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Rationalizing to Evaluate Limits | Study Mondo
Topics / Limits & Continuity / Rationalizing to Evaluate Limits Rationalizing to Evaluate Limits Use conjugate multiplication to handle limits with radicals
BC Written and reviewed by Brendan Cusack , Study Mondo Education Team • Last updated April 28, 2026 🎯 ⭐ INTERACTIVE LESSON
Try the Interactive Version! Learn step-by-step with practice exercises built right in.
Start Interactive Lesson → The Rationalizing Technique
When you have square roots or other radicals causing
💡 Study Tips✓ Work through examples step-by-step ✓ Practice with flashcards daily ✓ Review common mistakes 0 0
multiply by the conjugate
What's a Conjugate? For an expression with a radical:
Expression Conjugate x + a \sqrt{x} + a x + a x − a \sqrt{x} - a x − a x − a \sqrt{x} - a x − a x + a \sqrt{x} + a a + x a + \sqrt{x} a + x a − x a - \sqrt{x} a −
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 (a + b)(a - b) = a^2 - b^2 ( a + b ) ( a − b ) = a 2 − b 2
This eliminates the radical!
The Process
Identify which part has the radicalMultiply by the conjugate over itself (= 1)Expand using difference of squaresSimplify and cancelEvaluate the limit
Example 1: Basic Rationalization Find lim x → 0 x + 4 − 2 x \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} lim x → 0 x x + 4 − 2
Step 1: Try direct substitution
0 + 4 − 2 0 = 2 − 2 0 = 0 0 \frac{\sqrt{0 + 4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0} 0 0 + 4 − 2 = 0 2 − 2 = 0 0
Step 2: Multiply by the conjugate
The conjugate of x + 4 − 2 \sqrt{x + 4} - 2 x + 4 − 2 x + 4 + 2 \sqrt{x + 4} + 2 x + 4 + 2
lim x → 0 x + 4 − 2 x ⋅ x + 4 + 2 x + 4 + 2 \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} lim x → 0 x x + 4 − 2 ⋅ x + 4 + 2 x + 4
Step 3: Multiply the numerator (difference of squares)
( x + 4 − 2 ) ( x + 4 + 2 ) = ( x + 4 ) 2 − 2 2 = ( x + 4 ) − 4 = x (\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2) = (\sqrt{x + 4})^2 - 2^2 = (x + 4) - 4 = x ( x + 4 − 2 ) ( x + 4 + 2 ) = ( x + 4 ) 2 − 2 2 = ( x + 4 ) − 4 = x
lim x → 0 x x ( x + 4 + 2 ) \lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)} lim x → 0 x ( x + 4 + 2 ) x
lim x → 0 1 x + 4 + 2 \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2} lim x → 0 x + 4 + 2 1
Step 6: Direct substitution
= 1 4 + 2 = 1 2 + 2 = 1 4 = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4} = 4 + 2 1 = 2 + 2 1 = 4 1
Example 2: Conjugate in Denominator Find lim h → 0 5 25 + h − 5 \lim_{h \to 0} \frac{5}{\sqrt{25 + h} - 5} lim h → 0 25 + h − 5 5
Step 1: Check
5 25 − 5 = 5 0 \frac{5}{\sqrt{25} - 5} = \frac{5}{0} 25 − 5 5 = 0 5
Step 2: Multiply by conjugate
lim h → 0 5 25 + h − 5 ⋅ 25 + h + 5 25 + h + 5 \lim_{h \to 0} \frac{5}{\sqrt{25 + h} - 5} \cdot \frac{\sqrt{25 + h} + 5}{\sqrt{25 + h} + 5} lim h → 0 25 + h − 5 5 ⋅ 25 + h + 5 25 + h
Step 3: Simplify denominator
= lim h → 0 5 ( 25 + h + 5 ) ( 25 + h ) − 25 = \lim_{h \to 0} \frac{5(\sqrt{25 + h} + 5)}{(25 + h) - 25} = lim h → 0 ( 25 + h ) − 25 5 ( 25 + h + 5 )
= lim h → 0 5 ( 25 + h + 5 ) h = \lim_{h \to 0} \frac{5(\sqrt{25 + h} + 5)}{h} = lim h → 0 h 5 ( 25 + h + 5 )
Wait, this doesn't help directly. Let's think about what happens:
As h → 0 + h \to 0^+ h → 0 + 5 ( 5 + 5 ) = 50 5(5 + 5) = 50 5 ( 5 + 5 ) = 50 0 + 0^+ 0 +
This limit approaches + ∞ +\infty + ∞
When to Use This Technique ✓ Use rationalizing when:
You see square roots or radicals
Direct substitution gives 0 0 \frac{0}{0} 0 0
The radical is in the numerator or denominator
No radicals present (use factoring instead)
The radical isn't causing the problem
Key Formula to Remember ( a + b ) ( a − b ) = a − b (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b ( a + b ) ( a − 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 (a+b)(a-b) = a^2 - b^2 ( a + b ) ( a − b ) = a 2 − b 2
Simplify and evaluate
📚 Practice Problems
1 Problem 1medium ❓ Question:Evaluate lim x → 9 x − 9 x − 3 \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} lim x → 9 x − 3 x − 9
💡 Show Solution Step 1: Try direct substitution
9 − 9 9 − 3 = 0 0 \frac{9 - 9}{\sqrt{9} - 3} = \frac{0}{0} 9
2 Problem 2hard ❓ Question:Evaluate lim x → 0 1 + x − 1 − x x \lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} lim x → 0 x
