What is the global maximum of $x^{1/x}$ Let the following function be defined as such:
$$F_x: \Bbb R \to \Bbb C, x \mapsto x^{1/x}, \forall x \ne 0$$
What I want to know is 
$$\max_{x<0}\Re\left(F_x\right)=\,?$$
and
$$\max_{x<0}\Im\left(F_x\right)=\,?$$

Additional Questions


*

*Why does $F$ act weird when approaching $0$ form the left? For example $$\lim_{x\to0^-}F_x=\infty \text{ and }\lim_{x\to0^+}F_x=0$$

*What is with the oscillation when approaching?

*How are $\Re(F_x)$ and $\Im(F_x)$ related during the left approach?

*Are there any other functions with similar behaviors?

 A: For $x\gt0$, we have
$$
x^{1/x}=e^{\log(x)/x}
$$
As $x\to0^+$, $\log(x)/x\to-\infty$.
As $x\to+\infty$, $\log(x)/x\to0$.
In between, $\log(x)/x$ has a maximum of $1/e$ at $x=e$. We get this by looking at its derivative, which is $\frac{1-\log(x)}{x^2}$.
Therefore, for $x\gt0$, we have an infimum of $0$ as $x\to0^+$ and a maximum of $e^{1/e}$ at $x=e$.

For $x\lt0$, let $x=-t$. Then, taking $\log(x)=\log(-t)=\log(t)+i\pi$, we get
$$
\begin{align}
x^{1/x}
&=e^{-(\log(t)+i\pi)/t}\\
&=(\cos(\pi/t)-i\sin(\pi/t))e^{-\log(t)/t}
\end{align}
$$
As $t\to0^+$ (that is, $x\to0^-$), $-\log(t)/t\to+\infty$.
As $t\to+\infty$ (that is $x\to-\infty$), $-\log(t)/t\to0$.
In between, $-\log(t)/t$ has a minimum of $-1/e$ at $t=e$.
Therefore, for $t>0$, $e^{-\log(t)/t}$ have a minimum of $e^{-1/e}$ at $t=e$ and a supremum of $+\infty$ as $t\to0^+$.
However, since
$$
\begin{align}
\mathrm{Re}\left(x^{1/x}\right)&=\hphantom{-}\cos(\pi/t)e^{-\log(t)/t}\\
\mathrm{Im}\left(x^{1/x}\right)&=-\sin(\pi/t)e^{-\log(t)/t}
\end{align}
$$
the fact that $e^{-\log(t)/t}$ grows without bound and the $\sin(\pi/t)$ and $\cos(\pi/t)$ oscillate faster and faster between $+1$ and $-1$ as $t\to0^+$ means that
$$
\begin{align}
\limsup_{x\to0^-}\mathrm{Re}\left(x^{1/x}\right)
&=\limsup_{x\to0^-}\mathrm{Im}\left(x^{1/x}\right)=+\infty\\
\liminf_{x\to0^-}\mathrm{Re}\left(x^{1/x}\right)
&=\liminf_{x\to0^-}\mathrm{Im}\left(x^{1/x}\right)=-\infty\\
\end{align}
$$
which in turn imply
$$
\bbox[5px, border: 1px solid #C00000]{
\begin{align}
\sup_{x\lt0}\mathrm{Re}\left(x^{1/x}\right)
&=\sup_{x\lt0}\mathrm{Im}\left(x^{1/x}\right)=+\infty\\
\inf_{x\lt0}\mathrm{Re}\left(x^{1/x}\right)
&=\inf_{x\lt0}\mathrm{Im}\left(x^{1/x}\right)=-\infty\\
\end{align}
}
$$

Plots of $\mathrm{Re}\left(x^{1/x}\right)$ and $\mathrm{Im}\left(x^{1/x}\right)$:
$\hspace{3.2cm}$
$\hspace{3.2cm}$

Additional Answers
$\ \small\bullet$ The weird behavior when approaching $0$ from the left is two-fold:
First, for $x\lt0$, $\log(|x|)/x\to+\infty$ as $x\to0^-$; therefore, $\left|\,x^{1/x}\,\right|=e^{\log(|x|)/x}\to+\infty$.
Second, due to the $\pi i$ in $\log(x)$ for $x\lt0$, $\arg\left(x^{1/x}\right)=\pi/x$ forces both real and imaginary parts to oscillate positive and negative.
$\ \small\bullet$ For $x\lt0$, $\mathrm{Re}\left(x^{1/x}\right)^2+\mathrm{Im}\left(x^{1/x}\right)^2=e^{\log(|x|)/x}$ and $\frac{\displaystyle\mathrm{Im}\left(x^{1/x}\right)}{\displaystyle\mathrm{Re}\left(x^{1/x}\right)}=\tan(\pi/x)$.
A: Yes, of course. Now for those real values.
For x → 0 you have two cases: one is where the function has real values.  If you look at x values like -1/n Sr(-1/n) = (-1/n)-n = ± 1/nn which in absolute value → 0.  You'll get the same result with any sequence of x → 0- for which Sr(x) is real.
Now the case where Sr(x) is complex. Consider for example x = -8/9.  (-8/9)-9/8 = (8/9)-9/8(-1)-9/8.  The roots of -1 are all on the unit circle, so the absolute value cannot exceed |(-8/9)-9/8| =(9/8)9/8, but it could be close if the root you pick has absolute value near 1. Consider |Sr(-4/9)| is near (9/4)9/4, a larger base and a larger exponent.  Since the variance in the absolute value of the roots closest to 1 should be small, it looks as if this kind of sequence will → ∞ You have a few things to nail down to formalize this.  Once that is done you can trap other sequences with complex Sr within the sequences you know are → ∞.
However, you've got the problem that some of the subsequences are real and go to zero.  So I would say that generally you do not have a limit.  If you want the ∞ limit, you are going to have to restrict yourself to sequences where Sr(x) is complex.  
You have I think an additional problem, which is that if you pick the zeros of (-1) whose absolute values are closest to 0, the |Sr| may not → ∞.
Hope this helps.  
A: Correct me if im wrong, but is this not Steiner's problem concering the Euler number? 
He found it using an inequality, $e^{(x-e)/e} \geq 1 + \dfrac{x-e}{e}$.
