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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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    @MichaelHardy: I know what they are, I just could not see it at the moment.2011-12-17

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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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    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})$2011-12-17
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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?