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JEE Advanced 2025 Paper 1 · Mathematics · Limits · medium

Match each limit in List-I with its value in List-II. List-I (P) lim x→0(tan x-x)/(x^3 (Q) lim x→0(x-sin x)/(x^3 (R) lim x→0(1-cos xcos 2xcos…

Single correct option

Match each limit in List-I with its value in List-II.

List-I
(P) lim⁡x→0tan⁡x−xx3\displaystyle\lim_{x\to0}\frac{\tan x-x}{x^{3}}
(Q) lim⁡x→0x−sin⁡xx3\displaystyle\lim_{x\to0}\frac{x-\sin x}{x^{3}}
(R) lim⁡x→01−cos⁡xcos⁡2xcos⁡3xx2\displaystyle\lim_{x\to0}\frac{1-\cos x\cos 2x\cos 3x}{x^{2}}
(S) lim⁡x→02sin⁡x−sin⁡2xx3\displaystyle\lim_{x\to0}\frac{2\sin x-\sin 2x}{x^{3}}

List-II
(1) 16\dfrac16 (2) 13\dfrac13 (3) 77 (4) 22 (5) 11
  1. A. P→2, Q→1, R→3, S→5
  2. B. P→1, Q→2, R→3, S→5
  3. C. P→2, Q→1, R→5, S→3
  4. D. P→2, Q→3, R→1, S→5
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Answer: (A) P→2, Q→1, R→3, S→5

Use Maclaurin series. (P) tan⁡x=x+x33+⋯\tan x=x+\frac{x^3}{3}+\cdots, so the limit is 13\frac13: P→2. (Q) sin⁡x=x−x36+⋯\sin x=x-\frac{x^3}{6}+\cdots, so x−sin⁡x∼x36x-\sin x\sim\frac{x^3}{6} and the limit is 16\frac16: Q→1. (R) Near 00, cos⁡kx=1−k2x22+⋯\cos kx=1-\frac{k^2x^2}{2}+\cdots, so cos⁡xcos⁡2xcos⁡3x=1−(1+4+9)x22+⋯=1−7x2+⋯\cos x\cos2x\cos3x=1-\frac{(1+4+9)x^2}{2}+\cdots=1-7x^2+\cdots and the limit is 77: R→3. (S) 2sin⁡x=2x−x33+⋯2\sin x=2x-\frac{x^3}{3}+\cdots and sin⁡2x=2x−8x36+⋯=2x−4x33+⋯\sin2x=2x-\frac{8x^3}{6}+\cdots=2x-\frac{4x^3}{3}+\cdots. The difference is (−13+43)x3=x3\left(-\frac13+\frac43\right)x^3=x^3, so the limit is 11: S→5. The value 22 in item (4) is unused and arises if one wrongly takes sin⁡2x≈2sin⁡x\sin 2x\approx 2\sin x to higher order.
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Original question written in the JEE Advanced 2025 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