Let $a>0$ and $x_0>0$ and let $ax_n<2$.
The sequence $(x_n)$ is recursive defined by:
$$x_{n+1}:=x_n(1+\epsilon_n),\,\,\,\epsilon_n:=1-ax_n$$
Show that the sequence $(x_n)$ has the limit at $\frac{1}{a}$
Hint: Show at first that
$$\epsilon_{n+1}=\epsilon_n^2,\,\,\,\forall n\in \mathbb{N}$$
My attempt:
$\epsilon_{n+1}=1-ax_{n+1}=1-a(x_n(1+\epsilon_n))=1-a(x_n(1+1-ax_n))=1-2ax_n+(ax_n)^2=(1-ax_n)^2=\epsilon_n^2$
$ax_n<2\Longrightarrow|1-ax_n|<1\Longrightarrow \epsilon_0<1\Longrightarrow (b_n)$ with $b_n:=\epsilon_0^n$ has the limit $b_n\longrightarrow0$
and since $(b_n)=(\epsilon_n) \Longrightarrow \epsilon_n\longrightarrow 0$
Now $\epsilon_n:=1-ax_n\Longleftrightarrow x_n=\frac{1-\epsilon_n}{a}$
So we finally get:
$\lim\limits_{n\rightarrow\infty}x_n=\lim\limits_{n\rightarrow\infty}\frac{1-\epsilon_n}{a}=\frac{1-0}{a}=\frac{1}{a}$
$$x_n\longrightarrow \frac{1}{a}$$
$\Box$
It would be kind if someone could look over my work and give me feedback, if everything works out, and maybe also give some improving tips :) thank you