I want to find the following limit:


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?


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

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

  • $\begingroup$ Thanks, how is it clear? Both $e^x$ and $log|x|$ approach infinity so basically $+\infty\cdot+\infty=+\infty$? $\endgroup$ – Cesare Aug 10 '18 at 14:21
  • $\begingroup$ Yes. $(+\infty)\cdot (+\infty)=+\infty.$ $\endgroup$ – mfl Aug 10 '18 at 14:21
  • $\begingroup$ I'm not exactly sure how for $x\rightarrow-\infty$ the argument of the logarithm becomes $-x$ when it's encapsulated by an absolute $|x|$. Also not sure why $e^x$ becomes $e^{-x}$. Does having a limit that approaches $-\infty$ inverts all the variables? $\endgroup$ – Cesare Aug 10 '18 at 14:28
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    $\begingroup$ Note that $x<0\implies |x|=-x$ and $e^x=\dfrac{1}{e^{-x}}.$ $\endgroup$ – mfl Aug 10 '18 at 15:11

We have that


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

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


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