JEE Advanced: The Toughest Limit Problems Solved Step-by-Step
Master multi-concept limit problems with advanced techniques used in actual JEE Advanced papers.
Why These Limit Problems Are Crucial for JEE Advanced
Based on analysis of JEE Advanced papers from 2015-2024, these 8 problem types represent the most challenging limit questions that test multiple concepts simultaneously. Mastering these will give you:
- Multi-concept integration - combining limits with series, integrals, and inequalities
- Advanced techniques like Stolz-Cesàro, Taylor series, and Riemann sums
- Problem identification skills to quickly recognize which approach to use
- 4-8 marks secured in every JEE Advanced paper
Problem 1: Nested Radical Limit with Series
Evaluate: $$\lim_{n \to \infty} \left( \frac{1}{\sqrt{n}} \sum_{k=1}^{n} \frac{1}{\sqrt{2k-1 + \sqrt{4k^2-4k+1+n^2}}} \right)$$
Step-by-Step Solution:
Rationalize the denominator: Multiply numerator and denominator by $\sqrt{2k-1} - \sqrt{4k^2-4k+1+n^2}$
After simplification, we get: $\frac{\sqrt{2k-1} - \sqrt{4k^2-4k+1+n^2}}{n-2k+1}$
Recognize Riemann sum: Divide numerator and denominator by $n$ and set $x = \frac{k}{n}$
The expression becomes: $\frac{1}{\sqrt{n}} \sum_{k=1}^{n} \frac{\sqrt{\frac{2k}{n}-\frac{1}{n}} - \sqrt{4(\frac{k}{n})^2 - 4\frac{k}{n} + \frac{1}{n^2} + 1}}{1 - 2\frac{k}{n} + \frac{1}{n}} \cdot \frac{1}{n}$
As $n \to \infty$, this approaches: $\int_0^1 \frac{\sqrt{2x} - \sqrt{4x^2 - 4x + 1}}{1 - 2x} dx$
Simplify the integrand using algebraic identities to get: $\int_0^1 \frac{1}{\sqrt{2x} + \sqrt{4x^2-4x+1}} dx$
Final evaluation gives: $\boxed{\sqrt{2} - 1}$
Problem 2: Limit of Ratio of Sums
Evaluate: $$\lim_{n \to \infty} \frac{\sum_{k=1}^{n} \sqrt{k}}{\sum_{k=1}^{n} \sqrt{n^2 + k}}$$
Step-by-Step Solution:
Apply Stolz-Cesàro Theorem: For sequences $a_n = \sum_{k=1}^{n} \sqrt{k}$ and $b_n = \sum_{k=1}^{n} \sqrt{n^2 + k}$
The limit equals $\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim_{n \to \infty} \frac{\sqrt{n+1}}{\sum_{k=1}^{n+1} \sqrt{(n+1)^2 + k} - \sum_{k=1}^{n} \sqrt{n^2 + k}}$
Simplify denominator: $\sqrt{(n+1)^2 + (n+1)} + \sum_{k=1}^{n} [\sqrt{(n+1)^2 + k} - \sqrt{n^2 + k}]$
Use rationalization: $\sqrt{(n+1)^2 + k} - \sqrt{n^2 + k} = \frac{(n+1)^2 + k - (n^2 + k)}{\sqrt{(n+1)^2 + k} + \sqrt{n^2 + k}} = \frac{2n+1}{\sqrt{(n+1)^2 + k} + \sqrt{n^2 + k}}$
For large $n$, this approximates $\frac{2n}{2n} = 1$ for each term in the sum
Thus denominator $\approx \sqrt{n^2 + 3n + 2} + n \approx 2n$ for large $n$
Numerator $\sqrt{n+1} \approx \sqrt{n}$, so the ratio approaches $\frac{\sqrt{n}}{2n} = \frac{1}{2\sqrt{n}} \to 0$
Final answer: $\boxed{0}$
Problem 3: Exponential Limit with Definite Integral
Evaluate: $$\lim_{n \to \infty} n^2 \left( \int_0^1 e^{x^2} (1-x)^n dx - \frac{1}{n+1} \right)$$
Step-by-Step Solution:
Expand $e^{x^2}$ using Taylor series: $e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \cdots$
The integral becomes: $\int_0^1 (1 + x^2 + \frac{x^4}{2} + \cdots)(1-x)^n dx$
We know $\int_0^1 x^{2k} (1-x)^n dx = \frac{(2k)!n!}{(n+2k+1)!}$
For $k=0$: $\int_0^1 (1-x)^n dx = \frac{1}{n+1}$
For $k=1$: $\int_0^1 x^2 (1-x)^n dx = \frac{2!n!}{(n+3)!} = \frac{2}{(n+1)(n+2)(n+3)}$
The given expression becomes: $n^2 \left( \frac{1}{n+1} + \frac{2}{(n+1)(n+2)(n+3)} + O(\frac{1}{n^4}) - \frac{1}{n+1} \right)$
Simplify: $n^2 \cdot \frac{2}{(n+1)(n+2)(n+3)} + O(\frac{1}{n})$
As $n \to \infty$, this approaches: $\frac{2n^2}{n^3} = \frac{2}{n} \to 0$
Wait! We need to be more careful. Let's recompute more precisely...
Actually: $n^2 \cdot \frac{2}{(n+1)(n+2)(n+3)} = \frac{2n^2}{n^3(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})} \to 2$
Final answer: $\boxed{2}$
🚀 Advanced Limit Solving Techniques
For Series Limits:
- Riemann Sums: Recognize $\frac{1}{n}\sum f(\frac{k}{n}) \to \int_0^1 f(x)dx$
- Stolz-Cesàro: For sequences of sums $\frac{a_n}{b_n}$ when standard methods fail
- Telescoping: Create differences that cancel intermediate terms
- Comparison Tests: Sandwich between known convergent sequences
For Functional Limits:
- Taylor Expansion: Replace functions with polynomial approximations
- L'Hospital Rule: Multiple applications for indeterminate forms
- Substitution: Change variables to simplify the expression
- Asymptotic Analysis: Identify dominant terms for large values
Problems 4-8 Available in Full Version
Includes 5 more JEE Advanced level problems with multi-concept approaches:
- • Limit with trigonometric and logarithmic functions
- • Double limit with parameter dependence
- • Limit involving greatest integer function
- • Functional equation based limit
- • Limit with recursive sequence
📝 Quick Self-Test
Try these JEE Advanced level limit problems:
1. Evaluate: $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \sqrt{\frac{n}{n+k}}$
2. Evaluate: $\lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^{100}}$
3. Evaluate: $\lim_{n \to \infty} \left( \frac{1^1 + 2^2 + \cdots + n^n}{n^n} \right)$
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