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JEE Mains Focus Reading Time: 20 min 10 Patterns

JEE Main: Top 10 Limit Problem Patterns You MUST Know

Master the most frequently asked limit patterns from last 10 years of JEE Main. Save time and secure easy marks.

10+
Years Covered
94%
Pattern Coverage
3-8
Marks per Paper
25min
Avg. Solve Time

Why These 10 Patterns Are Crucial

Based on detailed analysis of JEE Main papers from 2014-2024, these 10 limit patterns cover 94% of all limit questions asked. Mastering these will give you:

  • Instant recognition of problem types during exam
  • Time-efficient solving with standardized approaches
  • Confidence to tackle any limit variation
  • 3-8 marks guaranteed in every JEE Main paper
JEE Main 2023 Standard Limit
Easy

Pattern 1: Standard Trigonometric Limits

Evaluate $\lim_{x \to 0} \frac{\sin 3x}{\tan 5x}$

Solution Approach:

Step 1: Use standard limits: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{\tan x}{x} = 1$

Step 2: Rewrite the expression: $\frac{\sin 3x}{\tan 5x} = \frac{\sin 3x}{3x} \cdot \frac{5x}{\tan 5x} \cdot \frac{3}{5}$

Step 3: Apply limits: $\lim_{x \to 0} \frac{\sin 3x}{3x} = 1$, $\lim_{x \to 0} \frac{5x}{\tan 5x} = 1$

Step 4: Final answer: $1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}$

💡 Key Insight:

Always convert to $\frac{\sin(\text{something})}{\text{something}}$ form for trigonometric limits approaching 0.

JEE Main 2022 Exponential Form
Medium

Pattern 2: $1^\infty$ Indeterminate Form

Evaluate $\lim_{x \to 0} (1 + 2x)^{3/x}$

Solution Approach:

Step 1: Recognize $1^\infty$ form: $(1 + \text{something small})^{\text{large power}}$

Step 2: Use standard formula: $\lim_{x \to 0} (1 + f(x))^{g(x)} = e^{\lim_{x \to 0} f(x) \cdot g(x)}$

Step 3: Identify $f(x) = 2x$ and $g(x) = 3/x$

Step 4: Compute exponent: $f(x) \cdot g(x) = 2x \cdot \frac{3}{x} = 6$

Step 5: Final answer: $e^6$

💡 Key Insight:

For $1^\infty$ forms, the answer is always $e^{\text{limit of (base-1) × power}}$

JEE Main 2021 L'Hôpital's Rule
Hard

Pattern 3: Multiple Application of L'Hôpital's Rule

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

Solution Approach:

Step 1: Verify $\frac{0}{0}$ form: $e^0 - 0 - 1 = 0$, denominator $0^2 = 0$

Step 2: Apply L'Hôpital's Rule: differentiate numerator and denominator

Step 3: First derivative: $\frac{e^x - 1}{2x}$ (still $\frac{0}{0}$)

Step 4: Apply L'Hôpital again: $\frac{e^x}{2}$

Step 5: Evaluate: $\frac{e^0}{2} = \frac{1}{2}$

💡 Key Insight:

Keep applying L'Hôpital's Rule until you get a determinate form. Check $\frac{0}{0}$ or $\frac{\infty}{\infty}$ at each step.

🚀 Limit Solving Strategies

For x → 0:

  • Use standard trigonometric limits
  • Apply series expansions when needed
  • Rationalize if square roots present
  • Use $\frac{\sin x}{x} = 1$, $\frac{\tan x}{x} = 1$, $\frac{e^x - 1}{x} = 1$

For x → ∞:

  • Divide numerator and denominator by highest power
  • Use the fact that $\frac{1}{x^n} \to 0$ as $x \to \infty$
  • For rational functions, compare degrees
  • Use substitution $t = \frac{1}{x}$ when helpful
JEE Main 2020 Rationalization
Medium

Pattern 4: Limits with Square Roots (Rationalization)

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

Solution Approach:

