Maximum zero converges to $\sqrt{2}$ In my research I came upon a recursively defined sequence, and I'm pretty sure it converges to $\sqrt{2}$ though I can't prove it easily. I don't think it is a difficult question but I'm not sure.
Consider the following sequence of functions over $\mathbb{R}$, where it makes sense:
$f_0(x)=0$,
$\displaystyle f_{n+1}(x)=\frac{1}{2(x-f_n(x))}\hspace{1cm}$ ($n\geq 0)$.
Now let us define the sequence $(x_n)$ by $\displaystyle x_n:=\max\left\{y\in[0,\sqrt{2}[\,:\quad y=\frac{1}{2y}+f_n(y)\right\}$.

Question: is is true that $x_n\rightarrow\sqrt{2}$ when $n\rightarrow +\infty$?

Numerical evidence strongly suggest that, and it completely makes sense with the problem it originated from. The issue is that the functions $f_n$ have more and more poles as $n$ grows, and there is no function it converges to. It looks like the set of the poles of $f_n$ tends to be dense in $[-\sqrt{2},\sqrt{2}]$ when $n\rightarrow +\infty$, and $f_n$ is always decreasing outside of the poles. The poles seem to accumulate more around $\pm\sqrt{2}$ than around $0$.
For visual reference, one can see a graph of $f_{10}$ here
In advance, thank you for your interest/time.
Edit: I added the fact that I'm only interested in the $y\in[0,\sqrt{2}[$. I don't care what happens outside the interval since then it's trivial.
 A: Modify the function so the equation becomes $f_n(x)=0$. That is
$$f_0(x)=\frac{1}{2 x}-x \quad\quad\quad f_n(x)=\frac{1}{2 \left(\frac{1}{2 x}-f_{n-1}(x)\right)}-x+\frac{1}{2 x}$$
Multiply both side of the equation by $-U_n(\sqrt{1/2}x)x$
$$f_0(x)=-U_0(\sqrt{1/2}x)x\left(\frac{1}{2 x}-x\right) \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{ } \\ f_n(x)=-U_n(\sqrt{1/2}x)x\left(\frac{1}{2 \left(\frac{1}{2 x}-\frac{f_{n-1}(x)}{-U_{n-1}(\sqrt{1/2}x)x}\right)}-x+\frac{1}{2 x}\right)$$
Substitute $x\to\sqrt{2}x$ and make common denominator
$$f_0(x)=U_2(x) \quad\quad\quad f_n(x)=\frac{U_n(x) \left(\left(1-4 x^2\right) f_{n-1}(x)+U_{n-1}(x)\right)}{f_{n-1}(x)+U_{n-1}(x)}$$
Replace $f_{n-1}$ by $U_{n+1}(x)$ which is valid for $n=1$ and hopefully more.
Also replace $U_{n-1}(x)$ using the recurrence relation for $U_{n+1}(x)$, and expand
$$f_0(x)=U_2(x) \quad\quad\quad f_n(x)=\frac{U_n(x) \left(\left(1-4 x^2\right) U_{n+1}(x)+2 x U_n(x)-U_{n+1}(x)\right)}{U_{n+1}(x)+2 x U_n(x)-U_{n+1}(x)}
\\f_0(x)=U_2(x) \quad\quad\quad f_n(x)=U_n(x)-2 x U_{n+1}(x)= U_{n+2}(x)\hspace{3.8cm}\text{ } $$
Now it's obvious that the maximal root approaches 1 which correspond to $\sqrt{2}$ because of the substitution.
The multiplication just cancel out singularities, but even if it did introduce additional roots, these would be in the range $(0,1)$ which doesn't affect the limit of the maximal root.
A: No need for any analysis: the roots are roots of shifted Tchebyshev
polynomials, all of the form $\sqrt{2}\cos(\pi/(2k))$ for suitable $k$
in arithmetic progression.
A: You have $f_n(x)=g_x^n(0)$, where $g_x(y)=1/(2(x-y))$.  The fixed points of $g_x$ are $\alpha_x=(x+\sqrt{x^2-2})/2$ and $\beta_x=(x-\sqrt{x^2-2})/2$.  If we put $m_x(y)=(y-\alpha_x)/(y-\beta_x)$ and 
$$ \gamma_x=\frac{\alpha_x}{\beta_x}=x^2-1+x\sqrt{x^2-2} $$ we find that $g_x(y)=m_x^{-1}(\gamma_x\,m_x(y))$.  This gives 
$$ f_n(x) = m_x^{-1}(\gamma_x^n\,m_x(0)) = m_x^{-1}(\gamma_x^{n+1}). $$ 
You want $f_n(y)=y-1/(2y)$ or equivalently $\gamma_y^{n+1}=m_y(y-1/(2y))$.  However, one can check that $m_y(1-1/(2y))=\gamma_y^{-2}$, so you want $\gamma_y^{n+3}=1$.  For $y>\sqrt{2}$ one can check that $\gamma_y$ is real and strictly greater than $1$, so there are no solutions, as required.
