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?