I understand that $\ln(1-x)$ can be approximated using a Taylor expansion at $x=0$ to yield that $\ln(1+x)$ is a bit more than $x$. However, in class I was recently given the following to show intuition for the log-loss function: $$\ln(1+e^{-x}) \approx e^{-x}$$ for $x \gg 0$ $$\ln(1+e^{-x}) \approx -x$$ for $x \ll 0$.
I'm not sure what this distinction is stemming from. In other words, I understand how Taylor expansion can prove the first line but not the second one.
The first one comes from the approximation of $\ln (1+y)$ for $y$ near $0$ and fact that $e^{-x} \to 0$ as $ x\to \infty$. For the second one note that $\ln (1+e^{-x})=\ln [(e^{-x} (1+e^{x})]=-x+\ln (1+e^{x})$ and $\ln (1+e^{x}) \to 0$ as $x \to -\infty$.