real analysis - Is really :$displaystyle lim _{xto-infty}e^x log x=0$ and what it does meant this in wolfram alpha?

I have tried to know more about limit calculations about the product of the function $f$ and it's inverse compositional $f^{-1}$ as example I have took the bellow example which mixed me in wolfram alpha , the function $e^x \log x$

defined over $(0,+\infty)$ and it has a limit equal's $0$ at $x= -\infty$ , as shown here in wolfram alpha , now my question here is

Question:
Is really $\displaystyle \lim _{x\to-\infty}e^x \log x=0$ and what it does meant this in wolfram alpha?

Answer

What does this mean? Well, if we let this be a complex valued function then we have that $\frac{d}{dz} \log(z) = \frac 1z$, and so

$$\lim_{x \to -\infty} \log(x)e^x$$
$$=\lim_{x \to \infty} \frac{\log(-x)}{e^x}$$
Where we let $x \to -x$. Now we apply L'Hôpital to get
$$=\lim_{x \to \infty} \frac{-1}{xe^x}$$
Clearly the denominator grows without bounds, and thus the fraction approaches zero.

Another approach is to let $z = re^{i \phi}$ and note that $\log(z) = \log(r) + i(\phi + 2k\pi)$. Here we can write this as $\log(-x) = \log(x) +(2k+1)\pi i$ because we have that $x$ is a positive real. By noticing the growth of each each term is less than $e^x$ we can conclude with a little work.

To answer the OP more directly perhaps, what is happening here is the analytic continuation of the function $\log(x)$ from $\mathbb{R}_{>0}$ to $\mathbb{R}$ (which we can further generalize to let $x$ be an element of $\mathbb{C}$)