When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+...}}}=0$? Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(x)$ is defined for some $n$. The unique root $x_0$ of the function $f$ satisfies $f(x_0)=0$, and it can be alternatively expressed as the limit of the roots of $f_0, f_1, f_2, ...$. See the graphs below:

Can anyone find an expression equal to this limit? I realize that the chances of something nice and closed-form are slim - can we find a series, integral, or even nested radical representation of the real root of $f(x)$?

$\color{brown}{\textbf{The task standing.}}$
Let $-r$ is the root of $f(x).$
Since $f(x)$ is increasing function and 
$$f(-r)=0,\quad f(-1) > f_2(-1) = 0,$$ 
then $r>1.$
At the same time,
\begin{cases}
f(x+1) = f^2(x)-x\\
f(-r)=0\\
f(-r+1) = r\\
f(-r+2) = r^2+r-1\\
f(-r+3)= (r^2+r-1)^2 + r-2 = (r-1)(r^4+3r^2+2r+1)\dots.\tag1
\end{cases}
Besides,
\begin{cases}
f'(x+1) = 2f(x)f'(x)-1\\
f'(-r) = \infty\\
f'(-r+2) = 2rf'(-r+1)-1\\
f'(-r+3) = 2(r^2-r+1)f'(-r+2)-1\dots\tag2
\end{cases}
Conditions $(1),(2)$ allow to define unknown parameters in the various parametric representations of $f(x).$
$\color{brown}{\textbf{Inverse polynomial model.}}$
Let the inverse function is $f^{-1}(x)=g(y),$ then 
\begin{cases}
g(0) = -r\\
g'(0)=0\\
g(r) = -r+1\\
g(r^2+r-1) = -r+2\\
g((r-1)(r^4+3r^2+2r+1)) = -r+3,\tag3
\end{cases}
and the coefficients of the cubic approximating polynomial in the form of
$$g(y)=-r+qy^2+py^3\tag4$$
can be obtained from the algebraic system
\begin{cases}
-r+1 = -r + pr^2 + qr^3\\
-r+2 = -r + p(r^2+r-1)^2 + q(r^2+r-1)^3\\
-r+3 = -r + p(r-1)^2(r^4+3r^2+2r+1)^2 + q(r-1)^3(r^4+3r^2+2r+1)^3,
\end{cases}
with the single positive solution
$$p\approx 0.0622998,\quad q\approx 0.606587,\quad r\approx1.21088$$
(see also Wolfram Alpha results).
Therefore, the cubic model estimation of root of $f(x)$ is $\color{brown}{\mathbf{-1.21088}}.$
$\color{brown}{\textbf{Power model.}}$
Taking in account conditions $(1),$ can be used the explicit power model
$$f(x) = r(x+r)^p,\tag5$$
where
\begin{cases}
r^2+r-1 = r\cdot 2^p\\
(r-1)(r^3+3r^2+2r+1) = r\cdot 3^p,
\end{cases}

The power model estimation of root of $f(x)$ is $\color{brown}{\mathbf{-1.21168}}.$
Plot of the inverse functions for the considered models see below.

Using of detalized models can improve approximations accuracy.
