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Limits & Continuity Reading Time: 15 min 6 Techniques

Rationalization Method: Conjugate Pairs for Limit Problems

Master the art of handling irrational functions and $\sqrt{a} - \sqrt{b}$ forms in limit problems using conjugate pairs.

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Why Rationalization is Crucial for JEE

When evaluating limits involving irrational expressions, we often encounter indeterminate forms like $\frac{0}{0}$. Rationalization using conjugate pairs is the most powerful technique to resolve these forms and find the actual limit value.

🎯 JEE Exam Pattern Insight

  • Rationalization problems appear in every JEE Main paper
  • Typically worth 3-4 marks in various forms
  • Essential for solving continuity and differentiability problems
  • Forms the basis for understanding derivatives from first principles

🧭 Quick Navigation

Concept Basic Examples Advanced Patterns Trig Forms Practice JEE Tips
Core Concept Fundamental

What is Rationalization Using Conjugate Pairs?

The conjugate of an expression $a + \sqrt{b}$ is $a - \sqrt{b}$, and vice versa.

$$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$$

This identity eliminates the irrational part from the denominator or numerator.

💡 When to Use Rationalization

  • Limits of the form $\frac{0}{0}$ with square roots
  • Expressions like $\sqrt{a} - \sqrt{b}$ where direct substitution fails
  • Functions with irrational components in denominator
  • Problems involving difference of square roots
Example 1 Easy

Basic Rationalization: $\frac{\sqrt{a} - \sqrt{b}}{x-c}$ Form

Evaluate: $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$

Step 1: Identify the Problem

Direct substitution gives $\frac{0}{0}$ (indeterminate form). We need to rationalize.

Step 2: Multiply by Conjugate

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

$$ \frac{\sqrt{x} - 2}{x - 4} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2} $$

Step 3: Simplify

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

Step 4: Cancel and Evaluate

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

Now substitute $x = 4$:

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

✅ Final Answer

$\frac{1}{4}$

Example 2 Medium

Advanced Pattern: Multiple Square Roots

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

Step 1: Identify Conjugate

The numerator $\sqrt{1+x} - \sqrt{1-x}$ has conjugate $\sqrt{1+x} + \sqrt{1-x}$

Step 2: Multiply and Simplify

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

Step 3: Cancel and Evaluate

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

Substitute $x = 0$:

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

✅ Final Answer

$1$

🎯 JEE Shortcut

For $\lim_{x \to 0} \frac{\sqrt{a+bx} - \sqrt{a+cx}}{x}$, the answer is always:

$$ \frac{b - c}{2\sqrt{a}} $$

Memorize this pattern for quick solving!

Example 3 Hard

Trigonometric + Irrational Combination

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

Step 1: Standard Rationalization

$$ \frac{\sqrt{1+\sin x} - \sqrt{1-\sin x}}{x} \times \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} + \sqrt{1-\sin x}} $$

Step 2: Simplify Numerator

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

Step 3: Separate Known Limit

$$ = \frac{2}{(\sqrt{1+\sin x} + \sqrt{1-\sin x})} \times \frac{\sin x}{x} $$

Step 4: Evaluate

Using $\lim_{x \to 0} \frac{\sin x}{x} = 1$:

$$ = \frac{2}{(\sqrt{1+0} + \sqrt{1-0})} \times 1 = \frac{2}{1 + 1} = 1 $$

✅ Final Answer

$1$

📊 Common Rationalization Patterns in JEE

Pattern Conjugate Used Result After Simplification
$\frac{\sqrt{x+a} - \sqrt{b}}{x-c}$ $\sqrt{x+a} + \sqrt{b}$ $\frac{x+a-b}{(x-c)(\sqrt{x+a}+\sqrt{b})}$
$\frac{\sqrt{a} - \sqrt{x}}{a-x}$ $\sqrt{a} + \sqrt{x}$ $\frac{a-x}{(a-x)(\sqrt{a}+\sqrt{x})} = \frac{1}{\sqrt{a}+\sqrt{x}}$
$\frac{\sqrt{x+h} - \sqrt{x}}{h}$ $\sqrt{x+h} + \sqrt{x}$ $\frac{1}{\sqrt{x+h} + \sqrt{x}}$
$\frac{\sqrt{ax+b} - \sqrt{cx+d}}{x}$ $\sqrt{ax+b} + \sqrt{cx+d}$ $\frac{a-c}{\sqrt{ax+b} + \sqrt{cx+d}}$

🎯 Practice Problems

Try these JEE-level problems to test your understanding:

1. $\lim_{x \to 1} \frac{\sqrt{2x} - \sqrt{3-x}}{x-1}$

Hint: Use conjugate of numerator

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

Hint: Rationalize and use binomial approximation

3. $\lim_{x \to 3} \frac{\sqrt{x+13} - 2\sqrt{x+1}}{x^2-9}$

Hint: Double rationalization needed

💡 Pro Tip

Always check if direct substitution gives $\frac{0}{0}$ before starting rationalization. This confirms you're on the right track!

🚀 JEE Exam Strategy

Time-Saving Techniques

  • Memorize common patterns and their results
  • Use binomial approximation for quick verification
  • Practice mental conjugation to speed up the process
  • Learn to identify when double rationalization is needed

Common Pitfalls to Avoid

  • Forgetting to multiply both numerator and denominator
  • Mishandling negative signs in conjugate pairs
  • Not simplifying completely before substitution
  • Overlooking domain restrictions after simplification

📝 60-Second Revision

Key Formulas

  • $(a+b)(a-b) = a^2 - b^2$
  • $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = a-b$
  • $\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x} = \frac{1}{2}$

When to Rationalize

  • $\frac{0}{0}$ forms with square roots
  • Difference of square roots
  • Irrational denominators
  • Before applying L'Hôpital's rule

Mastered Rationalization?

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