JEE Advanced 2023 Paper 2 · Mathematics · Limits · medium

Let L=lim x→0(1)/(x^4∫0^x(tan t-sin t) dt. The value of 1/L is

Numerical answer

Let

L=\displaystyle\lim_{x\to0}\frac{1}{x^4}\int_0^{x}(\tan t-\sin t)\,dt

. The value of

1/L

Show answer and solution

Answer: 8

Near

0

\tan t-\sin t=\frac{t^3}{2}+\frac{t^5}{8}+\dots

(since

\tan t=t+\frac{t^3}3+\dots

\sin t=t-\frac{t^3}{6}+\dots

, difference

\frac{t^3}{2}

). So the integral is

\frac{x^4}{8}+O(x^6)

and

L=\frac18

(L'Hopital on

\frac{\tan x-\sin x}{4x^3}\to\frac12\cdot\frac14

gives the same). Hence

1/L=8

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