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Could someone explain why

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

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Excuse me, I didn't understand your question. –  HipsterMathematician Sep 22 '12 at 21:27
i want to know how we got this result –  alan Sep 22 '12 at 21:32
Which result are you talking about? –  Nils Matthes Sep 22 '12 at 21:34
Limit[Log [1 + E^(-x)] + x, x -> -Infinity] = 0 –  alan Sep 22 '12 at 21:35
The first version of the question was dreadful but, after the revision by @PeterTamaroff, this became a bona fide mathematical question. Hence I disagree with the decision to close as not a real question. –  Did Sep 30 '12 at 10:06

3 Answers 3

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.

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yes exponential function –  alan Sep 22 '12 at 21:37

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

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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 $$\lim_{x\to \infty} \log(1+e^{-x})+x=\lim_{x\to \infty}(-x+x)=0$$

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