Q:

Why is $\lim\limits_{x \rightarrow \infty}x-\ln{(1+e^x)} = 0$?

How come $$\lim_{x \rightarrow \infty}x-\ln{(1+e^x)} = 0\quad ?$$
As I see it, when $x$ has a very very big value, $\ln{(1+e^x)}$ has a much lower value.
Why would the difference of those two values be $0$ when both those functions approach $\infty$?

A:

formally
$$x-\log(1+e^x)=\log(e^x)-\log(1+e^x)=\log(e^x/(1+e^x))=-\log(1+e^{-x})\to0$$
but you can think about $\log(1+e^x)$ being almost equal to $\log(e^x)=x$ for intuition.

