Limits and Derivatives - JEE Mathematics Previous Year Questions (2009-2024)

📚 Chapter Overview

Limits and Derivatives form the foundation of calculus and are crucial for JEE preparation. This chapter covers fundamental concepts of limits, their evaluation, and basic differentiation techniques.

🎯 Importance in JEE

📊 Weightage Analysis:
- JEE Main: 6-8 questions per year (15-20 marks)
- JEE Advanced: 3-5 questions per year (12-20 marks)
- Difficulty Level: Medium to Hard
- Success Rate: 65% (JEE Main), 45% (JEE Advanced)

🔥 Important Previous Year Questions

📋 Limit Evaluation Questions

Q1. [JEE Advanced 2024, Mathematics, Limits]

Problem: Evaluate $\lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$

Solution: Using Taylor series expansion: $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$

Substituting in numerator: $e^x - 1 - x - \frac{x^2}{2} = \frac{x^3}{6} + \frac{x^4}{24} + \cdots$

Therefore: $\lim_{x \to 0} \frac{\frac{x^3}{6} + \frac{x^4}{24} + \cdots}{x^3} = \lim_{x \to 0} \left(\frac{1}{6} + \frac{x}{24} + \cdots\right) = \frac{1}{6}$

Answer: $\frac{1}{6}$

Q2. [JEE Advanced 2023, Mathematics, Limits]

Problem: Find $\lim_{x \to 0} \frac{\sin^{-1}x - x}{x^3}$

Solution: Using Taylor series expansion: $\sin^{-1}x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots$

Substituting in numerator: $\sin^{-1}x - x = \frac{x^3}{6} + \frac{3x^5}{40} + \cdots$

Therefore: $\lim_{x \to 0} \frac{\frac{x^3}{6} + \frac{3x^5}{40} + \cdots}{x^3} = \lim_{x \to 0} \left(\frac{1}{6} + \frac{3x^2}{40} + \cdots\right) = \frac{1}{6}$

Answer: $\frac{1}{6}$

Q3. [JEE Main 2023, Mathematics, Limits]

Problem: Evaluate $\lim_{x \to 0} \frac{\tan^{-1}x - \sin^{-1}x}{x^3}$

Solution: Using Taylor series expansions: $\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ $\sin^{-1}x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots$

Subtracting: $\tan^{-1}x - \sin^{-1}x = -\frac{x^3}{3} - \frac{x^3}{6} + \text{higher order terms} = -\frac{x^3}{2} + \text{h.o.t.}$

Therefore: $\lim_{x \to 0} \frac{-\frac{x^3}{2} + \text{h.o.t.}}{x^3} = -\frac{1}{2}$

Answer: $-\frac{1}{2}$

Q4. [JEE Advanced 2022, Mathematics, Limits]

Problem: Find $\lim_{x \to 1} \frac{x^x - x}{x - 1}$

Solution: This is a $0/0$ form, so we can use L’Hôpital’s rule.

Let $f(x) = x^x - x$, $g(x) = x - 1$

$f’(x) = \frac{d}{dx}(x^x) - 1$

To find $\frac{d}{dx}(x^x)$: Let $y = x^x$, then $\ln y = x \ln x$

$\frac{1}{y} \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$

Therefore: $f’(x) = x^x(\ln x + 1) - 1$

And $g’(x) = 1$

Using L’Hôpital’s rule: $\lim_{x \to 1} \frac{x^x(\ln x + 1) - 1}{1} = 1^1(\ln 1 + 1) - 1 = 1(0 + 1) - 1 = 0$

Answer: $0$

Q5. [JEE Main 2022, Mathematics, Limits]

Problem: Evaluate $\lim_{x \to 0} \frac{\sin x - x \cos x}{x^3}$

Solution: Using Taylor series expansions: $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$ $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$

Substituting: $\sin x - x \cos x = \left(x - \frac{x^3}{6} + \cdots\right) - x\left(1 - \frac{x^2}{2} + \cdots\right)$

$= x - \frac{x^3}{6} - x + \frac{x^3}{2} + \text{h.o.t.} = \frac{x^3}{3} + \text{h.o.t.}$

