Let $f:[0,1] \to [0,1]$ be a smooth function and let $x_0\in[0,1]$ such that $$\left|\frac{\mathrm{d}f^n}{\mathrm{d}x}(x_0)\right| <e^{-2n}, \ \forall \ n\in\mathbb N,$$ where $$f^n := f \circ \ldots \circ f,\ n\ \text{times}.$$
I would like to know if the above conditions imply that there exists an open neighbourhood $U_{x_0}$ of $x_0,$ such that $$ y\in U_{x_0}\Rightarrow \lim_{n\to\infty}|f^n(x_0) - f^n(y)| = 0. $$
Or at least, that given $\varepsilon >0,$ $$ y\in U_{x_0}\Rightarrow |f^n(x_0) - f^n(y)|\leq \varepsilon,\ \forall \ n\geq n_0. $$
The only idea that I had was using Taylor expansion. For every $n\in\mathbb N$ we can write $$|f^n(x_0 + h) - f^n(x_0)| = | (f^n)'(x_0)| h + \mathcal O_n(h^2), $$ where $\mathcal O_n$ is function of order $2,$ that depends on $n.$ The problem is that
$$|f^n(x_0 + h) - f^n(x_0)| = | (f^n)'(x_0)| h + \mathcal O_n(h^2), $$ but I do not know why $\mathcal O_n(h^2),$ should be uniformly small on $n$.