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?

  • 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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

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, $$ -\frac{\ln(e^x-1)}{x}\xrightarrow[x\to\infty]{} -1- 0 = \boxed{-1}. $$


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: $$\lim_{x\to -\infty} \frac{\ln(e^{-x})}{x}=\lim_{x\to -\infty} \frac{-x}{x}=-1$$

  • $\begingroup$ 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

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