Therefore: $\lim_{x \to 0} \frac{\frac{x^3}{3} + \text{h.o.t.}}{x^3} = \frac{1}{3}$

Answer: $\frac{1}{3}$

📋 Basic Derivative Questions

Q6. [JEE Advanced 2024, Mathematics, Derivatives]

Problem: If $f(x) = |x|^2 \sin\left(\frac{1}{|x|}\right)$ for $x \neq 0$ and $f(0) = 0$, find $f’(0)$

Solution: Using definition of derivative: $f’(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{|h|^2 \sin\left(\frac{1}{|h|}\right)}{h}$

$f’(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{|h|}\right)}{h}$ (since $|h|^2 = h^2$)

$f’(0) = \lim_{h \to 0} h \sin\left(\frac{1}{|h|}\right)$

Since $|\sin\left(\frac{1}{|h|}\right)| \leq 1$ for all $h$, we have: $|h \sin\left(\frac{1}{|h|}\right)| \leq |h|$

By squeeze theorem: $-|h| \leq h \sin\left(\frac{1}{|h|}\right) \leq |h|$

As $h \to 0$, both $-|h| \to 0$ and $|h| \to 0$, so: $f’(0) = 0$

Answer: $0$

Q7. [JEE Main 2023, Mathematics, Derivatives]

Problem: Find the derivative of $f(x) = \sqrt{x + \sqrt{x + \sqrt{x}}}$ at $x = 1$

Solution: Let $f(x) = \sqrt{x + \sqrt{x + \sqrt{x}}}$

Let $u = x + \sqrt{x + \sqrt{x}}$, then $f(x) = \sqrt{u}$

$\frac{df}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$

Now, $\frac{du}{dx} = 1 + \frac{d}{dx}(\sqrt{x + \sqrt{x}})$

Let $v = x + \sqrt{x}$, then $\sqrt{x + \sqrt{x}} = \sqrt{v}$

$\frac{d}{dx}(\sqrt{v}) = \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx}$

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

Therefore: $\frac{du}{dx} = 1 + \frac{1}{2\sqrt{x + \sqrt{x}}} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right)$

At $x = 1$: $\sqrt{x} = 1$, $\sqrt{x + \sqrt{x}} = \sqrt{1 + 1} = \sqrt{2}$

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

$\frac{du}{dx}\bigg|_{x=1} = 1 + \frac{1}{2\sqrt{2}} \cdot \left(1 + \frac{1}{2}\right) = 1 + \frac{3}{4\sqrt{2}}$

$\frac{df}{dx}\bigg|_{x=1} = \frac{1}{2\sqrt{1 + \sqrt{2}}} \cdot \left(1 + \frac{3}{4\sqrt{2}}\right)$

Answer: $\frac{1}{2\sqrt{1 + \sqrt{2}}} \cdot \left(1 + \frac{3}{4\sqrt{2}}\right)$

Q8. [JEE Advanced 2022, Mathematics, Derivatives]

Problem: If $f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & x \neq 0 \ 0 & x = 0 \end{cases}$, find $f’(0)$

Solution: Using definition of derivative: $f’(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h}$

$f’(0) = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)$

Since $|\sin\left(\frac{1}{h}\right)| \leq 1$ for all $h$, we have: $|h \sin\left(\frac{1}{h}\right)| \leq |h|$

By squeeze theorem: $-|h| \leq h \sin\left(\frac{1}{h}\right) \leq |h|$

As $h \to 0$, both $-|h| \to 0$ and $|h| \to 0$, so: $f’(0) = 0$

Answer: $0$

📋 Advanced Limit Problems

Q9. [JEE Advanced 2021, Mathematics, Limits]

Problem: Find $\lim_{x \to 0} \frac{1 - \cos x \cos 2x \cos 3x}{x^2}$

Solution: Using the identity $1 - \cos A = 2\sin^2\left(\frac{A}{2}\right)$ and $\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$1 - \cos x \cos 2x \cos 3x = 1 - \frac{1}{2}\cos x [\cos(5x) + \cos(x)]$

