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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



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