|
|
Posted by Ioannis on 05/31/06 22:22
<julianmlp@gmail.com> wrote in message
news:1149111535.620793.202460@u72g2000cwu.googlegroups.com...
>
> btw, and sorry for the off topic.
>
> Do you know any site which explains why:
>
> d/dx[0] {lim x-->oo (1+1/x)^x} ^ x = 1
******************************************************
This is offtopic, but here goes. The limit does not affect the last "x"
(it's outside the limit brackets)
lim_{x->+oo}(1+1/x)^x = e, so your expression above reduces to
d/dx[0]{e^x}. Now,
d/dx{e^x} = e^x, so,
d/dx[0]{e^x} = e^0 = 1.
You'd be better off asking such questions in the newsgroup sci.math.
********************************************************
Anyway, thanks to all the responders for the answers about the Greek
letters. Unfortunately my site is over 25 MB and I really can't bother to do
many structural updates/changes. It has over 200 pages. I could conceivably
update the deprecated html I am using, but this would require editing all
the pages and reuploading them. I don't have time for such a thing.
> I mean, what is the relationship between the limit (1 + 1/x)^x and the
> base of the exponential function A <==> d/dx[0]A^x = 1 (of course, A
> has to be "2.7182....")
>
> Again, sorry for the off topic
>
> regards - julian
--
Ioannis
Navigation:
[Reply to this message]
|