Rationalizing to Evaluate Limits

The conjugate has the same terms but the opposite sign in the middle.

Why It Works

When you multiply conjugates, you get a difference of squares:

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

This eliminates the radical!

The Process

1. Identify which part has the radical
2. Multiply by the conjugate over itself (= 1)
3. Expand using difference of squares
4. Simplify and cancel
5. Evaluate the limit

Example 1: Basic Rationalization

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

Step 1: Try direct substitution $\frac{\sqrt{0 + 4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}$ ← Indeterminate!

Step 2: Multiply by the conjugate

The conjugate of $\sqrt{x + 4} - 2$ is $\sqrt{x + 4} + 2$

$\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}$

Step 3: Multiply the numerator (difference of squares)

$(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2) = (\sqrt{x + 4})^2 - 2^2 = (x + 4) - 4 = x$

Step 4: Rewrite

$\lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)}$

Step 5: Cancel x

$\lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2}$

Step 6: Direct substitution

$= \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$

Answer: $\frac{1}{4}$

Example 2: Conjugate in Denominator

Find $\lim_{h \to 0} \frac{5}{\sqrt{25 + h} - 5}$

Step 1: Check $\frac{5}{\sqrt{25} - 5} = \frac{5}{0}$ ← Undefined, but let's rationalize!

Step 2: Multiply by conjugate

$\lim_{h \to 0} \frac{5}{\sqrt{25 + h} - 5} \cdot \frac{\sqrt{25 + h} + 5}{\sqrt{25 + h} + 5}$

Step 3: Simplify denominator

$= \lim_{h \to 0} \frac{5(\sqrt{25 + h} + 5)}{(25 + h) - 25}$

$= \lim_{h \to 0} \frac{5(\sqrt{25 + h} + 5)}{h}$

Wait, this doesn't help directly. Let's think about what happens:

As $h \to 0^+$: numerator → $5(5 + 5) = 50$, denominator → $0^+$

This limit approaches $+\infty$!

When to Use This Technique

✓ Use rationalizing when:

• You see square roots or radicals
• Direct substitution gives $\frac{0}{0}$
• The radical is in the numerator or denominator

✗ Don't use it when:

• No radicals present (use factoring instead)
• The radical isn't causing the problem

Key Formula to Remember

$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$

This eliminates both radicals at once!

Practice Strategy

1. Spot the radical
2. Write down its conjugate
3. Multiply top and bottom
4. Use $(a+b)(a-b) = a^2 - b^2$
5. Simplify and evaluate

$\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} \cdot \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}$

Step 2: Multiply numerator

$(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})$ $= (\sqrt{1+x})^2 - (\sqrt{1-x})^2$ $= (1 + x) - (1 - x)$ $= 1 + x - 1 + x = 2x$

Step 3: Rewrite

$\lim_{x \to 0} \frac{2x}{x(\sqrt{1 + x} + \sqrt{1 - x})}$

Step 4: Cancel x

$\lim_{x \to 0} \frac{2}{\sqrt{1 + x} + \sqrt{1 - x}}$

Step 5: Direct substitution

$= \frac{2}{\sqrt{1} + \sqrt{1}} = \frac{2}{1 + 1} = \frac{2}{2} = 1$

Answer: 1