Answer
$$
\lim _{x \rightarrow 0} \frac{S(x)}{x^{3}}=\frac{\pi}{6}
$$
where
$$
S(x)=\int_{0}^{x} \sin \left(\pi t^{2} / 2\right) d t
$$
Work Step by Step
$$
\lim _{x \rightarrow 0} \frac{S(x)}{x^{3}},\quad\quad S(x)=\int_{0}^{x} \sin \left(\pi t^{2} / 2\right) d t
$$
notice that as $ \quad x \rightarrow 0 \quad $ we have $ \quad S(x) \rightarrow 0 \quad $ and $ \quad x^{3} \rightarrow 0 $
so l’Hospital’s Rule gives
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{S(x)}{x^{3}} &=\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin \left(\pi t^{2} / 2\right) d t}{x^{3}} \quad\quad \rightarrow \frac{0}{0}\\ & \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{\sin \left(\pi x^{2} / 2\right)}{3 x^{2}} \quad\quad \rightarrow \frac{0}{0}\\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{\pi x \cos \left(\pi x^{2} / 2\right)}{6 x} \\
&=\frac{\pi}{6} \cdot \cos 0 \\
&=\frac{\pi}{6}
\end{aligned}
$$
