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L'Hôpital's Rule: The Ultimate Guide for JEE Main & Advanced

Master when to apply L'Hôpital's Rule, understand all conditions, and learn multiple applications with JEE-level solved examples.

7
Indeterminate Forms
100%
JEE Relevance
15+
Solved Examples
35min
Practice Time

Why L'Hôpital's Rule is Crucial for JEE

L'Hôpital's Rule appears in 2-3 questions every JEE Main paper and is essential for solving complex limit problems in JEE Advanced. This powerful tool helps evaluate limits that yield indeterminate forms.

  • Saves time in solving complex limit problems
  • Essential for $0/0$ and $\infty/\infty$ forms
  • Multiple applications possible for stubborn limits
  • 4-6 marks secured in every JEE mathematics paper
Fundamental Concept Easy

What is L'Hôpital's Rule?

If $\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0$ or $\pm\infty$, and $\lim_{x \to c} \frac{f'(x)}{g'(x)}$ exists, then:

$$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$

Key Conditions:

  • Must get $0/0$ or $\infty/\infty$ form initially
  • $f(x)$ and $g(x)$ must be differentiable near $c$
  • $g'(x) \neq 0$ near $c$ (except possibly at $c$)
  • The limit of $f'(x)/g'(x)$ must exist or be $\pm\infty$
JEE Main 2023 Medium

7 Indeterminate Forms

L'Hôpital's Rule directly applies to $0/0$ and $\infty/\infty$, but other forms can be converted:

$0/0$
Direct Apply
$\infty/\infty$
Direct Apply
$0 \cdot \infty$
Rewrite
$\infty - \infty$
Combine
$0^0$
Log Method
$1^\infty$
Log Method
$\infty^0$
Log Method

Example: $0 \cdot \infty$ Form

Evaluate $\lim_{x \to 0^+} x \ln x$

Step 1: Identify form: $0 \cdot (-\infty)$

Step 2: Rewrite: $\lim_{x \to 0^+} \frac{\ln x}{1/x}$

Step 3: Now we have $\infty/\infty$ form

Step 4: Apply L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$

JEE Advanced 2022 Hard

Multiple Applications of L'Hôpital's Rule

Sometimes one application isn't enough. We may need to apply the rule multiple times until we get a determinate form.

Example: Multiple Applications

Evaluate $\lim_{x \to 0} \frac{e^x - x - 1}{x^2}$

Step 1: Check form: $\frac{0}{0}$ ✓

Step 2: First application: $\lim_{x \to 0} \frac{e^x - 1}{2x}$

Step 3: Still $\frac{0}{0}$ form

Step 4: Second application: $\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$

Step 5: Final answer: $\frac{1}{2}$

⚠️ Common Mistake:

Applying L'Hôpital's Rule when the form is not indeterminate. Always verify the form is $0/0$ or $\infty/\infty$ before applying.

🚀 L'Hôpital's Rule: Quick Decision Guide

When to USE L'Hôpital's Rule:

  • Direct substitution gives $0/0$ or $\infty/\infty$
  • Algebraic simplification seems difficult
  • Trigonometric limits with $0/0$ form
  • Exponential and logarithmic limits

When to AVOID L'Hôpital's Rule:

  • Form is not indeterminate
  • Simple algebraic factorization possible
  • Standard limits can be applied directly
  • Derivatives become too complicated

Advanced Concepts Available in Full Version

Includes 5 more essential L'Hôpital's Rule concepts with JEE Advanced level problems:

  • • L'Hôpital's Rule for $\infty - \infty$ forms
  • • Exponential indeterminate forms ($1^\infty$, $0^0$, $\infty^0$)
  • • Sequential applications with tricky derivatives
  • • When L'Hôpital's Rule fails to give an answer
  • • Comparison with other limit evaluation methods

📝 Quick Self-Test

Try these JEE-level problems to test your L'Hôpital's Rule skills:

1. $\lim_{x \to 0} \frac{\sin x - x}{x^3}$

2. $\lim_{x \to \infty} \frac{\ln x}{x}$

3. $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}$

4. $\lim_{x \to 0^+} x^x$ (Hint: Use logarithmic transformation)

📋 L'Hôpital's Rule: Quick Reference

Indeterminate Form Transformation Then Apply L'Hôpital
$0 \cdot \infty$ Rewrite as $\frac{0}{1/\infty}$ or $\frac{\infty}{1/0}$ Yes
$\infty - \infty$ Combine using common denominator Yes
$0^0$, $1^\infty$, $\infty^0$ Take natural log: $y = f(x)^{g(x)} \Rightarrow \ln y = g(x)\ln f(x)$ Yes, to find $\lim \ln y$

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