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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.2012-09-22
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    i want to know how we got this result2012-09-22
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    Which result are you talking about?2012-09-22
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    Limit[Log [1 + E^(-x)] + x, x -> -Infinity] = 02012-09-22
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    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*.2012-09-30
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    @did Well, yes, it was. I took me some time to understand what it was and edit accordingly. We can reopen.2012-09-30
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    How do I vote for reopen ? Or do I need more reputation for that ?2012-10-04

2 Answers 2

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Here is a one-line (full) explanation: $$\log(1+\mathrm e^{-x})+x=\log((1+\mathrm e^{-x})\cdot\mathrm e^{x})=\log(\mathrm e^{x}+1)\underset{x\to-\infty}{\longrightarrow}\log(0+1)=0$$

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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 function2012-09-22