$\lim\limits_{x\rightarrow 0 }\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}$ From the post Evaluating limit (iterated sine function)  and some discussions inside, one can collect the
following three limits 
\begin{eqnarray*}
\lim_{x\rightarrow 0 }\frac{\sin _{n}x}{x} &=&1 \\
\lim_{x\rightarrow 0}\frac{x-\sin _{n}x}{x^{3}} &=&\frac{n}{6} \\
\lim_{x\rightarrow 0}\frac{\sin _{n}x-x+\frac{n}{6}x^{3}}{x^{5}} &=&%
\frac{n^{2}}{24}-\frac{n}{30}.
\end{eqnarray*}
where $\sin _{n}x=\sin (\sin \cdots (\sin x)),\ n$ times composition.
So the next question would be, what is the following limit
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{%
n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}
\end{equation*}
and what are those corresponding limits after order 7 $?$ 
 A: Continuing in the same spirit as in my answer to the first question, we can probly admit that the limit could be expressed as a cubic polynomial in $n$. Going on with the expansions (tiring !!), I arrived to $${\sin _{n}(x)-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}=-\Big(\frac{5 n^3}{432}-\frac{n^2}{45}+\frac{41 n}{3780}\Big)x^7+\cdots$$ which gives for the fourth limit $$L_4=-\frac{5 n^3}{432}+\frac{n^2}{45}-\frac{41 n}{3780}$$ Continuing with the next limit (very tiring !!) $$L_5=\frac{35 n^4}{10368}-\frac{71 n^3}{6480}+\frac{67 n^2}{5670}-\frac{4 n}{945}$$
At this point, I give up (hoping that I did not make any mistake with all these developments).
A: Direct computation
$\def\wi{\subseteq}$
$\sin(x) \in x - \frac{1}{6} x^3 + \frac{1}{120} x^5 - \frac{1}{5040} x^7 + O(x^9)$ as $x \to 0$.
As $x \to 0$ and given function $f$ such that $f(x) \in x + a x^3 + b x^5 + c x^7 + O(x^9)$:
  $\sin(f(x))$
  $\ \in \left( x + a x^3 + b x^5 + c x^7 + O(x^9) \right)$
  $\hphantom{\ \in} - \frac{1}{6} \left( x + a x^3 + b x^5 + O(x^7) \right)^3$
  $\hphantom{\ \in} + \frac{1}{120} \left( x + a x^3 + O(x^5) \right)^5$
  $\hphantom{\ \in} - \frac{1}{5040} \left( x + O(x^3) \right)^7$
  $\hphantom{\ \in} + O(x^9)$
  $\ \wi x + (a-\frac{1}{6}) x^3 + (b-\frac{1}{2}a+\frac{1}{120}) x^5 + (c-\frac{1}{2}b-\frac{1}{2}a^2+\frac{1}{24}a^2-\frac{1}{5040}) x^7$
  [Thus we can set up recurrences for $a,b,c$ on iteration of applying $\sin$ and solve them to finish.]
Notes
As you can see, we only need to handle the coefficient sequences. I'm too lazy to figure out the computational complexity of this algorithm in general but it is quite bad because the numbers grow exponentially. Is there a faster method? I doubt so. No wonder Claude got tired! But at least this is implementable on a computer and we don't have to do it ourselves for high order approximations.
