"Easier Than It Looks"

$\begin{aligned} L & = \lim_{x \to 0} \frac {\int_0^x \sin t^3 \ dt}{x^4} & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies}} \\ & = \lim_{x \to 0} \frac {\sin x^3}{4x^3} & \small \blue{\text{Differentiate up and down w.r.t. }x} \\ & = \frac 14 = \boxed{0.25} & \small \blue{\text{Since }\lim_{u \to 0} \frac {\sin u}u = 1 \text{ by L'Hôpital's rule.}} \end{aligned}$

Reference: L'Hôpital's rule

Using the usual series expansion of sine, we have $\sin t^3 = t^3 + \mathcal{O} \left(t^9\right)$ . Hence the numerator is just $\frac14 x^4 + \mathcal{O} \left( x^{10} \right)$ . Dividing through by $x^4$ , we have to find the limit of $\frac14+\mathcal{O}\left(x^6\right)$ as $x \to 0$ , which is clearly $\boxed{\frac14}$ .