What is a proof of this limit of this nested radical? It seems as if $$\lim_{x\to 0^+} \sqrt{x+\sqrt[3]{x+\sqrt[4]{\cdots}}}=1$$
I really am at a loss at a proof here. This doesn't come from anywhere, but just out of curiosity. Graphing proves this result fairly well.
 A: For any $2 \le n \le m$, let $\phi_{n,m}(x) = \sqrt[n]{x + \sqrt[n+1]{x + \sqrt[n+2]{x + \cdots \sqrt[m]{x}}}}$. I will interpret the expression we have as following limit.
$$\sqrt{x + \sqrt[3]{x + \sqrt[4]{x + \cdots }}}\;
= \phi_{2,\infty}(x) \stackrel{def}{=}\;\lim_{m\to\infty} \phi_{2,m}(x)$$
For any $x \in (0,1)$, we have $\lim\limits_{m\to\infty}(1-x)^m = 0$. This implies
the existence of an $N$ so that for all $m > N$, we have
$$(1-x)^m < x \implies 1 - x < \sqrt[m]{x} \implies \phi_{m-1,m}(x) = \sqrt[m-1]{x + \sqrt[m]{x}} > 1$$
It is clear for such $m$, we will have $\phi_{2,m}(x) \ge 1$. 
Recall for any $k > 1$ and $t > 0$, $\sqrt[k]{1 + t} < 1 + \frac{t}{k}$.
Start from $\phi_{m,m}(x) = \sqrt[m]{x} \le 1$, we have
$$\begin{align}
& 
\phi_{m-1,m}(x) = \sqrt[m-1]{x + \phi_{m,m}(x)} 
\le \sqrt[m-1]{x + 1} \le 1 + \frac{x}{m-1}\\
\implies & 
\phi_{m-2,m}(x) = \sqrt[m-2]{x + \phi_{m-1,m}(x)} 
\le \sqrt[m-2]{x + 1 + \frac{x}{m-1}} \le 1 + \frac{1}{m-2}\left(1 + \frac{1}{m-1}\right)x\\
\implies & 
\phi_{m-3,m}(x) = \sqrt[m-3]{x + \phi_{m-2,m}(x)} 
\le 1 + \frac{1}{m-3}\left(1 + \frac{1}{m-2}\left(1 + \frac{1}{m-1}\right)\right)x\\
& \vdots\\
\implies &
\phi_{2,m}(x) \le 1 + \frac12\left( 1 + \frac13\left(1 + \cdots \left(1 + \frac{1}{m-1}\right)\right)\right)x \le 1 + (e-2)x
\end{align}
$$
Notice for fixed $x$ and as a sequence of $m$, $\phi_{2,m}(x)$ is monotonic increasing. By arguments above, this sequence is ultimately sandwiched between $1$ and $1 + (e-2)x$. As a result, $\phi_{2,\infty}(x)$ is defined for this $x$ and satisfies
$$1 \le \phi_{2,\infty}(x) \le 1 + (e-2) x$$
Taking $x \to 0^{+}$, we get
$$1 \le \liminf_{x\to 0^+} \phi_{2,\infty}(x) \le \limsup_{x\to 0^+}\phi_{2,\infty}(x) \le \limsup_{x\to 0^+}(1 + (e-2)x) = 1$$
This implies $\lim\limits_{x\to 0^+} \phi_{2,\infty}(x)$ exists and equal to $1$.
A: Edit: there was a silly mistake in the original post, this only proves an upper bound (on the limit, provided it exists), not the actual limit.
For $x > 0$, you have, for any integers $0 \leq n \leq m$,
$$
1 \leq x^{\frac{1}{m}} \leq x^{\frac{1}{n}}.
$$
Denoting your expression $f(x)$ (I'm sweeping under the rug the question of showing it is well-defined for $x> 0$, but that should be the very first step -- you are dealing with an infinite notation, after all):
$$
\sqrt{x} \leq f(x) \leq \sqrt{x+\sqrt{x+\sqrt{x+\dots}}} \stackrel{\rm def}{=} g(x).
$$
(where we used the above remark for the upper bound).
By squeezing, it is sufficient to show that $g(x)\xrightarrow[x\to0^+]{} 1$ [Not true -- one obviously needs a lower bound that also goes to 1]. But note that $g$ satisfies (under the same caveat -- how is it well-defined?) the functional equation
$$
g(x)^2-x = g(x)\qquad\forall x > 0
$$
Solving the quadratic equation (in $g(x)$), we get
$$
g(x) = \frac{1\pm \sqrt{1+4x}}{2}.
$$
Since one must have $g>0$ (by its definition as a square root), we can eliminate the spurious solution, getting 
$$
g(x) = \frac{1+ \sqrt{1+4x}}{2}, \quad\forall x >0.
$$
It only remains to show that $\lim_{x\to 0^+} g(x) = 1$, which is immediate.
