Is really :$\displaystyle \lim _{x\to-\infty}e^x \log x=0$ 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?


 A: 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} $) 
A: Mathematica evaluates $\log{x}$ for $x<0$ as $\pi i+\log{(-x)}$. Since $\lim_{x\rightarrow-\infty}\exp{(x)}=0$ and $\lim_{x\rightarrow-\infty}\exp{(x)}\log{(-x)}=0$, you have Mathematica claiming that $\lim_{x\rightarrow-\infty}\exp{(x)}\log{(x)}=0$. I am assuming wolfram alpha behaves in the similar way.
A: In THIS ANSWER, I showed that for the principal branch of the complex logarithm

$$|\text{Log}(z)|\le \sqrt{|z-1|^2+\pi^2}$$

for $|z|\ge 1$.  Then, we have for $x<-1$ 
$$\begin{align}
\left|e^x\log(x)\right|&\le e^x \sqrt{|x-1|^2+\pi^2}\\\\
&=\frac{\sqrt{|x-1|^2+\pi^2}}{e^{|x|}}\\\\
&\le \frac{\sqrt{|x-1|^2+\pi^2}}{|x|^2/2}\\\\
\end{align}$$
whereupon applying the squeeze theorem yields the coveted limit.
Note for any other branch of the logarithm $|\text{Log}(z)|\le \sqrt{|z-1|^2+(k\pi)^2}$ for any integer $k$.  And we arrive at the same result.
