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JEE Advanced 2022 Paper 1 · Mathematics · Limits · medium

Match the limits in List-I with their values in List-II. List-I (P) lim x→0(e^x^2-cos x)/(x^2 (Q) lim x→0(x-sin x)/(x^2tan x) (R) lim…

Single correct option

Match the limits in List-I with their values in List-II.

List-I
(P) lim⁡x→0ex2−cos⁡xx2\displaystyle\lim_{x\to0}\frac{e^{x^2}-\cos x}{x^2}
(Q) lim⁡x→0x−sin⁡xx2tan⁡x\displaystyle\lim_{x\to0}\frac{x-\sin x}{x^2\tan x}
(R) lim⁡n→∞∑k=1nkn2+k2\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2}
(S) lim⁡x→0(cos⁡x)1/x2\displaystyle\lim_{x\to0}(\cos x)^{1/x^2}

List-II
(1) 16\dfrac16
(2) 32\dfrac32
(3) ln⁡22\dfrac{\ln2}{2}
(4) e−1/2e^{-1/2}
(5) 22
  1. A. P→1, Q→2, R→3, S→4
  2. B. P→2, Q→1, R→3, S→4
  3. C. P→2, Q→1, R→5, S→3
  4. D. P→3, Q→1, R→2, S→4
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Answer: (B) P→2, Q→1, R→3, S→4

(P) ex2=1+x2+…e^{x^2}=1+x^2+\dots and cos⁡x=1−x22+…\cos x=1-\frac{x^2}2+\dots, so the limit is 1+12=321+\frac12=\frac32 (2). (Q) x−sin⁡x∼x36x-\sin x\sim\frac{x^3}6 and x2tan⁡x∼x3x^2\tan x\sim x^3, giving 16\frac16 (1). (R) ∑kn2+k2=1n∑k/n1+(k/n)2→∫01x1+x2dx=ln⁡22\sum\frac{k}{n^2+k^2}=\frac1n\sum\frac{k/n}{1+(k/n)^2}\to\int_0^1\frac{x}{1+x^2}dx=\frac{\ln2}2 (3). (S) ln⁡(cos⁡x)/x2→−12\ln(\cos x)/x^2\to-\frac12, so the limit is e−1/2e^{-1/2} (4). Hence P→2, Q→1, R→3, S→4.
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Original question written in the JEE Advanced 2022 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