Find the limit of $f(n)$ where $f(n)=(2-f(n+1))^2$

Let $f:mathbb{N}to[0,2]$ such that $f(1)=2$ and:

$$f(n)=(2-f(n+1))^2qquadforall ninmathbb{N}$$

Prove that the limit of $f(n)$ as $ntoinfty$ exists and show it is equal to $1$; namely:

$$lim_{ntoinfty}f(n)=1$$

I couldn’t figure out how to prove the limit exists. Usually when I encounter these recursion relations I try to prove that $f(n)$ is bounded and monotonic, but after substituting some values of $n$ I realized (mayble falsley) that $f(n)$ is not monotonic at all. I’d be happy to hear your thoughts.

• Assuming the limit exists, I think I know how to prove it is equal to $1$ – All I need to do is to solve the equation $L=(2-L)^2$. The solutions are $L=4$ (not possible since $f(n)leq2$) and $L=1$ (which is the solution).

Thanks

Edit: I did manage to show that if for some $kinmathbb{N}, f(k+1)geq f(k+2)$, then also $f(k+1)geq f(k)$. This is because:

$$f(k+1)geq f(k+2)iff2-sqrt{f(k)}geq2-sqrt{f(k+1)}iff f(k)leq f(k+1)$$

Similarly, if $f(k+1)leq f(k+2)$, then also $f(k+1)leq f(k)$. This means that for odd values of $n$, $f(n)$ is monotonically decreasing, and for even values of $n$, $f(n)$ is monotonically increasing. $f(n)$ is obviously bounded, therefore, both the sequences $f(2n)$ and $f(2n+1)$ have a limit. The only problem I have know, is that I have to show they approach the same limit. In order to prove that, I thought maybe I should solve the following equation for $L$:

$$f(n+2)=2-sqrt{2-sqrt{f(n)}}implies L=2-sqrt{2-sqrt{L}}$$

But I don’t think I can.