Solve $\lim_{x \rightarrow - \infty}\frac{\ln(e^{-x}-1)}{x}$ without L'Hopital or Taylor series

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?

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

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: $$\lim_{x\to -\infty} \frac{\ln(e^{-x})}{x}=\lim_{x\to -\infty} \frac{-x}{x}=-1$$
• 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) – Clement C. Mar 24 '17 at 18:16