Rationalizing to Evaluate Limits

Use conjugate multiplication to handle limits with radicals

The Rationalizing Technique

When you have square roots or other radicals causing 00\frac{0}{0}, multiply by the conjugate!

What's a Conjugate?

For an expression with a radical:

| Expression | Conjugate | |------------|-----------| | x+a\sqrt{x} + a | xa\sqrt{x} - a | | xa\sqrt{x} - a | x+a\sqrt{x} + a | | a+xa + \sqrt{x} | axa - \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)(ab)=a2b2(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 limx0x+42x\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}

Step 1: Try direct substitution 0+420=220=00\frac{\sqrt{0 + 4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0} ← Indeterminate!

Step 2: Multiply by the conjugate

The conjugate of x+42\sqrt{x + 4} - 2 is x+4+2\sqrt{x + 4} + 2

limx0x+42xx+4+2x+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)

(x+42)(x+4+2)=(x+4)222=(x+4)4=x(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2) = (\sqrt{x + 4})^2 - 2^2 = (x + 4) - 4 = x

Step 4: Rewrite

limx0xx(x+4+2)\lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)}

Step 5: Cancel x

limx01x+4+2\lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2}

Step 6: Direct substitution

=14+2=12+2=14= \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}

Answer: 14\frac{1}{4}

Example 2: Conjugate in Denominator

Find limh0525+h5\lim_{h \to 0} \frac{5}{\sqrt{25 + h} - 5}

Step 1: Check 5255=50\frac{5}{\sqrt{25} - 5} = \frac{5}{0} ← Undefined, but let's rationalize!

Step 2: Multiply by conjugate

limh0525+h525+h+525+h+5\lim_{h \to 0} \frac{5}{\sqrt{25 + h} - 5} \cdot \frac{\sqrt{25 + h} + 5}{\sqrt{25 + h} + 5}

Step 3: Simplify denominator

=limh05(25+h+5)(25+h)25= \lim_{h \to 0} \frac{5(\sqrt{25 + h} + 5)}{(25 + h) - 25}

=limh05(25+h+5)h= \lim_{h \to 0} \frac{5(\sqrt{25 + h} + 5)}{h}

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

As h0+h \to 0^+: numerator → 5(5+5)=505(5 + 5) = 50, denominator → 0+0^+

This limit approaches ++\infty!

When to Use This Technique

Use rationalizing when:

  • You see square roots or radicals
  • Direct substitution gives 00\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

(a+b)(ab)=ab(\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)(ab)=a2b2(a+b)(a-b) = a^2 - b^2
  5. Simplify and evaluate

📚 Practice Problems

1Problem 1medium

Question:

Evaluate limx9x9x3\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}

💡 Show Solution

Step 1: Try direct substitution

9993=00\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 x3\sqrt{x} - 3 is x+3\sqrt{x} + 3

limx9x9x3x+3x+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)

(x3)(x+3)=(x)232=x9(\sqrt{x} - 3)(\sqrt{x} + 3) = (\sqrt{x})^2 - 3^2 = x - 9

Step 4: Rewrite the expression

limx9(x9)(x+3)x9\lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9}

Step 5: Cancel (x9)(x - 9)

limx9(x+3)\lim_{x \to 9} (\sqrt{x} + 3)

Step 6: Direct substitution

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

Answer: 6

2Problem 2hard

Question:

Evaluate limx01+x1xx\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 1+x1x\sqrt{1 + x} - \sqrt{1 - x} is 1+x+1x\sqrt{1 + x} + \sqrt{1 - x}

limx01+x1xx1+x+1x1+x+1x\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

(1+x1x)(1+x+1x)(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x}) =(1+x)2(1x)2= (\sqrt{1+x})^2 - (\sqrt{1-x})^2 =(1+x)(1x)= (1 + x) - (1 - x) =1+x1+x=2x= 1 + x - 1 + x = 2x

Step 3: Rewrite

limx02xx(1+x+1x)\lim_{x \to 0} \frac{2x}{x(\sqrt{1 + x} + \sqrt{1 - x})}

Step 4: Cancel x

limx021+x+1x\lim_{x \to 0} \frac{2}{\sqrt{1 + x} + \sqrt{1 - x}}

Step 5: Direct substitution

=21+1=21+1=22=1= \frac{2}{\sqrt{1} + \sqrt{1}} = \frac{2}{1 + 1} = \frac{2}{2} = 1

Answer: 1