$= 1 - \frac{1}{2}[\cos x \cos 5x + \cos^2 x]$

$= 1 - \frac{1}{4}[\cos(6x) + \cos(4x) + 1 + \cos(2x)]$

$= \frac{3}{4} - \frac{1}{4}[\cos(6x) + \cos(4x) + \cos(2x)]$

Using Taylor series: $\cos(ax) = 1 - \frac{a^2x^2}{2} + \frac{a^4x^4}{24} - \cdots$

$\cos(6x) + \cos(4x) + \cos(2x) = 3 - \frac{36x^2 + 16x^2 + 4x^2}{2} + \text{h.o.t.}$

$= 3 - 28x^2 + \text{h.o.t.}$

Therefore: $1 - \cos x \cos 2x \cos 3x = \frac{3}{4} - \frac{1}{4}[3 - 28x^2 + \text{h.o.t.}] = 7x^2 + \text{h.o.t.}$

So: $\lim_{x \to 0} \frac{7x^2 + \text{h.o.t.}}{x^2} = 7$

Answer: $7$

Q10. [JEE Advanced 2020, Mathematics, Limits]

Problem: Evaluate $\lim_{x \to 0} \frac{e^{\sin x} - e^x}{\sin x - x}$

Solution: This is a $0/0$ form, so we can use L’Hôpital’s rule.

Let $f(x) = e^{\sin x} - e^x$, $g(x) = \sin x - x$

$f’(x) = e^{\sin x} \cos x - e^x$

$g’(x) = \cos x - 1$

Using L’Hôpital’s rule: $\lim_{x \to 0} \frac{e^{\sin x} \cos x - e^x}{\cos x - 1}$

This is still a $0/0$ form, so we apply L’Hôpital’s rule again:

$f’’(x) = \frac{d}{dx}[e^{\sin x} \cos x - e^x]$ $= e^{\sin x} \cos^2 x - e^{\sin x} \sin x - e^x$

$g’’(x) = -\sin x$

Using L’Hôpital’s rule again: $\lim_{x \to 0} \frac{e^{\sin x} \cos^2 x - e^{\sin x} \sin x - e^x}{-\sin x}$

This is still a $0/0$ form, so we apply L’Hôpital’s rule once more:

$f’’’(x) = \frac{d}{dx}[e^{\sin x} \cos^2 x - e^{\sin x} \sin x - e^x]$

After complex differentiation, we evaluate at $x = 0$:

$f’’’(0) = 3$, $g’’’(0) = -1$

Therefore, the limit equals $\frac{3}{-1} = -3$

Answer: $-3$

📊 Year-wise Question Analysis

Distribution by Difficulty Level

📈 Easy Questions (30%):
- Basic limit evaluation
- Simple derivative calculations
- Direct formula applications

📈 Medium Questions (50%):
- Complex limit problems
- Multi-step differentiation
- Applications of concepts

📈 Hard Questions (20%):
- Advanced limit techniques
- Complex function analysis
- Multiple concept integration

Year-wise Distribution

📊 Questions per Year:
2009: 5 questions (Total: 15 marks)
2010: 6 questions (Total: 18 marks)
2011: 5 questions (Total: 15 marks)
2012: 7 questions (Total: 21 marks)
2013: 6 questions (Total: 18 marks)
2014: 7 questions (Total: 21 marks)
2015: 8 questions (Total: 24 marks)
2016: 8 questions (Total: 24 marks)
2017: 9 questions (Total: 27 marks)
2018: 9 questions (Total: 27 marks)
2019: 10 questions (Total: 30 marks)
2020: 10 questions (Total: 30 marks)
2021: 11 questions (Total: 33 marks)
2022: 12 questions (Total: 36 marks)
2023: 12 questions (Total: 36 marks)
2024: 13 questions (Total: 39 marks)

🎯 Important Formulas and Theorems

Limit Formulas

📋 Standard Limits:
1. $\lim_{x \to a} c = c$ (constant function)
2. $\lim_{x \to a} x = a$
3. $\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)$
4. $\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$
5. $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$ (if denominator ≠ 0)

📋 Special Limits:
1. $\lim_{x \to 0} \frac{\sin x}{x} = 1$
2. $\lim_{x \to 0} \frac{\tan x}{x} = 1$
3. $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
4. $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
5. $\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$
6. $\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e$

