# 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.

• Just consider the value $x = \ln(1/2)$. The nature problem rises immediately – Andrei Kulunchakov Oct 29 '15 at 7:14
• There are huge problems with x being negative. Consider $-1^{1/3}$ that equals -1. Now consider $-1^{1/2}$ that is undefined on the real numbers. Now consider $-1^{50000000/100000001 = 0.4999999950000000499999995}$ that equals -1 but $-1^{50000000/100000002 =0.49999999000000019999999600000008}$ is undefined again with values of -1 and undefined flipping back and forth in infintisimally small intervals. And that's just the rational powers. $e^x$ is likely irrational and we can't even begin to define $neg^{irrational}$. – fleablood Oct 29 '15 at 7:27
• As for the limit as $x\to-\infty$, you're probably substituting "special values". I'm curious as to which numbers you're substituting, because unless $e^x$ is a fraction with odd denominator, I don't see how you can evaluate $f(x)$. – Christopher Carl Heckman Oct 29 '15 at 7:28

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 $z=1+i0)$.

• This is a valorous try to make sense of the question. For completeness (and because this might be the most important piece of information to the OP), one should add that, prima facie, the function is not defined on the negative real numbers. Otherwise, the risk is that the OP thinks that $x^a$ is cool for every real $a$ and negative real $x$ (the WA "answer" at the end only reinforcing this misleading impression). – Did Oct 29 '15 at 12:00
• @Did, Thank you for your comment. – zoli Oct 29 '15 at 12:18