Limit with Integral in it $$\lim_{x \to 0} \dfrac{\displaystyle \int_0^x \sin \left(\pi t^2/2\right) dt}{x^3}$$
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?
 A: Yes. You will have to take the integral first and then take the limit. However, note that the limit is of the form $\dfrac{0}0$, when you plug in $x=0$. Hence, you can use L'Hospital rule.
Use L'Hospital rule along with Fundamental theorem of Calculus. We then get
$$\lim_{x \to 0} \dfrac{\sin(\pi x^2/2)}{3x^2}$$
Now you should be able to finish this off.
A: 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*}
A: Hint: You can use L'Hospital's Rule
A: Hint: Use L'Hospital's Rule, using the Fundamental Theorem of Calculus to evaluate the derivative of the top.
Remark: The function $\sin x^2$ does not have an elementary antiderivative, so a strategy based on evaluating the integral will not succeed. 
But there is a good alternative to L'Hospital's Rule. Use the power series expansion of $\sin x$ to find the power series expansion of $\sin(\pi t^2/2)$. Then integrate term by term. The answer to your limit problem will pop out.
The first term in the power series expansion of $\sin x$ is $x$. Substitute $\pi t^2/2$, and integrate from $0$ to $x$. We get $\pi x^3/6$. The remaining terms of the power series are negligible in comparison for $x$ near $0$. So the limit of the quotient is $\pi/6$.
