Let $f: \mathbb{R} \to \mathbb{R}$ and $\exists \ \ b \in \mathbb{R} : f(x+b)=\sqrt{f(x)-f^2(x)}+\frac{1}{2}$ 
Let $f: \mathbb{R} \to \mathbb{R}$ and $\exists  \ \ b \in \mathbb{R} : f(x+b)=\sqrt{f(x)-f^2(x)}+\frac{1}{2}$
  then find the :
$$\lim_{x \to \infty} f(x)=?$$

My Try : $f^2(x+b)+\frac{1}{4}-f(x+b)=f(x)-f^2(x)$
and let $b=0$ then $f^2(x)+\frac{1}{4}-f(x)=f(x)-f^2(x)$ so $b \neq 0$
now what do i do ?

Rewrite the starting equation like this:
$$\left(f(x+b)-\frac{1}{2}\right)^2={1\over 4}-{1\over 4}+{f(x)-f^2(x)}= {1\over 4}- \left(f(x)-{1\over 2}\right)^2$$
so we have also 
$$\left(f(x)-\frac{1}{2}\right)^2= {1\over 4}- \left(f(x-b)-{1\over 2}\right)^2$$
Combining both equations we get $$\left(f(x+b)-\frac{1}{2}\right)^2 = \left(f(x-b)-{1\over 2}\right)^2$$
and so $$\left|f(x+b)-\frac{1}{2}\right| = \left|f(x-b)-{1\over 2}\right|$$
But $f(x)\geq 1/2$ for all $x$ so we have $$f(x+b)-\frac{1}{2} = f(x-b)-{1\over 2}$$
which means that $f$ is periodic with period $2b$ and thus the limit doesn't exist unless $f$ is constant.
