Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- To: mathgroup at smc.vnet.net
- Subject: [mg118580] Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Tue, 3 May 2011 05:48:09 -0400 (EDT)
On 2 May 2011, at 12:51, Richard Fateman wrote: > No, but it should say what the program does. > What are the valid inputs to Limit ? Is ComplexInfinity not allowed? > What else? Where did you get this strange idea? ComplexInfinity is certainly allowed. Take any complex function that is continuous at ComplexInfinity, for example: f[z_] := Exp[1/z^2] Then: Limit[f[z], z -> ComplexInfinity] 1 All directional limits will also return the same value: Limit[f[z], z -> ComplexInfinity, Direction -> I] 1 Since the function is continuous and has a finite limit at ComplexInfinity, there is no "conflict" between the two compactification models. To see the conflict, you need a function that is analytic at ComplexInfinity but tends to infinity as the modulus of its argument increases. Such a function is, for example. g[z_] := (z^2 + 1)/(z + 2) In this case Limit[g[z], z -> ComplexInfinity] ComplexInfinity This is exactly right - if one uses the one point compactification model. The "disk" model gives different answers: Limit[g[z], z -> ComplexInfinity, Direction -> I] DirectedInfinity[I] Limit[g[z], z -> ComplexInfinity, Direction -> 1] Infinity This is all perfectly valid and useful. I admit that there is a slight problem. In complex analysis one would like to say that the function g is continuous on the Riemann Sphere and has the value ComplexInfinity there, but Mathematica considers it "Indeterminate": g[ComplexInfinity] During evaluation of In[97]:= Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered. >> Indeterminate I guess I would prefer the answer ComplexInfinity but this is not necessary since, for example, Sin[x]/x also returns Indeterminate rather than 1 at 0. This is the kind od thing any user can decide for himself by explicitly defining g[ComplexInfinity]=ComplexInfinity. There is absolutely nothing here that in any way disagrees with any mathematics I know of. Andrzej Kozlowski