Rationalizing to Evaluate Limits

Use conjugate multiplication to handle limits with radicals

The Rationalizing Technique

When you have square roots or other radicals causing $\frac{0}{0}$ , multiply by the conjugate!

What's a Conjugate?

For an expression with a radical:

| Expression | Conjugate | |------------|-----------| | $\sqrt{x} + a$ | $\sqrt{x} - a$ | | $\sqrt{x} - a$ | $\sqrt{x} + a$ | | $a + \sqrt{x}$ | $a - \sqrt{x}$ |

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

Identify which part has the radical
Multiply by the conjugate over itself (= 1)
Expand using difference of squares
Simplify and cancel
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

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

📚 Practice Problems

1Problem 1medium

❓ Question:

Evaluate $\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}$

💡 Show Solution

Step 1: Try direct substitution

$\frac{9 - 9}{\sqrt{9} - 3} = \frac{0}{0}$

Indeterminate form - we need to rationalize!

Step 2: Multiply by the conjugate of the denominator

The conjugate of $\sqrt{x} - 3$ is $\sqrt{x} + 3$

$\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3}$

Step 3: Multiply denominator (difference of squares)

$(\sqrt{x} - 3)(\sqrt{x} + 3) = (\sqrt{x})^2 - 3^2 = x - 9$

Step 4: Rewrite the expression

$\lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9}$

Step 5: Cancel $(x - 9)$

$\lim_{x \to 9} (\sqrt{x} + 3)$

Step 6: Direct substitution

$= \sqrt{9} + 3 = 3 + 3 = 6$

Answer: 6

2Problem 2hard

❓ Question:

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

💡 Show Solution

This one has radicals in both terms! Let's rationalize.

Step 1: Multiply by conjugate

Conjugate of $\sqrt{1 + x} - \sqrt{1 - x}$ is $\sqrt{1 + x} + \sqrt{1 - x}$

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

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

3Problem 3easy

❓ Question:

Evaluate lim(x→0) (√(x + 1) - 1)/x

💡 Show Solution

Step 1: Check direct substitution: (√1 - 1)/0 = 0/0 (indeterminate)

Step 2: Multiply by conjugate: Conjugate of (√(x + 1) - 1) is (√(x + 1) + 1)

Step 3: Multiply numerator and denominator: [√(x + 1) - 1]/x · [√(x + 1) + 1]/[√(x + 1) + 1]

Step 4: Use difference of squares: Numerator: (√(x + 1))² - 1² = (x + 1) - 1 = x Denominator: x(√(x + 1) + 1)

Step 5: Simplify: x/[x(√(x + 1) + 1)] = 1/(√(x + 1) + 1) for x ≠ 0

Step 6: Evaluate: lim(x→0) 1/(√(x + 1) + 1) = 1/(√1 + 1) = 1/2

Answer: 1/2

4Problem 4medium

❓ Question:

Find lim(h→0) (√(4 + h) - 2)/h

💡 Show Solution

Step 1: Direct substitution gives 0/0

Step 2: Multiply by conjugate: [√(4 + h) - 2]/h · [√(4 + h) + 2]/[√(4 + h) + 2]

Step 3: Apply difference of squares to numerator: (√(4 + h))² - 2² = (4 + h) - 4 = h

Step 4: Rewrite: h/[h(√(4 + h) + 2)]

Step 5: Cancel h: 1/(√(4 + h) + 2) for h ≠ 0

Step 6: Evaluate: lim(h→0) 1/(√(4 + h) + 2) = 1/(2 + 2) = 1/4

Answer: 1/4

5Problem 5hard

❓ Question:

Evaluate lim(x→9) (x - 9)/(√x - 3)

💡 Show Solution

Step 1: Direct substitution: 0/0

Step 2: Rationalize by multiplying by conjugate: (x - 9)/(√x - 3) · (√x + 3)/(√x + 3)

Step 3: Expand numerator: (x - 9)(√x + 3)

Step 4: Expand denominator using difference of squares: (√x)² - 3² = x - 9

Step 5: Simplify: [(x - 9)(√x + 3)]/(x - 9) = √x + 3 for x ≠ 9

Step 6: Evaluate: lim(x→9) (√x + 3) = √9 + 3 = 3 + 3 = 6

Answer: 6

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