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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?

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    $x<0$? what about $\log x$? – 2017-01-26
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    log is not defined for $ x <0$ so the limit you wrote does not exists. – 2017-01-26
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    pleas my problem in wolfram alpha , try to check the link above – 2017-01-26
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    WA treats $\log$ as a complex-valued function. When $x<0$, you get complex numbers (this is how Mathematica's `Log` function works by design). – 2017-01-26
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    Mathematica (and probably Wolfram Alpha as well) assumes variables range over the complex numbers, unless you tell it otherwise. For example, you can try $(-8)^{1/3}$ and you do not get $-2$. – 2017-01-26
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    @zeraouliarafik You can see exactly how WA interpretes your input by trying [for example](https://www.wolframalpha.com/input/?i=lim+e%5Ex+ln(x),+x+to+-2): $$\lim_{x \to -2} e^x \log(x) = \frac{i π + \log{2}}{e^2}$$ – 2017-01-26

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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} $)

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    your last sentence convince me and letting me better of care – 2017-01-26
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    @zeraouliarafik glad to hear it :) you do have to be careful with Wolfram Alpha and Mathematica... They will switch automatically to complex numbers if you begin doing operations over complex numbers and don't explicitly tell the program to assume only real number output (and thus throw an error, e.g. in this case when $\log(x) $ receives negative values of $x$) – 2017-01-26
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    The problem what will happen with the student users of wolfarm alpha when they were asked to do some calculations there , i think it's negative thing for them because they ignor that as what happen now to me – 2017-01-26
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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.

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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.