RE: Re: Re: One to the power Infinity
- To: mathgroup at smc.vnet.net
- Subject: [mg36015] RE: [mg35978] Re: [mg35940] Re: One to the power Infinity
- From: "DrBob" <majort at cox-internet.com>
- Date: Mon, 12 Aug 2002 03:34:32 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
The other one is easy too, via L'Hospital's rule: Limit[(1 + 1/n)^n, n -> Infinity] == Exp[Limit[n Log[1 + 1/n], n -> Infinity]] == Exp[Limit[D[Log[1 + 1/n], n]/D[1/n, n], n -> Infinity]] == Exp[Limit[n/(1 + n), n -> Infinity]] == Exp[1] Lim -----Original Message----- From: DrBob [mailto:majort at cox-internet.com] To: mathgroup at smc.vnet.net Subject: [mg36015] RE: [mg35978] Re: [mg35940] Re: One to the power Infinity I think you meant Limit[(1 + 1/n)^n, n -> Infinity], which is EXACTLY e. The limit you wrote about below is 1, and it IS easy because, after all, (1 + 1/n)n // Simplify 1+n Bobby -----Original Message----- From: Jonathan Rockmann [mailto:MTheory at msn.com] To: mathgroup at smc.vnet.net Subject: [mg36015] [mg35978] Re: [mg35940] Re: One to the power Infinity 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: [mg36015] [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