The function $f(x)=\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x...}}}}}\qquad$ quickly approaches $f(x) =x$.
$$f(x)=\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x...}}}}}\\=\sqrt{xf(x)}\\ \left[f(x)\right]^2=xf(x)\\\left[f(x)\right]^2-xf(x)=0\\f(x)[f(x)-x]=0\\ f(x)=0\\ f(x)=x$$
To say $2=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}\qquad$ makes intuitive sense. What do I do with $0=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}\qquad$?
On
You can also talk about stable and unstable fixed points, so in this setting, $x$ is a stable fixed point of the dynamical system defined by $f:z\mapsto \sqrt{xz}$ : there is a neighbourhood $V$ of $x$ such that for $y$ in $V$, the sequence defined by $a_0 = y$ and $a_{n+1} = f(a_n)$ for any $n\in \mathbb{N}$, converges to $x$. In contrast, $0$ is an unstable fixed point, for any neighbourhood $W$ of $0$, all the sequences starting at a point of $W - \{0\}$ will not have $0$ as an accumulation point.
For binary $\circ$, the usual convention is to define $a\circ(a\circ(a\circ\cdots))$ as $\lim_{n\to\infty}a_n$ for $a_0:=a,\,a_{n+1}:=a\circ a_n$, not simply any fixed point of $x\mapsto a\circ x$. So you don't need to "do anything with" fixed points other than the resulting limit, which is why @soupless called such apparent alternatives extraneous. (Finally, if the limit doesn't exist, the expression is undefined.)