MathGroup Archive 2002

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: One to the power Infinity

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35964] Re: One to the power Infinity
  • From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
  • Date: Fri, 9 Aug 2002 16:05:37 -0400 (EDT)
  • References: <aitfon$cfu$1@smc.vnet.net> <aj01qv$18l$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Selwyn Hollis <slhollis at earthlink.net> wrote:
> Because it depends on how you get there. For instance,
> Limit[(1-x)^(1/x), x->0] gives 1/E, while Limit[(1-x)^(1/x^2), x->0]
> gives 0.

You are correct. However, that last limit claim is somewhat deceptive.
Based on it, one might conclude, incorrectly, that the direction of
approach to 0 is immaterial. Note that, although
Limit[(1-x)^(1/x^2), x->0, Direction-> -1] yields 0,
Limit[(1-x)^(1/x^2), x->0, Direction-> +1] yields Infinity.
Thus, when Mathematica says that Limit[(1-x)^(1/x^2), x->0] is 0, it is
making a hidden assumption of direction of approach.

> In fact, you can construct similar examples in which
> "1^Infinity" = anything you like.

Well, yes, as long as you don't like negative values.

  David

> Matthias.Bode at oppenheim.de wrote:
> > 1^\[Infinity] => Indeterminate, unexpected. Naively expected: 1.
> >
> > For which reason(s) is 1^\[Infinity] defined as Indeterminate?

-- 
-------------------- http://NewsReader.Com/ --------------------
                    Usenet Newsgroup Service


  • Prev by Date: Out of memory
  • Next by Date: Scope problem
  • Previous by thread: Re: One to the power Infinity
  • Next by thread: Re: Re: One to the power Infinity