I'm trying to find out if this limit exists I'm trying to calculate the following limit but I can't wrap my head,  around it. Can you guys give me some hints:
$$\lim_{x\to0^+}\frac{{\int_0^{x^2}\sin{\sqrt{t}}}~ dt}{x^3}$$
 A: Letting $t=u^2$, we get
$$
2\int_0^{x} u \sin u\, du\ .
$$
Integrating by parts,
$$
\int_0^{x} u \sin u \, du = -x \cos x + \int_0^x \cos u\, du = \sin x -x\cos x.
$$
Hence we get, by applying de l'Hôpital's rule,
$$
\lim_{x\to0}\frac{2}{x^3}\left( \sin x - x\cos x \right)=
\lim_{x\to0}\frac{2x \sin x}{3 x^2}=\frac{2}{3}.
$$
A: You can also just use L'Hôpital's rule directly and the 2nd fundamental theorem of calculus: since
$$
\frac{d}{dx} \int_0^{x^2} \sin \sqrt{t} \ dt = \sin(x)\cdot2x, 
$$
the indeterminate limit is equal to
$$
\lim_{x \to 0} \frac{\sin(x)\cdot2x}{3x^2} = \frac{2}{3} \lim_{x \to 0} \frac{\sin(x)}{x}.
$$
The latter limit is known to be $1$ (from the standard proof that the derivative of the sine function is the cosine function).
A: $$\begin{align*}
\lim_{x \to 0^+} \frac{1}{x^3} \int_0^{x^2} \sin \sqrt{t} \ \mathrm dt
&= \lim_{x \to 0^+} \frac{1}{x^3} \int_0^{x^2} \sqrt{t} \left( 1 + O(t) \right) \ \mathrm dt \\
&= \lim_{x \to 0^+} \frac{1}{x^3} \left( \frac{2}{3} t^{3/2} + O(t^{5/2}) \right)_0^{x^2} \\
&= \lim_{x \to 0^+} \frac{1}{x^3} \left( \frac{2}{3} x^3 + O(x^5) \right) \\
&= \lim_{x \to 0^+} \left(\dfrac23+O(x^2)\right) \\
&= \frac{2}{3}
\end{align*}$$
