L'Hospital Rule with Trigonometry

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, $$ \lim_{x\to0}\frac{f_k(x)-x}{x^3}=-\frac{k}{6} $$

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:

$$x=\arcsin y$$

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 $$ \begin{align} &\lim_{x\to0}\frac{\sin(\sin(\sin(x)))-x}{x^3}\\ &=\lim_{x\to0}\frac{[\sin(\sin(\sin(x)))-\sin(\sin(x))]+[\sin(\sin(x))-\sin(x)]+[\sin(x)-x]}{x^3}\\ &=\lim_{x\to0}\frac{\sin(\color{#C00}{\sin(\sin(x))})-\color{#C00}{\sin(\sin(x))}}{\color{#C00}{\sin(\sin(x))}^3}\lim_{x\to0}\frac{\sin(\color{#C00}{\sin(x)})^3}{\color{#C00}{\sin(x)}^3}\lim_{x\to0}\frac{\sin(x)^3}{x^3}\\ &+\lim_{x\to0}\frac{\sin(\color{#C00}{\sin(x)})-\color{#C00}{\sin(x)}}{\color{#C00}{\sin(x)}^3}\lim_{x\to0}\frac{\sin(x)^3}{x^3}\\ &+\lim_{x\to0}\frac{\sin(x)-x}{x^3}\\ &=-\frac16\cdot1\cdot1-\frac16\cdot1-\frac16\\ &=-\frac12 \end{align} $$