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