Show that $x_{n+1}=x_n(2-ax_n)$ converges and find the limit 
Let $a>0$ and $x_0\in I=\left (\frac{1}{2a}, \frac{3}{2a}\right )$. Show that the sequence $(x_n)$, $n\geq 0$, $$x_{n+1}=x_n(2-ax_n), \quad n \geq 0$$ converges. Which is the limit? Hint: Consider $\phi (x)=x(2-ax)$ and show that $\phi (I)\subset \left [\frac{3}{4a}, \frac{1}{a}\right ]$.


So that I understand  that correctly, with the hint we want to show that $\phi(x)$ is bounded and monotonic, which means that the sequence converges?
We have the following: \begin{align*}\frac{1}{2a}<x< \frac{3}{2a} &\Rightarrow a\cdot \frac{1}{2a}<ax<a\cdot  \frac{3}{2a}\Rightarrow \frac{1}{2}<ax< \frac{3}{2} \Rightarrow -\frac{3}{2}<-ax< -\frac{1}{2}\\ & \Rightarrow 2-\frac{3}{2}<2-ax< 2-\frac{1}{2}\Rightarrow \frac{1}{2}<2-ax< \frac{3}{2} \\ & \Rightarrow x\cdot \frac{1}{2}<x(2-ax)< x\cdot \frac{3}{2}\\ & \Rightarrow \frac{1}{2a}\cdot \frac{1}{2}<x\cdot \frac{1}{2}<x(2-ax)< x\cdot \frac{3}{2}< \frac{3}{2a}\cdot \frac{3}{2}\\ & \Rightarrow \frac{1}{4a}<x(2-ax)< \frac{9}{4a}\end{align*}
That is not the interval that we want to get. Do we have to do that maybe using derivatives?
 A: Complete the square on the right side to find
$$
1-ax_{n+1}=(1-ax_n)^2
$$
so that the convergence of $y_n=1-ax_n$ is very easy to discuss (subsequence of a geometric sequence).
A: 
P1. $\phi(x)$ has a maximum at $x_m=\frac{1}{a}$ and $\phi(x_m)=x_m$. Thus $\phi(x)\leq \frac{1}{a}$.



P2. $\phi(x)$ is ascending on $\left(-\infty,\frac{1}{a}\right]$ and descending on $\left(\frac{1}{a},\infty\right)$.

Since $\phi'(x)=2-2ax$


P3. For $x\in\left[\frac{1}{2a},\frac{3}{2a}\right] \Rightarrow
\phi(x)\in\left[\frac{3}{4a},\frac{1}{a}\right]$

Since $\frac{1}{2a}<\frac{1}{a}<\frac{3}{2a}$ and considering Pr2

*

*if $x\in\left[\frac{1}{2a},\frac{1}{a}\right] \Rightarrow \phi(x)\geq\phi\left(\frac{1}{2a}\right)=\frac{3}{4a}$

*if $x\in\left[\frac{1}{a},\frac{3}{2a}\right] \Rightarrow \phi(x)\geq\phi\left(\frac{3}{2a}\right)=\frac{3}{4a}$
And from Pr1, $\phi(x)\in\left[\frac{3}{4a},\frac{1}{a}\right]$.


Pr4. $\left[\frac{3}{4a},\frac{1}{a}\right]\subset\left[\frac{1}{2a},\frac{3}{2a}\right]$

Since
$\frac{1}{2a}<\frac{3}{4a}<\frac{1}{a}<\frac{3}{2a}$.


Pr5. $\phi(x)$ is a contraction map on $\left[\frac{3}{4a},\frac{1}{a}\right]$.

Since $$\forall x\in\left[\frac{3}{4a},\frac{1}{a}\right] \overset{Pr4}{\Rightarrow}
x\in\left[\frac{1}{2a},\frac{3}{2a}\right] \overset{Pr3}{\Rightarrow} 
\phi(x)\in\left[\frac{3}{4a},\frac{1}{a}\right]$$
and $\exists \epsilon$ in between $x,y$ such that
$$|\phi(y)-\phi(x)|=|\phi'(\epsilon)|\cdot|y-x|=
|2-2a\epsilon|\cdot|y-x|\leq \color{red}{\frac{1}{2}}\cdot|y-x|$$

Summary. Although $x_0$ may not fall inside $\left[\frac{3}{4a},\frac{1}{a}\right]$ (as seen in Pr4), $x_1=\phi(x_0)$ does (as seen in Pr3). And so does any $x_n$ for $\forall n\geq 1$ (as seen in Pr5). Applying Banach fixed-point theorem, the limit exists and it's a solution of
$$L=L(2-aL)$$
