We have the function $\phi:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ with $\phi(x) = \dfrac{2x+2}{x+2}$ being strictly increasing .
The only positive fixed point of $\phi$ is $\overline{x} = \sqrt{2}$. We consider now the sequence defined by $u_0=1$ and $u_{n+1} = \phi(u_n)$. We can note that the limit of this sequence is $\sqrt{2}$.
We have to express $(u_{n+1} - \overline{x})$ according to $(u_n-\overline{x})$ and $u_n$. Then, we have to deduce an equivalent of $(u_{n+1}-\overline{x})$ when $n \rightarrow +\infty$.
Could someone help me ? Thank you in advance.
** just a hint**
$$\phi (x)=2-\frac {2}{x+2} $$
$$u_{n+1}-\bar {x}=\phi (u_n)-\phi (\bar {x}) $$ $$=2\frac {u_n-\bar {x}}{(u_n+2)(\bar {x}+2)}$$