<s>[INST] How do i solve this limit: $\lim_{x\to 0}{x-\sin(\sin(...(\sin x)))\over x^{3}}$ I have the next limit :
$$\large \lim_{x\to 0}{x-\sin(\sin(\overbrace {\cdot \ \cdot \ \cdot }^n(\sin(x))\overbrace {\cdot \ \cdot \ \cdot }^n))\over x^{3}}$$
$\sin(\sin(...(\sin(x))...))$-is n times.
I have no idea. Someone can help me? Thank you!
 [/INST] Let us rewrite
\begin{align}
\frac{x-\sin^{[n]}x}{x^3}&=\frac{x-\sin x+\sin x-\sin\sin x+\ldots+\sin^{[n-1]} x-\sin^{[n]} x}{x^3}=\\
&=\frac{x-\sin x}{x^3}+\frac{\sin x-\sin\sin x}{x^3}+\ldots+\frac{\sin^{[n-1]}x-\sin^{[n]}x}{x^3}
\end{align}
and calculte the limit of each fraction separately. 


*

*Since
$$
\sin t=t-\frac{t^3}{6}+o(t^3)
$$
we have the first fraction just as
$$
\frac{x-\sin x}{x^3}=\frac{\frac{x^3}{6}-o(x^3)}{x^3}=\frac16+o(x)\to\frac16.
$$

*Similarly with notation $y=\sin^{[k-1]}x$ the $k$-th fraction is
$$
\frac{\sin^{[k-1]}x-\sin^{[k]}x}{x^3}=\frac{y-\sin y}{x^3}=\frac{y-\sin y}{y^3}\cdot\frac{y^3}{x^3}=\left(\frac16+o(y)\right)\cdot\left(\frac{y}{x}\right)^3\to\frac16\cdot 1^3=\frac16
$$
because 
$$
\frac{y}{x}=\frac{\sin^{[k-1]}x}{x}=\frac{\sin^{[k-1]}x}{\sin^{[k-2]}x}\cdot\frac{\sin^{[k-2]}x}{\sin^{[k-3]}x}\cdot\ldots\cdot\frac{\sin x}{x}\to1\cdot 1\cdot\ldots\cdot 1=1.
$$ 

*Finally the limit is
$$
\underbrace{\frac16+\frac16+\ldots+\frac16}_{n\text{ times}}=\frac{n}{6}.
$$

</s>