finding the limit of the integral of an unknown function I have a question from an exam written 18 years ago, and I can't solve it (I'm solving old questions to prepare for my own exam).
The Question
The question goes like so:
let  g be a function, which has a derivative defined $\forall x\in [-1,1]$, and $g(0)=0,g'(0)=1$.
Does the limit $$\lim_{x\rightarrow0}\frac{\sin\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)}{x^{8}}$$ converge?
What have I tried?
I thought about using L'Hôpital's rule, and F.T.C. so I wrote:
$$\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)'=2x\int\limits _{0}^{x^{2}}g\left(s^{2}\right)ds-3x^{2}\int\limits _{0}^{x^{3}}g\left(s^{2}\right)ds$$
but that didn't help much. I can even show that the limit is of the form $"\frac{0}{0}"$ to use the rule.
I tested this on $g(x)=x$ and it seems like this limit does converge (to 0 if I'm not mistaken)
I also thought Taylor was somehow related, but I really doubt it. I am also unsure of where the information we have about $g(0),g'(0)$ comes in.
BTW
this is what I got after applying L'Hôpital once
$$\lim_{x\rightarrow0}\frac{\left(2\int\limits _{0}^{x^{2}}g\left(s^{2}\right)ds-3x\int\limits _{0}^{x^{3}}g\left(s^{2}\right)ds\right)\cos\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)}{8x^{6}}$$
 A: Let us write
\begin{equation}g (x) = x h (x)\end{equation}
so that $h (x) \rightarrow  1$ when $x \rightarrow  0$. Let us first substitute $s = t u$ in the integral
\begin{equation}I \left(x\right) \colon  = \int_{{x}^{3}}^{{x}^{2}}\left(\int_{0}^{t}g \left({s}^{2}\right) d s\right) d t = \int_{{x}^{3}}^{{x}^{2}}\left(\int_{0}^{t}{s}^{2} h \left({s}^{2}\right) d s\right) d t = \int_{{x}^{3}}^{{x}^{2}}{t}^{3} \left(\int_{0}^{1}u^2h \left({t}^{2} {u}^{2}\right) d u\right) d t\end{equation}
Now let us substitute $t = {x}^{2} v$, it follows that
\begin{equation}I \left(x\right) = {x}^{8} \int_{x}^{1}v^3\left(\int_{0}^{1} u^2h \left({x}^{4} {u}^{2} {v}^{2}\right) d u\right) d v \sim  {x}^{8} \int_{0}^{1}{v}^{3} \left(\int_{0}^{1}u^2 d u\right) d v = \frac{{x}^{8}}{12}\end{equation}
As $\sin  \left({\theta}\right) \sim  {\theta}$ when ${\theta} \rightarrow  0$ it follows that the limit is $1/12$.
A: This is an answer based on my comments.
Let $$G(t) =\int_{0}^{t}g(s^2)\,ds,F(u)=\int_{0}^{u}G(t)\,dt$$ Then we have $$F'(u) =G(u), G'(t) =g(t^2)$$ and hence by two applications of LHospital Rule we get $$\lim_{u\to 0}\frac{F(u)}{u^4}=\lim_{u\to 0}\frac{G(u)}{4u^3}=\lim_{u\to 0}\frac{g(u^2)}{12u^2}=\frac {1}{12}$$ The limit in question is $$\lim_{x\to 0}\frac{\sin(F(x^2)-F(x^3))}{x^8}$$ Since $F(x^2)\to 0,F(x^3)\to 0$ we can write the expression under limit as $$\frac {\sin(F(x^2)-F(x^3))}{F(x^2)-F(x^3)}\cdot \frac {F(x^2)-F(x^3)}{x^8}$$ and the first factor tends to $1$ so the limit equals the limit of second factor.
Now we can write $$\frac{F(x^2)-F(x^3)}{x^8}=\frac {F(x^2)}{x^8}-x^4\cdot\frac{F(x^3)}{x^{12}}$$ which tends to $$\frac {1}{12}-0\cdot\frac {1}{12}=\frac {1}{12}$$

LHospital Rule is a very powerful tool for evaluation of limits, but it has become infamous due to the crappy ways in which it is used frequently by beginners.
It is possible to avoid LHospital and use $\epsilon, \delta$ and integrating inequalities.
