$x_1$ such that given sequence $(x_n)$ converges Let $(x_n)_{n \in \mathbb N}$ be a sequence in $\mathbb R$ such that $|4-x_{n+1}|&lt;q|4-x_n|^2$, $\ q&gt;0$

Under what circumstances for the initial value $x_1$ can we guarantee that $(x_n)_{n\in \mathbb N}$ converges?
One can show by induction that
$$ \forall n\in\mathbb{N},\,|4-x_n|\leqslant q^n|4-x_0|^{2^n} $$
If $x_0\in]3,5[$, then $|4-x_0|&lt;1$ and $\lim\limits_{n\rightarrow +\infty}x_n=4$, if $x_0\in\{3,5\}$, then $|4-x_0|=1$ and $(x_n)$ converges if $q&lt;1$, in this case the limit is $4$. If $x_0\notin[3,5]$, then $|x_0-4|&gt;1$ and $\lim\limits_{n\rightarrow +\infty}{|x_0-4|^{2^n}}=+\infty$, we can't say anything.
One can show by induction that
$$ \forall n\in\mathbb{N},\,|4-x_n|\leqslant q^n|4-x_0|^{2^n} $$
If $x_0\in]3,5[$, then $|4-x_0|&lt;1$ and $\lim\limits_{n\rightarrow +\infty}x_n=4$, if $x_0\in\{3,5\}$, then $|4-x_0|=1$ and $(x_n)$ converges if $q&lt;1$, in this case the limit is $4$. If $x_0\notin[3,5]$, then $|x_0-4|&gt;1$ and $\lim\limits_{n\rightarrow +\infty}{|x_0-4|^{2^n}}=+\infty$, we can't say anything.