Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\infty$?

share|improve this question
4  
$\ln(1+e^x)\approx x$. –  David Mitra Dec 17 '11 at 17:55
5  
$\ln(1+e^x) = \ln(e^x(e^{-x} +1)) = x\ln(1+e^{-x})$ –  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
1  
@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
1  
@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
show 1 more comment

2 Answers

up vote 11 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.

share|improve this answer
    
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
add comment

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?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.