A question about limits

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Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

A question about limits

Postby Hom » Fri Oct 07, 2011 11:49 pm

Please help evaluate the following expression. It seems that it is 0. But I have no clue how that happens.
(log is based on e. don't know about the constraints about a and b. Let's assume they are not 0s).

\lim_\infty {  } b\cdot log(e^{ax}+1)-a\cdot (e^{bx}+1)

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: A question about limits

Postby owlpride » Sat Oct 08, 2011 6:32 am

I assume that the second term has a log in front of it as well?

The quickest solution is probably to observe that \lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax, which shows that the limit you care about is zero.

Alternatively, you could start by rewriting the expression as the log of a fraction, interchange the limit and the log and then show that the limit of the fraction is 1:

\lim_{x \rightarrow \infty }log \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }= log \lim_{x \rightarrow \infty } \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }

Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

Re: A question about limits

Postby Hom » Sat Oct 08, 2011 8:38 am

owlpride wrote:I assume that the second term has a log in front of it as well?

The quickest solution is probably to observe that \lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax, which shows that the limit you care about is zero.

Alternatively, you could start by rewriting the expression as the log of a fraction, interchange the limit and the log and then show that the limit of the fraction is 1:

\lim_{x \rightarrow \infty }log \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }= log \lim_{x \rightarrow \infty } \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }


Thanks for helping me out again, owlpride. I should have put a big parentheses for these two terms.
The key lies in this thing \lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax. Is this a common rule or something that obvious? That's where I got stuck. Could you also show some intermediate steps?

Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

Re: A question about limits

Postby Hom » Sat Oct 08, 2011 11:24 pm

Hom wrote:
owlpride wrote:I assume that the second term has a log in front of it as well?

The quickest solution is probably to observe that \lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax, which shows that the limit you care about is zero.

Alternatively, you could start by rewriting the expression as the log of a fraction, interchange the limit and the log and then show that the limit of the fraction is 1:

\lim_{x \rightarrow \infty }log \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }= log \lim_{x \rightarrow \infty } \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }


Thanks for helping me out again, owlpride. I should have put a big parentheses for these two terms.
The key lies in this thing \lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax. Is this a common rule or something that obvious? That's where I got stuck. Could you also show some intermediate steps?


Sorry, my bad. There is another condition, a>0 and b>0. in this case, it's obvious \lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax.

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: A question about limits

Postby DDswife » Sat Aug 23, 2014 3:50 pm

I think that you meant lim ax, not ax




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