We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Show that $\lim_{n\rightarrow+\infty}\left(\frac{1}{\varepsilon_{n+1}}-\frac{1}{\varepsilon_n}\right)=\frac{f''(1)}{2}$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Show that $\lim_{n\rightarrow+\infty}\left(\frac{1}{\varepsilon_{n+1}}-\frac{1}{\varepsilon_n}\right)=\frac{f''(1)}{2}$.