TITLE: L'Hospital Rule with Trigonometry
QUESTION [6 upvotes]: 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 $$\lim_{x\to 0} \frac{\sin(\sin(\sin x)) - x}{x^3}\qquad?$$

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