Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- To: mathgroup at smc.vnet.net
- Subject: [mg118383] Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 25 Apr 2011 07:29:05 -0400 (EDT)
On 24 Apr 2011, at 23:22, Andrzej Kozlowski wrote: > > On 24 Apr 2011, at 15:25, Richard Fateman wrote: > >> On 4/23/2011 10:53 AM, Leonid Shifrin wrote: >>> Richard, >>> >>> On Sat, Apr 23, 2011 at 4:49 AM, Richard Fateman >>> <fateman at cs.berkeley.edu <mailto:fateman at cs.berkeley.edu>> wrote: >>> >>> Let c=ComplexInfinity >>> >>> then Mathematica (7.0) "knows" that Sin[c] is Indeterminate >>> >>> but Limit[Sin[x],x->c] is not evaluated. >>> >>> I would have thought that if f[a] is known, Limit[f[x],x->a] is >>> also known. >>> >>> >>> Not necessarily. The limiting procedure has nothing to do with the >>> value of the function at a given point, it has to do with values of >>> the function in the neighborhood of that point. >>> >>> Regards, >>> Leonid >>> >>> >>> RJF >>> >>> >> >> I agree not necessarily, but sine() is continuous and differentiable >> everywhere. >> I think a computer algebra system should do better. >> >> >> I poked around some more.. >> >> Limit[1 - Exp[I x], x -> 0] yields 0 >> Limit[1/x, x->0] yields Infinity >> >> Limit[1/(1 - Exp[I x]), x -> 0] yields I*Infinity. ?? >> I would think this would be a place for ComplexInfinity. >> >> All in Mathematica 7.0 > > Mathematica uses DirectedInfinity[z] for representing various limits in the complex plane computed along straight lines and uses this concept when computing limits but not when computing function values. The answer CompexInfinity is not used when computing limits because it is considered essentially useless - it carries no information while directed infinities do. The answer DirectedInfinity[I] is, for example, justified by the fact that: > > In[17]:= Limit[ComplexExpand[Abs[1/(1 - Exp[I*x])]], x -> 0] > > Out[17]= Infinity > > Limit[ComplexExpand[Arg[1/(1 - Exp[I*x])]], x -> 0] > > Pi/2 > > > In other words, the limit has infinite modulus and argument Pi/2 - so it is exactly DirectedInfinity[I]. > > So ComplexInfinity is used for values (e.g. 1/(I 0) ) but DirectedInfinity for limits > > In[18]:= Limit[1/(I*x), x -> 0] > > Out[18]= DirectedInfinity[-I] > > I don't myself know of any mathematical theory of "directed infinities" (although I have in the past indicated how I thought such a theory could be constructed) and there has been a discussion of this on this forum involving (among others) Maxim Rytin and myself. > > In fact this is precisely a case of computer algebra "doing better". One just has to understand what it is doing and why. > > Andrzej Kozlowski > > I forgot to add that the function Sin is certainly not continuous at infinity (Sin[1/z] has an essential singularity at 0) so there is no reason why it's value at there should agree with its limit. In fact, it definitely should not do so. Andrzej Kozlowski