<s>[INST] 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?
 [/INST] 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).
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