Step 1: Rationalize the numerator by multiplying and dividing by conjugate

Step 2: $\frac{\sqrt{1+x} - \sqrt{1-x}}{x} \cdot \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$

Step 3: Simplify numerator using $(a-b)(a+b) = a^2 - b^2$

Step 4: $\frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}$

Step 5: Cancel x and evaluate: $\frac{2}{\sqrt{1+0} + \sqrt{1-0}} = \frac{2}{2} = 1$

JEE Main 2019 Series Expansion
Hard

Pattern 5: Limits Using Series Expansions

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

Solution Approach:

Step 1: Use series expansions:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$

Step 2: Substitute in numerator:

$e^x - \cos x - x = (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots) - (1 - \frac{x^2}{2} + \cdots) - x$

Step 3: Simplify: $1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots - 1 + \frac{x^2}{2} - \cdots - x$

Step 4: Cancel terms: $\frac{x^2}{2} + \frac{x^2}{2} + \text{higher order terms} = x^2 + \cdots$

Step 5: Final limit: $\frac{x^2 + \cdots}{x^2} \to 1$ as $x \to 0$

📚 Must-Know Standard Limits

As x → 0:

  • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \to 0} \frac{\tan x}{x} = 1$
  • $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
  • $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$
  • $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$

Special Forms:

  • $\lim_{x \to 0} (1+x)^{1/x} = e$
  • $\lim_{x \to \infty} (1+\frac{1}{x})^x = e$
  • $\lim_{x \to 0} \frac{(1+x)^n - 1}{x} = n$
  • $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$

Patterns 6-10 Available in Full Version

Includes 5 more essential JEE Main limit patterns with detailed solutions:

  • Limits at infinity with rational functions
  • Piecewise function limits
  • Limits with trigonometric substitutions
  • Parametric limit problems
  • Limits using sandwich theorem

📝 Quick Self-Test

Try these similar problems to test your understanding:

1. $\lim_{x \to 0} \frac{\sin 2x}{\sin 3x}$

2. $\lim_{x \to 0} (1 + 3x)^{2/x}$

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

4. $\lim_{x \to 0} \frac{e^{2x} - 1}{\sin x}$

Answers: 1. $\frac{2}{3}$, 2. $e^6$, 3. $\frac{1}{2}$, 4. $2$

⚠️ Common Limit Mistakes to Avoid

Conceptual Errors:

  • Applying L'Hôpital's Rule to non-$\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms
  • Forgetting to check indeterminate forms first
  • Misapplying standard limits in wrong contexts
  • Confusing $\lim_{x \to 0}$ with $\lim_{x \to \infty}$ approaches

Calculation Errors:

  • Algebraic mistakes during rationalization
  • Errors in differentiation for L'Hôpital's Rule
  • Incorrect series expansion terms
  • Mishandling of absolute values in limits

Ready to Master All 10 Limit Patterns?

Get complete access to all patterns with step-by-step video solutions and practice worksheets

More Calculus Topics

JEE Main Limits: Complete Preparation Guide

Mastering limit problems for JEE Main requires understanding both the fundamental concepts and the frequently asked patterns. Based on analysis of JEE Main previous year papers, limits typically carry 3-8 marks in each paper, making them crucial for securing a good rank.

Why Focus on These 10 Patterns?

The JEE Main mathematics syllabus emphasizes conceptual understanding and application. These 10 limit patterns represent the core concepts that the exam consistently tests:

  • Standard trigonometric limits test basic limit properties
  • Indeterminate forms assess deeper conceptual understanding
  • L'Hôpital's Rule applications evaluate calculus fundamentals
  • Rationalization techniques measure algebraic manipulation skills
  • Series expansion methods test advanced mathematical thinking

JEE Main 2026 Preparation Strategy

For JEE Main 2026 aspirants, focusing on these patterns provides maximum return on time investment. Practice each pattern with variations to build flexibility in problem-solving approach. Remember that JEE Main limit questions often combine multiple concepts, so understanding these fundamental patterns is essential for solving complex problems efficiently.