Fundamental Theorem of Calculus for $\lim\limits_{x\to 0}\frac{\int_0^x(x-t)\sin t^2\ dt}{x\sin^3x}$ How to integrate this?

Evaluate
  $$\lim_{x\to0}\frac{\displaystyle\int_0^x(x-t)\sin t^2\ dt}{x\sin^3x}.$$

I've had difficulty in using L'Hopital rule. At the same time failed to really understand how to differentiate or evaluate the limit of the $x\sin^3x$ and $\int_0^x(x-t)\sin t^2\ dt$
Would like to appreciate your help
 A: Use repeatedly $\sin u=u\,{\rm sinc}(u)$, whereby $\lim_{u\to0}{\rm sinc}(u)={\rm sinc}(0)=1$. We have
$$\int_0^x (x-t)\sin(t^2)\>dt=x^4\int_0^1(1-\tau)\,\tau^2{\rm sinc}(x^2\tau^2)\>d\tau$$
and $$\>x\sin^3 x=x^4\>{\rm sinc}^3(x)\ ,$$ so that
$${\int_0^x (x-t)\sin(t^2)\>dt \over x\sin^3 x}={\int_0^1(1-\tau)\,\tau^2\bigl(1+r(x,\tau)\bigr)\>d\tau\over 1+\bar r(x)}\ ,$$
whereby $\lim_{x\to0}r(x,\tau)=0$ uniformly in $\tau$, and $\lim_{x\to0}\bar r(x)=0$ as well. It follows that the limit in question is
$$\int_0^1(1-\tau)\,\tau^2\>d\tau={1\over12}\ .$$
A: Write the numerator as
$$
f(x)=x\int_0^x\sin t^2\,dt-\int_0^x t\sin t^2\,dt
$$
Then the FTC and the product rule tell you that
$$
f'(x)=\int_0^x\sin t^2\,dt+x\sin x^2-x\sin x^2=\int_0^x\sin t^2\,dt
$$
Therefore
$$
f''(x)=\sin x^2
$$
Thus, applying l'Hôpital twice,
$$
\lim_{x\to0}\frac{f(x)}{x^4}=
\lim_{x\to0}\frac{f'(x)}{4x^3}=
\lim_{x\to0}\frac{f''(x)}{12x^2}=\frac{1}{12}
$$
Hence
$$
\lim_{x\to0}\frac{f(x)}{x\sin^3x}=
\lim_{x\to0}\frac{f(x)}{x^4}\frac{x^3}{\sin^3x}=\frac{1}{12}
$$
A: Let
$f(x)=x\int_0^x\sin(t^2)dt-\int_0^xt\sin(t^2)dt$
and
$g(x)=x\sin^3(x)$
$g(x)$ is equivalent to $h(x)=x^4$ and we will have the same limit if we replace
$g(x)$by $h(x)$.
we will compute
$\lim_{x\to0}\frac{f(x)}{h(x)}\frac{h(x)}{g(x)}$
and use
$\lim_{x\to 0}\frac{h(x)}{g(x)}=1$.
$f'(x)=\int_0^x\sin(t^2)dt+x\sin(x^2)-x\sin(x^2)=x\int_0^x\sin(t^2)dt $
thus your limit is $\color{red}{\frac{1}{12}}$ after a second l' Hopital rule.
A: Let $$f(x) = \int_{0}^{x}(x - t)\sin t^{2}\,dt$$ then it can be easily proved (using integration by parts) that $$f''(x) = \sin x^{2}, f(0) = f'(0) = 0$$ and therefore we can use L'Hospital's Rule twice to get the answer easily.
We have
\begin{align}
L &= \lim_{x \to 0}\frac{f(x)}{x\sin^{3}x}\notag\\
&= \lim_{x \to 0}\frac{f(x)}{x^{4}}\cdot\frac{x^{3}}{\sin^{3}x}\notag\\
&= \lim_{x \to 0}\frac{f(x)}{x^{4}}\notag\\
&= \lim_{x \to 0}\frac{f'(x)}{4x^{3}}\text{ (via L'Hospital's Rule)}\notag\\
&= \lim_{x \to 0}\frac{f''(x)}{12x^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{12}\lim_{x \to 0}\frac{\sin x^{2}}{x^{2}}\notag\\
&= \frac{1}{12}
\end{align}

More generally if $$F(x) = \frac{1}{(n - 1)!}\int_{a}^{x}(x - t)^{n - 1}f(t)\,dt$$ then $$F^{(n)}(x) = f(x), F(a) = F'(a) = F''(a) = F^{(n - 1)}(a) = 0$$ Further note that it is never a good idea to jump to L'Hospital Rule directly (unless the problem is too simple) and it is better to transform the expression into a form where application of L'Hospital's Rule is easy. Thus in the solution provided above we have effectively replaced the expression $x\sin^{3}x$ in denominator with $x^{4}$ and then applied L'Hospital's Rule. This considerably simplifies the process of L'Hospital's Rule.
A: An idea with Taylor expansions:
$$(x-t)\sin t^2=(x-t)\left(t^2-\frac{t^6}6+\ldots\right)=xt^2-\frac{xt^6}6-t^3+\frac{t^7}6+\ldots\implies$$
$$\int_0^x(x-t)\sin t^2\,dt=\left.\left(\frac x3t^3-\frac14t^4+\ldots\right)\right|_0^x=\frac{x^4}3-\frac{x^4}4+\ldots=\frac{x^4}{12}+\ldots$$
Observe that the $\;\ldots\;$ above is an expression which, upon substituting $\;t\to x\;$, has an exponent higher than $\;4\;$ . Because of the following this won't contribute anything in the limit.
We also have
$$x\sin^3x=x\left(x-\frac{x^3}6+\ldots\right)^3=x(x^3+\ldots)=x^4+\ldots$$
Thus our limit is
$$\lim_{x\to0}\frac{\frac{x^4}{12}}{x^4}=\frac1{12}$$
