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 $\lim_{n \to \infty} x_n$.
