What is the limit of $x^{e^{x}}$ if $x$ tends to $-\infty$? I was looking at the function $f(x)=x^{e^{x}}$ and was curious as to why its domain is defined for $x\geq0$, when it looks like there are no problems with negative values of $x$. Also, when plugging in negative values for $x$, I noticed that it looks like $f(x)$ is approaching $-1$, but when doing the actual limit, this is the result: $$\lim_{x\to-\infty} f(x) = 1$$
Could someone give me an explanation as to why these are the results? I feel really lost. I want to say that it has to do with trying to take $\ln({f(x)})$, but I am not really sure if that has anything to do with it. Any help/insight is greatly appreciated. Thanks.
 A: First of all:
The function at stake is not defined on the negative reals.
However, if one considers complex numbers then a negative real $(x<0)$ as a complex number can be written as $$z=-|x|+i0,\,\,\text{ or } \,\, z=|x|e^{i\pi}.$$
With this in mind
$$z^{e^z}=e^{\ln\left(z^{e^z}\right)}=e^{e^z\ln(z)}=e^{e^z\ln\left(|x|e^{i\pi}\right)}=e^{e^z\ln(|x|)+e^xi\pi}=e^{e^{-|x|}\ln(|x|)}e^{e^{-|x|}i\pi}. \tag 1$$
Since
$$\lim_{x\to -\infty}e^{-|x|}\ln(|x|)=0$$
and
$$\lim_{x\to -\infty}e^{-|x|}i\pi=0$$
the original limit is, indeed $1=1+i0$.
However, $e^{e^{-|x|}i\pi}=(-1)^{e^{-|x|}}$, the second term in $(1)$, goes through the path shown below, while $x$ goes through the interval  $[0,\infty)$.

The other factor in $(1)$ is real. So, for the time being I don't understand how to get negative numbers by substituting negative reals in  $x^{e^x}$ -- let alone that substituting negative reals here is not comme il faut.
BTW, here is how Wolfram $\alpha$ plots our function:

(Here the blue line represents the real part and the orange line represents the imaginary part.) For negative $x$'s the result is never real. The limit, of course, is $z=1+i0)$.
