TITLE: Limit with Integral in it
QUESTION [6 upvotes]: $$\lim_{x \to 0} \dfrac{\displaystyle \int_0^x \sin \left(\pi t^2/2\right) dt}{x^3}$$
I am having trouble trying to figure out how to compute the limit.
Do I have to take the integral first and then take the limit?

REPLY [2 votes]: You also can use $\sin x=x+O(x^3)$ to get the limit. In fact
\begin{eqnarray*}
\lim_{x\to 0}\frac{\int_0^x\sin(\pi t^2/2)dt}{x^3}=
\lim_{x\to 0}\frac{\int_0^x(\pi t^2/2+O(t^6))dt}{x^3}=\lim_{x\to 0}\frac{\pi x^3/6+O(x^7)}{x^3}=\pi/6.
\end{eqnarray*}