Evaluating the limit $\lim_{x\to0}\frac{1}{x^3}\int_{0}^{x}\sin(\sin(t^2))dt$ $$\lim_{x\to0}\frac{1}{x^3}\int_{0}^{x}\sin(\sin(t^2))dt$$
This is a compound question from me.


*

*I don't know how to begin evaluating this limit. My guess would be that I would have to find the value of this Riemann's integral and then plug the result into the limit. Is this the right direction to head?


Which brings me to...


*I am also stuck trying to resolve the integral. I tried integrating by substitution, trying with both $u = t^2$ and $u = \sin(t^2)$, but both have lead me to finding that $t$ or $dt$ popping back into the equation sooner or later and I'm not quite sure how to handle that. Any hints as to how I can integrate that function?


Thank you.

The expression screams L'Hopital's Rule:
\begin{align}\lim_{x\to0}\frac1{x^3}\int_0^x \sin\left(\sin(t^2)\right)\,dt&\overset{\text{L'H}}=\lim_{x\to0}\frac{\frac{d}{dx}\int_0^x\sin\left(\sin(t^2)\right)\,dt}{\frac{d}{dx}x^3}\\
&=\lim_{x\to0}\frac{\sin\left(\sin(x^2)\right)}{3x^2}\\
&\overset{\text{L'H}}=\lim_{x\to0}\frac{\cos\left(\sin (x^2)\right)\cdot\cos(x^2)\cdot2x}{6x}\\
&=\frac13
\end{align}
