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Re: One to the power Infinity

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35943] Re: [mg35911] One to the power Infinity
  • From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
  • Date: Fri, 9 Aug 2002 05:17:51 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Presumably the argument is something like this. Take any positive 
numbers a and s. Then a = (a^(1/s))^s. Now make s go to infinity so 
a^(1/s)->1. Thus you can "make" 1^Infinity equal to any positive a, 
which is exactly the kind of situation in which  Indeterminate is used 
for the answer.

Best regards

Andrzej

Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/


On Thursday, August 8, 2002, at 07:06  PM, Matthias.Bode at oppenheim.de 
wrote:

> Dear Colleagues,
>
> when fooling around with MATHEMATICA I found:
>
> 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?
>
> Best regards,
>
> Matthias Bode.
>
>
>



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