Maclaurin expansion of Sin(Sin(x)) I want to calculate the limit,
$$ \lim_{x\to 0} \frac{sin(sin x) - sin x}{x^3}$$
and doing so using Maclaurin expansion.
Now $sin x$ expands to $x -\frac{x^3}{3!}x^3 + O(x^5)$
Which would give $sin(sinx)= (x -\frac{1}{3!}x^3 + O(x^5)) -\frac{1}{3!}(x -\frac{x^3}{3!}x^3 + O(x^5))^3 + O(x^5)$
This expression should simplify to (I have the answer) $x - \frac{1}{3}x^3 + O(x^5)$ But i cannot see how this step comes about. Particulary, how do I handle the term $(x -\frac{x^3}{3!}x^3 + O(x^5))^3$?
Thanks a lot!
Alexander

You are on the right track. 
Since $\sin x=x-\frac{x^3}{6}+O(x^5)$, we have
$$
\begin{align}
\sin(\sin x)&=x-\frac{x^3}{6}+O(x^5)-\frac{1}{6}(x-\frac{x^3}{6}+O(x^5))^3+O((x-\frac{x^3}
{6}+O(x^5))^5)\\
&\text{Expanding $(x-\frac{x^3}{6}+O(x^5))^3=x^3+O(x^5)$, you get}\\
&=x-\frac{x^3}{6}-\frac{1}{6}x^3+O(x^5).
\end{align}
$$
Now I believe you can continue.
