrt{x} + 3}

\dfrac{1}{3 + 3} = \dfrac{1}{6}

\dfrac{(x + 4) - 4}{x(\sqrt{x + 4} + 2)} = \dfrac{x}{x(\sqrt{x + 4} + 2)} = \dfrac{1}{\sqrt{x + 4} + 2}

Substitute

x = 0

\dfrac{1}{2 + 2} = \dfrac{1}{4}

\frac{( x + 1 ) - 4}{( x - 3 ) ( x + 1 + 2 )} = \frac{x - 3}{( x - 3 ) ( x + 1 + 2 )} \frac{1}{x + 1 + 2}

Substitute

x = 3

\dfrac{1}{2 + 2} = \dfrac{1}{4}

\frac{1 + x - 1}{x ( 1 + x + 1 )} = \frac{x}{x ( 1 + x + 1 )} = \frac{1}{1 + x + 1}

Substitute

x = 0

\dfrac{1}{1 + 1} = \dfrac{1}{2}

\dfrac{(x + 4 - 4)(\sqrt{x + 1} + 1)}{(x + 1 - 1)(\sqrt{x + 4} + 2)} = \dfrac{x(\sqrt{x + 1} + 1)}{x(\sqrt{x + 4} + 2)} = \dfrac{\sqrt{x + 1} + 1}{\sqrt{x + 4} + 2}

Substitute

x = 0

\dfrac{1 + 1}{2 + 2} = \dfrac{1}{2}