QJEEM21Mad450: JEE Main 2021 Mathematics - Class 12 JEEM21Mad450 Solution

Question

Verified Solution • 2025–26 Session moderate

\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}

is equal to :

Accepted Answer

\mathop {\lim }\limits_{x \to {0^ + }} {{\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \over {{x^3}}}

This is in

{0 \over 0}

form, so use L' Hospital rule

= \mathop {\lim }\limits_{x \to {0^ + }} {{{d \over {dx}}\left( {\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \right)} \over {{d \over {dx}}\left( {{x^3}} \right)}}

= \mathop {\lim }\limits_{x \to {0^ + }} {{\sin x.2x - 0} \over {3{x^2}}}

(applying Leibnitz rule)

= {2 \over 3}\mathop {\lim }\limits_{x \to {0^ + }} {{\sin x} \over x}

= {2 \over 3}

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