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JEE Advanced 2021 Paper 1 · Mathematics · Limits · medium

Limits has (have) the stated value?

One or more options correct

Which of the following limits has (have) the stated value?
  1. A. lim⁡n→∞∑k=1nkn2+k2=12ln⁡2\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2}=\frac12\ln2
  2. B. lim⁡n→∞∑k=1n1n2+k2=ln⁡(1+2)\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}=\ln\left(1+\sqrt2\right)
  3. C. lim⁡n→∞((2n)!n! nn)1/n=e4\displaystyle\lim_{n\to\infty}\left(\frac{(2n)!}{n!\,n^{n}}\right)^{1/n}=\frac{e}{4}
  4. D. lim⁡x→0(cos⁡x)1/x2=e−1\displaystyle\lim_{x\to0}(\cos x)^{1/x^{2}}=e^{-1}
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Answer: A, B

(A) Write the sum as 1n∑k/n1+(k/n)2→∫01x1+x2dx=12ln⁡2\frac1n\sum \frac{k/n}{1+(k/n)^2}\to\int_0^1\frac{x}{1+x^2}dx=\frac12\ln2. True. (B) Similarly 1n∑11+(k/n)2→∫01dx1+x2=ln⁡(1+2)\frac1n\sum\frac1{\sqrt{1+(k/n)^2}}\to\int_0^1\frac{dx}{\sqrt{1+x^2}}=\ln(1+\sqrt2). True. (C) (2n)!n!nn=∏k=1nn+kn\frac{(2n)!}{n!n^n}=\prod_{k=1}^n\frac{n+k}{n}, so the logarithm of the nn-th root tends to ∫01ln⁡(1+x)dx=2ln⁡2−1\int_0^1\ln(1+x)dx=2\ln2-1, and the limit is 4e\frac4e, not e4\frac e4. False. (D) ln⁡cos⁡x≈−x22\ln\cos x\approx-\frac{x^2}{2}, so ln⁡cos⁡xx2→−12\frac{\ln\cos x}{x^2}\to-\frac12 and the limit is e−1/2e^{-1/2}. False.
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Original question written in the JEE Advanced 2021 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake