Asymptotic behavior of $n$-fold iterate near the origin

Question

Suppose $f:\mathbb{R}\to\mathbb{R}$ and let $f^{n}$ be its $n$-fold iterate. If $f(x)\sim x$ as $x\to 0$, is $f^{n}(x)\sim f^{n-1}(x)$?
If not, what is a counterexample and what conditions on $f$ are sufficient?

This problem is motivated by the recent question $\lim_{x\to 0}\frac{\sin(\sin(\sin(\sin x)))}{\tan(\tan(\tan(\tan x)))}$. This computation reduced to arguing that $\sin(f(x))\sim f(x)$ where $f(x)$ is some iterate of $x\mapsto \sin x$ (and similarly for $x\mapsto\tan x$.) But while that reasoning makes sense for $x\mapsto \sin x$ and $x\mapsto \tan x$ (both of which are analytic near the origin) it's not obvious to me in what generality this claim actually holds.

Answer 1

You don't say what you mean by $\sim$.

If by $f(x)\sim x$ you mean $f(x)=x+o(x^2)$, then $$ f(f(x))=f(x+o(x^2))=x+o(x^2)+o(x^2)=x+o(x^2), $$ so $f^n(x)\sim x$ for any $n$.

If by $f(x)\sim x$ you mean $\lim_{x\to0}\frac{x}{f(x)}=1$, then letting $g(x)=\frac x{f(x)}-1$ you have $\lim_{x\to0}g(x)=0$, and $$ f(x)=\frac{x}{1+g(x)}. $$ Then $$ \frac{f(x)}{f(f(x))}=\frac{\frac{x}{1+g(x)}}{\frac{f(x)}{1+g(f(x))}}=\frac{x}{f(x)}\,\frac{1+g(f(x))}{1+g(x)}\xrightarrow[x\to0]{}1, $$ so $f^2(x)\sim f(x)$. Now you can iterate.

Answer 2

The answer is always positive, I think.

$$\lim_{x\to0}\frac{f(f(x))}{f(x)} = \lim_{x\to0}\frac{f(\tfrac{f(x)}{x}x)}{f(x)} \overset{(1)}{=} \lim_{x\to0}\frac{\frac{f(x)}{x}x}{f(x)}=1, $$

and $(1)$ is true, because $\lim_{x\to0}\frac{f(x)}{x}x=1\cdot0=0$.

Now, use induction.