What the minimum of infinite tetration divided by $\sqrt{x}$?

Question

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?

Answer 1

$\def\W{\operatorname W}$ The asker’s plot and numerical minimum do not match, but the following does match their plot:

$$\min\left(\frac{x^{x^{\unicode{x22F0} }}}{\sqrt x}\right)=0.902858\dots,x=0.579687\dots$$

Rewriting the infinite tetration with Lambert W and using $x=e^{-we^w}$:

$$-\frac d{dx}\frac{\W(-\ln(x))}{\sqrt x\ln(x)}=\frac{\W(-\ln(x))(\ln(x)+\W(-\ln(x))(\ln(x)+2))}{2(\W(-\ln(x))+1)x^\frac32\ln^2(x)}=0\implies e^w(w+1)=2$$

Therefore the minimum is at:

$$\left(e^{\frac2{\W(2e)}-2},\frac{\W\left(2-\frac2{\W(2e)}\right)\W(2e)}{2 e^{\frac1{\W(2e)}-1}(\W(2e)-1)}\right)$$

shown here: