$\lim\limits_{x\rightarrow 0 }\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}$ 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 $?$ 

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).
