<s>[INST] Evaluate $\lim\limits \frac{x-\sin\sin\cdots\sin x}{x^3}.$ Problem
Evaluate $$\lim\limits_{x \to 0} \frac{x-\sin\sin\cdots\sin x}{x^3},$$where $\sin\sin\cdots\sin x$ denotes n-fold composite sine function.
Solution
Consider applying Taylor's Formula with 3-order at $x=0$. We may obtain $$f_n(x)=\sin\sin\cdots\sin x=x-\frac{n}{6}x^3+\mathcal{O}(x^3).\tag{*}$$ To prove this, we can apply the mathematical induction. Let $n=1,$ then $$f(x)=\sin x=x-\frac{1}{6}x^3+\mathcal{O}(x^3),$$  It's true and shows that $(*)$ holds for $n=1$. Assume that $(*)$ holds for $n=k$. Then $$\begin{align*}f_{k+1}(x)&=\sin(f_k(x))\\&=x-\frac{k}{6}x^3+\mathcal{O}(x^3)-\frac{1}{6}\left(x-\frac{k}{6}x^3+\mathcal{O}(x^3)\right)^3+\mathcal{O}(x^3)\\&=x-\frac{k+1}{6}x^3+\mathcal{O}(x^3)\end{align*}.$$ This shows that $(*)$ holds for $n=k+1$. As a result, $(*)$ holds for all $n=1,2,\cdots.$
Now,let's go back to deal with the problem. $$\lim\limits_{x \to 0} \frac{x-\sin\sin\cdots\sin x}{x^3}=\lim\limits_{x \to 0} \dfrac{x-\left(x-\dfrac{n}{6}x^3+\mathcal{O}(x^3)\right)}{x^3}=\frac{n}{6}.$$
Please correct me if I'm wrong. Hope to see other solutions. Thanks.
 [/INST] I would like to share a more tricky method with you.
Use the same notation, and assume $f_0(x) = x$, then we can prove that
$$
\lim_{x \to 0} \frac{x - f(x)}{x^3} = \frac{1}{6}
$$
and
$$
\lim_{x \to 0} \frac{f_k(x)}{x} = 1, ~ \forall k \in \mathbb{N}.
$$
Thus we have
$$
\begin{aligned}
\lim_{x \to 0} \frac{f_0(x) - f_k(x)}{x^3} &= \lim_{x \to 0}\sum_{i = 1}^k \frac{f_{i - 1}(x) - f_{i}(x)}{x^3}\\
&=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f_{i}(x)}{x^3}\\
&=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f(f_{i-1}(x))}{(f_{i-1}(x))^3}\cdot \frac{(f_{i-1}(x))^3}{x^3}\\
&=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f(f_{i-1}(x))}{(f_{i-1}(x))^3} \cdot \lim_{x \to 0}\frac{(f_{i-1}(x))^3}{x^3}\\
&= \sum_{i = 1}^k \frac{1}{6}\\
&= \frac{k}{6}.
\end{aligned}
$$
</s>