TITLE: Finding $\lim_{x\rightarrow 1}\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$
QUESTION [6 upvotes]: Finding

$$\lim_{x\rightarrow 1}\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$$

REPLY [2 votes]: For fixed $a\ge0 $ and $0\le x\le a$, let $f_a(x)=\sqrt{ a-x}$. Define $f_a^{\circ0}(x)=x$ and recursively $f_a^{\circ(n+1)}(x)=f_a(f_a^{\circ n}(x))$.
We want to define
$$ F(a)=\lim_{n\to\infty}f_a^{\circ n}(0).$$
(At least, this seems to be the most acceptable way to define the nested root expression).
The basic questions are: 

For which values of $a$ are all terms $f_a^{\circ n}(0)$ defined (i.e., a real number)?
And for which of these $a$ does the limit exist?

For $a=0$, we have $f_a^{\circ n}(0)=0$ for all $n$, hence no problem with definedness and the limit.
For $0<a<1$, we have $\sqrt a>a$, hence already $f_a^{\circ 2}(0)=f_a(f_a(0))=\sqrt{a-\sqrt a}$ is not real.
For $a=1$, the values of $f_a^{\circ n}(0)$ flip between $0$ and $1$, hence all terms are defined, but the limit does not exist.
Note that $f_a^{\circ 2}$ is strictly increasing on $[0,a]$. As certainly $f_a^{\circ n}(0)\le \sqrt a$ for all $n$, we conclude that $f_a^{\circ 2}(0)$ converges to some limit $G(a)$.
Then $G(a)$ must be a fixed point of $f_a^{\circ 2}$, i.e., a solution of
$$ (a-X^2)^2=a-X,$$
transformed:
$$ \tag1(X^2+X-a)(X^2-X-a+1)=0.$$
The first factor in $(1)$ corresponds to fixed points of $f_a$, the second factor to points of order $2$.
The latter are interchanged by $f_a$. But for $a>1$, one of them is negative (their product $-a+1$ is negative). Since neither $G(a)$ nor $f_a(G(a))$ can be negative, we can ignore the second factor and are left with $X^2+X-a$, which has two real roots $\frac{-1\pm\sqrt{4a+1}}{2}$.
Again, one of the roots is negative, so that the only possibility is $G(a)=\frac{-1+\sqrt{4a+1}}{2}$, and, as this is a fixed point of $f_a$, also $F(a)=\frac{-1+\sqrt{4a+1}}{2}$.
We conclude 
$$\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x\ldots}}}}=\begin{cases}0&\text{for $x=0$}\\\frac{-1+\sqrt{4x+1}}{2}&\text{for $x>1$}\\\text{undefined}&\text{otherwise}\end{cases} $$
and consequently the right-sided limit at $1$ is 
$$\lim_{x\to 1^+}\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x\ldots}}}}=\lim_{x\to1^+}\frac{-1+\sqrt{4x+1}}{2}=\frac{\sqrt 5-1}{2}.$$