Find $\lim_{x\rightarrow\pm\infty}e^x\log|x|$

Question

I want to find the following limit:

$$\lim_{x\rightarrow\pm\infty}e^x\log|x|$$

For $x\rightarrow+\infty$, $e^x\rightarrow+\infty$ and $\log|x|\rightarrow+\infty$ but this is a multiplication between two factors so I'm not able to draw conclusions by using asymptotics. I guess I need to use the method which changes the variable $x$ into something like $t$ or $1/t$ but I'm not exactly sure how.

I'll figure what happens when $x\rightarrow-\infty$ after I understand what happens when the limit approaches $+\infty$.

Any hints?

Answer 1

It is clear that $$\lim_{x\to \infty} e^x\log x=\infty.$$ Now,

$$\lim_{x\to -\infty} e^x\log (-x)=\lim_{x\to -\infty} \dfrac{\log (-x)}{e^{-x}}=\lim_{x\to -\infty} \dfrac{\dfrac{1}{x}}{-e^{-x}}=\lim_{x\to -\infty} \dfrac{-e^x}{x}=0.$$

Answer 2

We have that

$$\lim_{x\rightarrow\infty}e^x\log|x|=\infty$$

and by $y=-x \to \infty$

$$\lim_{x\rightarrow-\infty}e^x\log|x|=\lim_{y\rightarrow\infty}\frac{\log y}{e^y}\to 0$$

indeed $\log y <y$ and $e^y>y^2$ therefore

$$0\le \frac{\log y}{e^y}\le \frac{y}{y^2}=\frac1 y \to 0$$