Maximum zero converges to $\sqrt{2}$

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.

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.