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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 $\infty$?

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$\ln(1+e^x)\approx x$. – David Mitra Dec 17 '11 at 17:55
$\ln(1+e^x) = \ln(e^x(e^{-x} +1)) = x\ln(1+e^{-x})$ – user20266 Dec 17 '11 at 17:56
Pick an $x$ which is not too large, like $x=20$, so your calculator won't blow up. Calculate $\ln(1+e^x)$. – André Nicolas Dec 17 '11 at 17:59
@AndréNicolas: yes, correct, thanks. Sorry :-) But that will help even better :-) And, you see, my comment got upvoted nonetheless :-) -- to all those upvoters of my comment: check your calculus scills and undo!! – user20266 Dec 17 '11 at 18:16
@Dimme : If you don't notice that $\ln(1+e^x)$ is nearly $x$ but slightly bigger, then you should work on your understanding of what the natural logarithmic and natural exponential functions are. – Michael Hardy Dec 17 '11 at 18:43
up vote 12 down vote accepted

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.

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Thank you, $x=\ln{(e^x)}$ answered my question. – Dimme Dec 17 '11 at 18:05
You can also pull a $\log(e^x)=x$ out: $$x-\log(1+e^x)=x-\log(e^x)-\log(e^{-x}+1)=-\log(1+e^{-x})$$ – robjohn Dec 17 '11 at 19:57

Note that $$\ln (1 + {e^x}) \sim \ln ({e^x}) = x$$ for large $x$. So $$x - \ln (1 + {e^x}) \sim x - \ln ({e^x}) = x - x$$ where $\sim $ denotes asymptotic equality. Think about it. When $x$ is large will adding 1 to ${e^x}$ make any difference?

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