JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 6)
$$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}$$ is equal to :
$${1 \over {15}}$$
0
$${2 \over 3}$$
$${3 \over 2}$$
Explanation
$$\mathop {\lim }\limits_{x \to {0^ + }} {{\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \over {{x^3}}}$$
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)
$$ = {2 \over 3}\mathop {\lim }\limits_{x \to {0^ + }} {{\sin x} \over x}$$
$$ = {2 \over 3}$$
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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