1
$\begingroup$

I tried:

$$\lim_{x \rightarrow - \infty}\frac{\ln(e^{-x}-1)}{x} = \\ \lim_{x \rightarrow \infty}\frac{\ln(e^{x}-1)}{-x} = \\ \frac{\ln(e^{x}-1)}{-x} \cdot \frac{e^{x}-1}{e^{x}-1} = \\ \frac{\ln(e^{x}-1)}{e^{x}-1} \cdot -\frac{e^{x}-1}{x} = \\ 0 \cdot -\infty$$

What did I do wrong? How do I solve this?

$\endgroup$
  • 1
    $\begingroup$ What's your definition of $e^x$? $\endgroup$ – Chappers Mar 24 '17 at 17:44
  • $\begingroup$ @Chappers What do you mean? $\endgroup$ – Mark Read Mar 24 '17 at 17:45
  • 1
    $\begingroup$ How was the exponential function first introduced and defined to you? By a series, a differential equation, something else? $\endgroup$ – Clement C. Mar 24 '17 at 17:46
  • 1
    $\begingroup$ hint: $\log(e^x-1)=x+\log(1-e^{-x})=x+e^{-x}+\mathcal{O}(e^{-2x})$. Also $e^{-x}/x\rightarrow 0$ as $x\rightarrow \infty$ $\endgroup$ – tired Mar 24 '17 at 17:49
  • 2
    $\begingroup$ @tiredNo need for the $O(\cdot)$ (which is tantamount to a first step in a Taylor expansion). The fact that $\log(1+e^{-x})\to 0$ suffices. $\endgroup$ – Clement C. Mar 24 '17 at 17:52
4
$\begingroup$

You did nothing wrong, but end up with an indeterminate form; so you cannot conclude with this approach.

Now, let us start from your second step: for $x> 0$, $$\begin{align} -\frac{\ln(e^x-1)}{x} &=-\frac{\ln(e^x(1-e^{-x}))}{x} =-\frac{\ln(e^x)+\ln(1-e^{-x})}{x} =-\frac{x+\ln(1-e^{-x})}{x}\\ &= -1 - \frac{\ln(1-e^{-x})}{x} \end{align}$$ It only remains to show the second term goes to $0$. When $x\to\infty$, we have $1-e^{-x}\xrightarrow[x\to\infty]{} 1$, so by continuity $\ln(1-e^{-x})\xrightarrow[x\to\infty]{} \ln 1=0$. It follows that $$ \frac{\ln(1-e^{-x})}{x}\xrightarrow[x\to\infty]{} 0. $$ Putting it all together, $$ -\frac{\ln(e^x-1)}{x}\xrightarrow[x\to\infty]{} -1- 0 = \boxed{-1}. $$

$\endgroup$
2
$\begingroup$

Let's start by writing: $$\ln(e^{-x}-1)= \ln(e^{-x})+\ln(1-e^{x})$$ Therefore: $$\lim_{x\to -\infty} \frac{\ln(e^{-x}-1)}{x}=\lim_{x\to -\infty} \frac{\ln(e^{-x})+\ln(1-e^{x})}{x}$$ Since $\ln(1-e^x)\to 0$ as $x \to -\infty$, we have just: $$\lim_{x\to -\infty} \frac{\ln(e^{-x})}{x}=\lim_{x\to -\infty} \frac{-x}{x}=-1$$

$\endgroup$
  • $\begingroup$ Not that I want to nitpick, but isn't it nearly the same proof as in my answer? (starting from the first step of the OP's question, instead of the second) $\endgroup$ – Clement C. Mar 24 '17 at 18:07
  • $\begingroup$ @ClementC. I figured that this way may make it more visually apparent (My solve is in a slightly different order than yours). Besides that, I upvoted your answer. $\endgroup$ – projectilemotion Mar 24 '17 at 18:13
  • $\begingroup$ I was just wondering about the rationale (or if you actually were typing it before, which may happen). Rather ironically, I went for the limit as $x\to\infty$ because the $x\to-\infty$ case always strikes me as less intuitive and harder to "see" -- and more prone to errors. (In any case, +1) $\endgroup$ – Clement C. Mar 24 '17 at 18:16
  • $\begingroup$ @ClementC. Thank you. I actually was typing this before you posted your answer, however I noticed that I misread the limit so I had to make slight modifications. $\endgroup$ – projectilemotion Mar 24 '17 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.