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Re: Limit[f[x], x->a] vs. f[a]. When are they equal?

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  • 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







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