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.

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.
