Why is $\lim\limits_{x\to-\infty}\log [1 + \exp(-x)]+x=0$?

Question

Could someone explain why

$$\lim\limits_{x\to-\infty}\log [1 + \exp(-x)]+x=0$$

Answer 1

$$\log(1+\mathrm e^{-x})+x=\log((1+\mathrm e^{-x})\cdot\mathrm e^{x})=\log(\mathrm e^{x}+1)\underset{x\to-\infty}{\longrightarrow}\log(0+1)=0$$

Answer 2

I think you may have meant the exponential function, which is Exp[-x].

See: http://www.wolframalpha.com/input/?i=Limit%5BLog%5B1%2BExp%5B-x%5D%5D%20%2B%20x%2C%20x%20-%3E%20Infinity%5D&t=crmtb01

Do you see why it is infinity, just by inspection?

The exponential approaches zero, Log[1] is zero and the remaining terms grows to infinity.

If this is acceptable, you should accept the answer.

Answer 3

Being completely non-rigorous for a moment, we can say that for x large and negative, $\log(1+e^{-x}) \approx \log(e^{-x})=-x$
So we have $$\lim_{x\to \infty} \log(1+e^{-x})+x=\lim_{x\to \infty}(-x+x)=0$$