3 Problem 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
4 Problem 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
5 Problem 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
Explain using: 📝 Simple words 🔗 Analogy 🎨 Visual desc. 📐 Example 💡 Explain
📋 AP Calculus AB — Exam Format Guide⏱ 3 hours 15 minutes 📝 51 questions 📊 4 sections
Section Format Questions Time Weight Calculator Multiple Choice (No Calculator) MCQ 30 60 min 33.3% 🚫 Multiple Choice (Calculator) MCQ 15 45 min 16.7% ✅ Free Response (Calculator) FRQ 2 30 min 16.7% ✅ Free Response (No Calculator) FRQ 4 60 min 33.3% 🚫
💡 Key Test-Day Tips✓ Show all work on FRQs✓ Use proper notation✓ Check units✓ Manage your time⚠️ Common Mistakes: Rationalizing to Evaluate LimitsAvoid these 4 frequent errors
1 Forgetting the constant of integration (+C) on indefinite integrals
▾ 2 Confusing the Power Rule with the Chain Rule
▾ 3 Not checking continuity before applying the Mean Value Theorem
▾ 4 Dropping negative signs when differentiating trig functions
▾ 🌍 Real-World Applications: Rationalizing to Evaluate LimitsSee how this math is used in the real world
⚙️ Optimizing Package Design
Engineering
▾ 🏥 Predicting Drug Dosage Decay
Medicine
▾ 🔬 Calculating Distance from Velocity
Physics
▾ 💰 Revenue Optimization
Finance
▾
📝 Worked Example: Related Rates — Expanding CircleProblem: A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 2 2 10 10 10
1 Identify the known and unknown rates Click to reveal →
2 Write the relationship between variables
3 Differentiate both sides with respect to time
🧪 Practice Lab Interactive practice problems for Rationalizing to Evaluate Limits
▾ 📌 Related Topics in Limits & Continuity❓ Frequently Asked QuestionsWhat is Rationalizing to Evaluate Limits?▾ Use conjugate multiplication to handle limits with radicals
How can I study Rationalizing to Evaluate Limits effectively?▾ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Rationalizing to Evaluate Limits?▾ Rationalizing to Evaluate Limits is part of the AP Calculus AB course on Study Mondo, specifically in the Limits & Continuity section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Rationalizing to Evaluate Limits?▾ Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
x
+
a
+
2
+
5
−
3
9 − 9
=
0 0
Indeterminate form - we need to rationalize!
Step 2: Multiply by the conjugate of the denominator
The conjugate of x − 3 \sqrt{x} - 3 x − 3 x + 3 \sqrt{x} + 3 x + 3
lim x → 9 x − 9 x − 3 ⋅ x + 3 x + 3 \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} lim x → 9 x − 3 x − 9 ⋅ x + 3 x
Step 3: Multiply denominator (difference of squares)
( x − 3 ) ( x + 3 ) = ( x ) 2 − 3 2 = x − 9 (\sqrt{x} - 3)(\sqrt{x} + 3) = (\sqrt{x})^2 - 3^2 = x - 9 ( x − 3 ) ( x + 3 ) = ( x ) 2 − 3 2 = x − 9
Step 4: Rewrite the expression
lim x → 9 ( x − 9 ) ( x + 3 ) x − 9 \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} lim x → 9 x − 9 ( x − 9 ) ( x + 3 )
Step 5: Cancel ( x − 9 ) (x - 9) ( x − 9 )
lim x → 9 ( x + 3 ) \lim_{x \to 9} (\sqrt{x} + 3) lim x → 9 ( x + 3 )
Step 6: Direct substitution
= 9 + 3 = 3 + 3 = 6 = \sqrt{9} + 3 = 3 + 3 = 6 = 9 + 3 = 3 + 3 = 6
💡 Show Solution This one has radicals in both terms! Let's rationalize.
Step 1: Multiply by conjugate
Conjugate of 1 + x − 1 − x \sqrt{1 + x} - \sqrt{1 - x} 1 + x − 1 − x 1 + x + 1 − x \sqrt{1 + x} + \sqrt{1 - x} 1 + x + 1 − x
lim x → 0 1 + x − 1 − x x ⋅ 1 + x + 1 − x 1 + x + 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}} lim x → 0 x
Step 2: Multiply numerator
( 1 + x − 1 − x ) ( 1 + x + 1 − x ) (\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x}) ( 1 + x −
Step 3: Rewrite
lim x → 0 2 x x ( 1 + x + 1 − x ) \lim_{x \to 0} \frac{2x}{x(\sqrt{1 + x} + \sqrt{1 - x})} lim x → 0 x ( 1 + x
Step 4: Cancel x
lim x → 0 2 1 + x + 1 − x \lim_{x \to 0} \frac{2}{\sqrt{1 + x} + \sqrt{1 - x}} lim x → 0 1 + x
Step 5: Direct substitution
= 2 1 + 1 = 2 1 + 1 = 2 2 = 1 = \frac{2}{\sqrt{1} + \sqrt{1}} = \frac{2}{1 + 1} = \frac{2}{2} = 1 = 1 +
Answer: 1
+
3
⋅
1 − x
)
(
+
= ( 1 + x ) 2 − ( 1 − x ) 2 = (\sqrt{1+x})^2 - (\sqrt{1-x})^2 = ( 1 + x ) 2 − ( 1 − x ) 2 = ( 1 + x ) − ( 1 − x ) = (1 + x) - (1 - x) = ( 1 + x ) − ( 1 − x ) = 1 + x − 1 + x = 2 x = 1 + x - 1 + x = 2x = 1 + x − 1 + x = 2 x
+
)
2 x
+
2
1
2
=
1 + 1 2 =
2 2 =
1