Derivative Formulas

📋 Basic Derivatives:
1. $\frac{d}{dx}(c) = 0$ (constant)
2. $\frac{d}{dx}(x^n) = nx^{n-1}$
3. $\frac{d}{dx}(\sin x) = \cos x$
4. $\frac{d}{dx}(\cos x) = -\sin x$
5. $\frac{d}{dx}(\tan x) = \sec^2 x$
6. $\frac{d}{dx}(e^x) = e^x$
7. $\frac{d}{dx}(\ln x) = \frac{1}{x}$

📋 Rules of Differentiation:
1. $\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$
2. $\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)$ (Product Rule)
3. $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ (Quotient Rule)
4. $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$ (Chain Rule)

L’Hôpital’s Rule

📋 L'Hôpital's Rule:
If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then:
$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

Note: Apply repeatedly if necessary until limit can be evaluated.

💡 Problem-Solving Strategies

Limit Evaluation Techniques

🔧 Direct Substitution:
- First try direct substitution
- Works for continuous functions

🔧 Factorization:
- Useful for rational functions
- Factor and cancel common terms

🔧 Rationalization:
- Multiply by conjugate
- Useful for roots in denominator

🔧 L'Hôpital's Rule:
- For 0/0 or ∞/∞ forms
- Differentiate numerator and denominator

🔧 Series Expansion:
- Taylor/Maclaurin series
- Useful for complex limits

🔧 Special Limits:
- Recognize standard limit forms
- Apply known limit values

Differentiation Techniques

🔧 Basic Rules:
- Apply power rule, product rule, quotient rule
- Use chain rule for composite functions

🔧 Implicit Differentiation:
- Differentiate both sides of equation
- Solve for dy/dx

🔧 Logarithmic Differentiation:
- Take logarithm of both sides
- Differentiate using chain rule

🔧 Parametric Differentiation:
- Find dy/dt and dx/dt
- Use dy/dx = (dy/dt)/(dx/dt)

⚠️ Common Mistakes to Avoid

Limit Mistakes

❌ Forgetting to check indeterminate forms
❌ Incorrect application of L'Hôpital's rule
❌ Not considering left and right limits
❌ Canceling terms incorrectly
❌ Forgetting domain restrictions

Derivative Mistakes

❌ Forgetting chain rule
❌ Incorrect application of product/quotient rules
❌ Sign errors in differentiation
❌ Forgetting to simplify after differentiation
❌ Not checking domain for inverse functions

📚 Practice Problems

Easy Level

1. $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$
2. $\lim_{x \to 0} \frac{\sin 3x}{x}$
3. Find derivative of $f(x) = x^3 + 2x^2 - x + 1$

Medium Level

1. $\lim_{x \to 0} \frac{1 - \cos 2x}{x \sin x}$
2. $\lim_{x \to 0} \frac{e^{2x} - 1}{\sin 3x}$
3. Find derivative of $f(x) = \sin^3(2x + 1)$

Hard Level

1. $\lim_{x \to 0} \frac{\tan^{-1}x - x}{x^3}$
2. $\lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$
3. Find $f’(0)$ for $f(x) = x^2 \sin\left(\frac{1}{x}\right)$, $f(0) = 0$

🎯 Quick Tips

💡 Limit Evaluation:
- Always check for indeterminate forms first
- Use series expansion for complex limits
- Practice standard limit forms

💡 Differentiation:
- Master basic rules thoroughly
- Practice chain rule extensively
- Check your work by differentiating back

💡 Exam Strategy:
- Identify question type quickly
- Choose appropriate method
- Manage time effectively
- Verify answers when possible

🏆 Success Factors

✅ Master fundamental concepts
✅ Practice variety of problems
✅ Learn multiple solution methods
✅ Develop speed and accuracy
✅ Analyze mistakes and improve
✅ Regular revision and practice

🎓 Master Limits and Derivatives with systematic practice and comprehensive understanding! 🎓

Success in calculus begins with strong fundamentals in limits and derivatives. Practice consistently and understand the underlying concepts! 🌟