Solve the equation $\lim_{n\to\infty}\sqrt{1+\sqrt{x+\sqrt{x^2+\dots+\sqrt{x^n}}}}=2$ 
Solve the equation 
  $$\lim_{n\to\infty}\sqrt{1+\sqrt{x+\sqrt{x^2+\dots+\sqrt{x^n}}}}=2$$

I think there can only prove the existence of a solution. And evaluate the root indicating a segment.
 A: For any $0 \le n \le m < \infty$, let $$g_{mn}(x) \stackrel{def}{=} \sqrt{x^n + \sqrt{x^{n+1} + \cdots \sqrt{x^{m-1} + \sqrt{x^m}}}}
\quad\text{ and }\quad
g_{\infty n}(x) \stackrel{def}{=} \lim_{m\to\infty} g_{mn}(x)
$$
The question at hand is equivalent to finding a $x$ such that $g_{\infty 0}(x)$ exists and equals to $2$. 
We claim that $x = 4$ is the unique solution.
For simplicity of presentation, we will use $g_{mn}$ as a shorthand of $g_{mn}(4)$.
It is clear $g_{mn} \ge 2^n$ for any admissible pair of $m,n$. Notice
$$\begin{align}
 g_{mm} &= 2^m < 2^m +1\\
\implies g_{m,m-1} &= \sqrt{4^{m-1} + g_{mm}} < \sqrt{4^{m-1}+ 2^m+1} = 2^{m-1} + 1\\
\implies g_{m,m-2} &= \sqrt{4^{m-2} + g_{m,m-1}} < \sqrt{4^{m-2} + 2^{m-1} + 1} = 2^{m-2}+1\\
&\;\vdots \\
\implies g_{m0} &= \sqrt{4^0 + g_{m1}} < \sqrt{1 + 2^1 + 1} = 2\\
\end{align}
$$
We find $2^n \le g_{mn} \le 2^n + 1$ in general.
For any fixed $m$, let $\epsilon_n = 2^n + 1 - g_{mn} \in [0,1]$. 
Notice for $1 \le n \le m$, we have
$$\begin{align}
\epsilon_{n-1} &= 2^{n-1} + 1 - \sqrt{ 4^{n-1} + g_{mn}}
= 2^{n-1}+1 - \sqrt{(2^{n-1}+1)^2 - \epsilon_n}\\
&= \frac{\epsilon_n}{
2^{n-1}+1 + \sqrt{(2^{n-1}+1)^2 - \epsilon_n}}
\le \frac{\epsilon_n}{2^n+1} \le \frac{\epsilon_n}{3}
\end{align}$$
This implies 
$$\epsilon_{0} \le \frac{\epsilon_m}{3^m} = \frac{1}{3^m}
\quad\implies\quad |g_{m0} - 2| \le \frac{1}{3^m}
$$
As a result, $g_{\infty 0}(4) = \lim\limits_{m\to\infty} g_{m0} = 2$ and $x = 4$ is a solution.
To see $x = 4$ is the unique solution. Let $X \subset [0,\infty)$ be the 
collection of $x$ where $g_{\infty 0}(x)$ exists as a limit. For any $x \in X$, it is clear the existence of $g_{\infty 0}(x)$ implies the existence of $g_{\infty n}(x)$ for any $n$. 
For any $u, v \in X$ with $u < v$, notice
$$\begin{align}
& g_{m2}(u) < g_{m2}(v), \text{ for all } m \ge 2\\
\implies & g_{\infty 2}(u) \le g_{\infty 2}(v)\\
\implies & g_{\infty 0}(u) = \sqrt{1 + \sqrt{ u + g_{\infty 2}(u)}} < \sqrt{1 + \sqrt{ v + g_{\infty 2}(v)}} = g_{\infty 0}(v)
\end{align}
$$
$g_{\infty 0}(\cdot)$ is strictly monotonic on $X$ where it is defined. As a result, any equation of the form $g_{\infty 0}(x) = y$ has at most one solution.
A: It is an increasing sequence as $n$ increases, bounded above (if $x>1$) by 
$$\sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+...\sqrt{x^{2^n}}}}}}}$$
If I replace the final term $\phi^2x^{2^n}$, I think my upper bound becomes $\sqrt{1+\phi\sqrt{x}}$, with the golden ratio.
Then we have bounds on $x$, namely
$$\sqrt{1+\sqrt{2x}}<2<\sqrt{1+\phi\sqrt{x}}\\ \frac92>x>\frac9{\phi^2}$$
