Re: One to the power Infinity
- To: mathgroup at smc.vnet.net
- Subject: [mg36001] Re: One to the power Infinity
- From: Selwyn Hollis <slhollis at earthlink.net>
- Date: Sun, 11 Aug 2002 05:14:07 -0400 (EDT)
- References: <aitfon$cfu$1@smc.vnet.net> <aj01qv$18l$1@smc.vnet.net> <aj17ad$2kh$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
David W. Cantrell wrote: > 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. Actually it seems you've discovered a bug in Limit. If we allow complex values, then Limit[(1-x)^(1/x^2), x->0, Direction-> +1] should be 0, which is evident from the graph of Abs[(1-x)^(1/x^2)]. Mathematica gives Infinity for Limit[Abs[(1-x)^(1/x^2)], x->0, Direction-> +1] as well. If we don't allow complex values, then Limit[(1-x)^(1/x^2), x->0, Direction-> +1] can't exist at all (even as Infinity or -Infinity). >>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 Some of my favorite numbers are negative, such as (1 + x I)^(I Pi/Log[1 + x I]) = -1 Note that this has the form 1^Infinity as x->0. > > >>Matthias.Bode at oppenheim.de wrote: >> >>>1^\[Infinity] => Indeterminate, unexpected. Naively expected: 1. >>> >>>For which reason(s) is 1^\[Infinity] defined as Indeterminate? >> >