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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






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