Yatir Halevi
Posted on Tuesday, 04 March, 2003 - 07:35 pm:
Does anybody have an example of when 1f(x) doesn't tend to 1 as f(x)®¥ as x® ¥?
(There might be a simple example, I haven't given it much thought yet...)

Yatir
Dan Goodman
Posted on Tuesday, 04 March, 2003 - 07:42 pm:
1x =1 for all x...
Colin Prue
Posted on Tuesday, 04 March, 2003 - 08:16 pm:
perhaps this is cheating, but:

13/2=(e2pi)3/2=-1

Colin Prue
Posted on Tuesday, 04 March, 2003 - 08:17 pm:
...therefore if you are careful how you define things you can have a divergent sequence (1,-1,1,-1,1...etc)
David Loeffler
Posted on Wednesday, 05 March, 2003 - 12:30 am:
Yes. It's cheating. You are trying to tell me that 1 = -1. From this I can validly deduce that I am the Pope.

David
Hauke Worpel
Posted on Wednesday, 05 March, 2003 - 08:36 am:
How is it cheating, your Holiness? ;) 13/2 =(13 )1/2

13 =1 all the time, no disagreement there, but what is 11/2 ? Either 1 or -1. Is this a semantic quibble or a genuine mathematical point?
David Loeffler
Posted on Wednesday, 05 March, 2003 - 09:37 am:
It is not the case for general complex numbers that (ab )c = abc ; the values of the latter will be a subset, in some cases a proper subset, of the values of the former.

-----

Yatir, can I check exactly what question you were asking? Did you really mean to ask,

"Do there exist real functions a(x) and b(x) such that a(x) -> 1 but a(x)b(x) -> infinity as x -> infinity?"

In this case the answer is yes, and there is a fairly easy counterexample.

David
Yatir Halevi
Posted on Wednesday, 05 March, 2003 - 05:46 pm:
David, I know that a(x) exists. But a lot of things in mathematics are very un-intuitional, and I wanted to know if maybe here lies one of them.

Yatir