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