For some values of $x$ the limit of infinite tetration converges. For example when $x=\sqrt{2}$ this is fixed point
$$\lim_{n\rightarrow \infty} \sqrt{2} \uparrow \uparrow n = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdot^{\cdot^{\cdot}}}}} = \sqrt{2}$$
This limit converges for all $e^{-e} \le x \le e^{1/e}$. What is the minimum of this function?
$$f(x) = \frac{\lim_{n\rightarrow \infty} \sqrt{x} \uparrow \uparrow n}{\sqrt{x}} $$
Numerically the answer looks to have a single minima at $f(x\approx 0.579690102929) = 0.687413037245$, is there an exact solution?
