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.

Without L'Hopital:
\begin{align*}
\dfrac{1}{x^{3}}\int_{0}^{x}\sin(\sin t^{2})dt=\dfrac{1}{x^{3}}\left(x\sin(\sin x^{2})-\int_{0}^{x}t\cos(\cos t^{2})2tdt\right).
\end{align*}
Note that
\begin{align*}
\dfrac{1}{x^{3}}(x\sin(\sin x^{2}))=\dfrac{\sin(\sin x^{2})}{\sin x^{2}}\dfrac{\sin x^{2}}{x^{2}}\rightarrow 1.
\end{align*}
On the other hand,
\begin{align*}
\int_{0}^{x}t\cos(\cos t^{2})2tdt=\dfrac{2}{3}x^{3}\cos(\cos x^{2})-\dfrac{2}{3}\int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt.
\end{align*}
And we have
\begin{align*}
-\dfrac{\dfrac{2}{3}x^{3}\cos(\cos x^{2})}{x^{3}}=-\dfrac{2}{3}\cos(\cos x^{2})\rightarrow-\dfrac{2}{3},
\end{align*}
whereas fot the integral, by the change of variable $u=t^{4}$, we obtain that
\begin{align*}
\int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt=\dfrac{1}{2}\int_{0}^{x^{4}}u\sin(\sin u^{1/2})\dfrac{du}{u^{3/4}}=\dfrac{1}{2}\int_{0}^{x^{4}}u^{1/4}\sin(\sin u^{1/2})du,
\end{align*}
and that
\begin{align*}
\dfrac{1}{x^{3}}\int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt&=\dfrac{1}{2}\cdot x\cdot\dfrac{1}{x^{4}}\int_{0}^{x^{4}}u^{1/4}\sin(\sin u^{1/2})du\\
&=\dfrac{1}{2}\cdot x\cdot\eta_{x}^{1/4}\sin(\sin\eta_{x}^{1/2})\\
&\rightarrow 0,
\end{align*}
where $\eta_{x}\in[0,x]$ is chosen by Integral Mean Value Theorem, therefore the whole limit is $1-2/3=1/3$.
