L'Hospital Rule with Trigonometry I've been working on this question for the final exam, yet I could not come to a closure. I'd appreciate if you can help me.

What is $$\lim_{x\to 0} \frac{\sin(\sin(\sin x)) - x}{x^3}\qquad?$$

 A: Consider the function
$$
f_n(x)=\underbrace{\sin\sin\dotsb\sin}_{n\text{ times}}\,x
$$
where $f_0(x)=x$. With a recursive formula, $f_{k+1}(x)=\sin(f_k(x))$; note that $f_k(0)=0$ and $f_k$ is continuous and invertible in a neighborhood of $0$.
Then your limit can be written as
$$
\lim_{x\to0}\left(
\frac{f_3(x)-f_2(x)}{x^3}+
\frac{f_2(x)-f_1(x)}{x^3}+
\frac{f_1(x)-f_0(x)}{x^3}
\right)
$$
so we may as well ask what's
$$
\lim_{x\to0}\frac{\sin(f_k(x))-f_k(x)}{x^3}=
\lim_{x\to0}\frac{f_k(x)-f_k(x)^3/6+o(f_k(x)^3)-f_k(x)}{x^3}
$$
Thus we just need to check what's
$$
\lim_{x\to0}\frac{f_k(x)}{x}
$$
The limit is $1$ for $k=0$; suppose we know that
$$
\lim_{x\to0}\frac{f_k(x)}{x}=1
$$
Then
$$
\lim_{x\to0}\frac{f_{k+1}(x)}{x}=
\lim_{x\to0}\frac{\sin(f_k(x))}{f_k(x)}\frac{f_k(x)}{x}=1
$$
using the fact that $\lim_{x\to0}\frac{\sin x}{x}=1$.
Then your limit is
$$
-\frac{1}{6}-\frac{1}{6}-\frac{1}{6}=-\frac{1}{2}
$$
More generally,
$$
\lim_{x\to0}\frac{f_k(x)-x}{x^3}=-\frac{k}{6}
$$
A: Do not use l’Hospital here!
Let’s use $$\sin {(x)} =x-\frac{x^3}6+o(x^3)$$
As an alternative you could try with the substitution:
$$x=\arcsin y$$
A: This answer tries to clarify egreg's answer by using the function at hand instead of general functions.
If we can use that $\lim\limits_{x\to0}\frac{\sin(x)-x}{x^3}=-\frac16$ and $\lim\limits_{x\to0}\frac{\sin(x)}{x}=1$, then
$$
\begin{align}
&\lim_{x\to0}\frac{\sin(\sin(\sin(x)))-x}{x^3}\\
&=\lim_{x\to0}\frac{[\sin(\sin(\sin(x)))-\sin(\sin(x))]+[\sin(\sin(x))-\sin(x)]+[\sin(x)-x]}{x^3}\\
&=\lim_{x\to0}\frac{\sin(\color{#C00}{\sin(\sin(x))})-\color{#C00}{\sin(\sin(x))}}{\color{#C00}{\sin(\sin(x))}^3}\lim_{x\to0}\frac{\sin(\color{#C00}{\sin(x)})^3}{\color{#C00}{\sin(x)}^3}\lim_{x\to0}\frac{\sin(x)^3}{x^3}\\
&+\lim_{x\to0}\frac{\sin(\color{#C00}{\sin(x)})-\color{#C00}{\sin(x)}}{\color{#C00}{\sin(x)}^3}\lim_{x\to0}\frac{\sin(x)^3}{x^3}\\
&+\lim_{x\to0}\frac{\sin(x)-x}{x^3}\\
&=-\frac16\cdot1\cdot1-\frac16\cdot1-\frac16\\
&=-\frac12
\end{align}
$$
A: $$\lim_{x\rightarrow0}\frac{\sin\sin\sin{x}-x}{x^3}=\lim_{x\rightarrow0}\frac{\cos\sin\sin{x}\cos\sin{x}\cos{x}-1}{3x^2}=$$
$$=\lim_{x\rightarrow0}\tfrac{-\sin\sin\sin{x}\cos^2\sin{x}\cos^2x-\cos\sin\sin{x}\sin\sin{x}\cos^2x-\cos\sin\sin{x}\cos\sin{x}\sin{x}}{6x}=$$
$$=\lim_{x\rightarrow0}\tfrac{-\tfrac{\sin\sin\sin{x}}{\sin\sin{x}}\cdot\tfrac{\sin\sin{x}}{\sin{x}}\cdot\tfrac{\sin{x}}{x}\cdot\cos^2\sin{x}\cos^2x-\cos\sin\sin{x}\cdot\tfrac{\sin\sin{x}}{\sin{x}}\cdot\tfrac{\sin{x}}{x}\cdot\cos^2x-\cos\sin\sin{x}\cos\sin{x}\cdot\tfrac{\sin{x}}{x}}{6}=$$
$$=\frac{-1-1-1}{6}=-\frac{1}{2}.$$
