JEE Main 2021 · 24 Feb · Shift 1 · Mathematics Q66

MathematicsSingle correct+4 / -1Question 66 of 90

$\displaystyle\lim_{x\to0}\frac{\int_0^{x^2}\left(\sin\sqrt t\right)dt}{x^3}$ is equal to :

\dfrac23

\dfrac32

\dfrac1{15}

Answer and worked solution

Official answer: A

L'Hôpital. $\dfrac{2x\sin|x|}{3x^2}\to\dfrac23$ (equivalently $\sin\sqrt t\approx\sqrt t$ , integral $\frac23x^3$ ).

(A)

Topics:L'Hospital's rule Leibniz rule

JEE Main 2021 · 24 Feb · Shift 1, Mathematics question 66