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

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


Let
$$
a_n=\overbrace{\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-\dots}}}}}^{n\ x\text{s}}\tag{1}
$$
If $x\lt1$, then $a_2^2=x-\sqrt{x}\lt0$. If $x=1$, then $a_n$ oscillates between $0$ and $1$.  So we will assume that $x\gt1$.

Since $a_n^2=x-a_{n-1}$, we have $a_n^2-a_{n-1}^2=a_{n-2}-a_{n-1}$. Therefore,
$$
a_n-a_{n-1}=-\frac{a_{n-1}-a_{n-2}}{a_n+a_{n-1}}\tag{2}
$$
Since $f(a)=a+\sqrt{x-a}$ is convex, $f(0)=\sqrt{x}$, and $f(x)=x$, we have that 
$$
f(a)\ge\sqrt{x}\tag{3}
$$
Therefore,
$$
a_n+a_{n-1}=f(a_{n-1})\ge\sqrt{x}\gt1\tag{4}
$$
Combining $(2)$ and $(4)$, we get
$$
\left|a_n-a_{n-1}\right|\le\frac{\left|a_{n-1}-a_{n-2}\right|}{\sqrt{x}}\tag{5}
$$
Since $x\gt1$, the sequence $a_n$ converges.

For a given $x\gt1$, let
$$
a(x)=\lim_{n\to\infty}a_n\tag{6}
$$
Then taking the limit of
$$
a_n^2=x-a_{n-1}\tag{7}
$$
we get that $a(x)^2+a(x)-x=0$; that is,
$$
a(x)=\frac{-1+\sqrt{1+4x}}2\tag{8}
$$
Therefore,
$$
\lim_{x\to1^+}a(x)=\frac{-1+\sqrt5}2\tag{9}
$$
