An exercise says: if the real sequence $(x_n)$ is defined by $x_{n+1}=F(x_n)$ for some $F$, $x_n\to x$ and $F'(x)=0$, prove that $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$.
The textbook says assume $F$ is continuously differentiable and apply the mean-value theorem. It's straightforward: $$\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n}=\frac{F(x_{n+1})-F(x_n)}{x_{n+1}-x_n}=F'(y_n)$$ for $x_n How can it be proved without assuming the continuity of the derivative? It holds (by differentiability and continuity of $F$ at $x$) that $$0\gets \frac{F(x_n)-F(x)}{x_n-x} = \frac{x_{n+1}-x}{x_n-x}$$ Can the limit of $$\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n}$$ be obtained from here?