_x0_0^x^2 t,dtx^3 is equal to:

Question

$\displaystyle\lim_{x	o0}\frac{\displaystyle\int_0^{x^2}\sin\sqrt t\,dt}{x^3}$ 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)

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

$$ = {2 \over 3}$$

Solutions · 1