Re: One to the power Infinity
- To: mathgroup at smc.vnet.net
- Subject: [mg35948] Re: One to the power Infinity
- From: Selwyn Hollis <slhollis at earthlink.net>
- Date: Fri, 9 Aug 2002 05:17:56 -0400 (EDT)
- References: <aitfon$cfu$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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. In fact, you can construct similar examples in which "1^Infinity" = anything you like. ---- Selwyn Hollis slhollis2mac.com 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. >