TITLE: Evaluating limit (iterated sine function)
QUESTION [11 upvotes]: The limit is 
$$\lim_{x\rightarrow0} \frac{x-\sin_n(x)}{x^3},$$
where $\sin_n(x)$ is the $\sin(x)$ function composed with itself $n$ times:
$$\sin_n(x) = \sin(\sin(\dots \sin(x)))$$
For $n=1$ the limit is $\frac{1}{6}$, $n=2$, the limit is $\frac{1}{3}$ and so on.
Can we define a recurrent relation upon that given hypothesis? Also, how do I involve $n$ into calculation, because the final limit will depend on it?
Any suggestions on how to tackle this?
Thank you!

REPLY [4 votes]: This is not an answer but it is too long for a comment.
The problem being really interesting, I tried to have an extra information : how is approached the limit ?
Doing almost the same as in Paramanand Singh's and Vim's answers and comments, I used one extra term for the Taylor series of $\sin(x)$ and arrive to $$sin_n(x)=x-\frac n 6 x^3 -\big(\frac n{30}-\frac{n^2}{24}\big) x^5+\cdots$$ which means that $$ \frac{x-\sin_n(x)}{x^3}=\frac n 6+\big(\frac n{30}-\frac{n^2}{24}\big) x^2+\cdots$$