Re: Re: One to the power Infinity
- To: mathgroup at smc.vnet.net
- Subject: [mg35978] Re: [mg35940] Re: One to the power Infinity
- From: "Jonathan Rockmann" <MTheory at msn.com>
- Date: Sun, 11 Aug 2002 05:13:36 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
As a follow up to David's example: Some students seem to think that lim n->0 (1+(1/n))n = 1. Their reasoning is this: "When n->0, then 1+(1/n) -> 1. Now compute lim n->0 1 n = 1." This reasoning is just too simplistic. You have to deal with both of the n's in the expression (1+(1/n))n at the same time -- i.e., the y both go to infinity simultaneously; you can't figure that one goes to infinity and then the other goes to infinity. In fact, if you let the other one go to infinity first, you'd get a different answer: lim n->0 (1+0.0000001)n = 0. So evidently the answer lies somewhere between 1 and 0. Easy methods do not work on this problem. The correct answer is a number that is near 2.718. (It's an important constant, known to mathematicians as "e" = aprrox. 2.718281828...) There's no way you could get that by an easy method. Jonathan Rockmann mtheory at msn.com ----- Original Message ----- From: David W. Cantrell To: mathgroup at smc.vnet.net Subject: [mg35978] [mg35940] Re: One to the power Infinity Matthias.Bode at oppenheim.de wrote: > 0^\[Infinity] => 0, as expected; > 0.9^\[Infinity] => 0, as expected; > 2^\[Infinity] => Infinity, as expected; > 1^\[Infinity] => Indeterminate, unexpected. Naively expected: 1. > > For which reason(s) is 1^\[Infinity] defined as Indeterminate? Presumably because, as a limit form, 1^Infinity is indeterminate. In other words, if f(x) approaches 1 and g(x) increases without bound, f(x)^g(x) need not approach 1. Consider the limit of (1+1/x)^x as x increases without bound, for example. David