# Show that the sequence $x_{n+1}=x_n(2-x_n)$ is convergent

QUESTION: The sequence $(x_n)$ is given by $x_{n+1}=x_n(2-x_n)$ and $0<x_1<1$. Show that the sequence is convergent.

MY ATTEMPT: As per the question, we have $x_{n+1}=x_n(2-x_n)$. But I can do nothing with this recursion. Is there any way that I can transform the recursion to a relation for $x_n$ in terms of $n$? Or is there some other way to prove the convergence?

P.S. Proof by monotone convergence theorem is preferred.

We have $$x_{n+1} = x_n(2-x_n) = 2x_n-x_n^2$$ This gives us $$1-x_{n+1} = 1-2x_n + x_n^2 = (1-x_n)^2$$ Setting $a_n=1-x_n$, we have the sequence $$a_{n+1} = a_n^2$$ where $a_1 \in (0,1)$. Now prove that $a_n<1$ and that $a_n$ is monotone decreasing and bounded below by $0$.
Now conclude that $\lim_{n \to \infty} a_n$ exists and compute the limit. Therefore compute $\lim_{n \to \infty} x_n$.
Let $f(x)=x(2-x)$. $[0,1]$ is stable by $f$, $f$ is strictly increasing on $[0,1]$ and over $y=x$, hence the sequence converges to $f(1)=1$