Re: One to the power Infinity
- To: mathgroup at smc.vnet.net
- Subject: [mg35940] Re: One to the power Infinity
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Fri, 9 Aug 2002 05:17:49 -0400 (EDT)
- References: <aitfon$cfu$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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