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

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  • Subject: [mg118411] Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
  • From: Andrzej Kozlowski <akozlowski at gmail.com>
  • Date: Wed, 27 Apr 2011 05:38:37 -0400 (EDT)

On 26 Apr 2011, at 11:57, Andrzej Kozlowski wrote:

>
> On 26 Apr 2011, at 10:43, Richard Fateman wrote:
>
>> Hm. continuity  (excerpt from wikipedia, but whatever)
>>
>> The limit of f(x) as x approaches c through domain of f does exist and
>> is equal to f(c); in mathematical notation, \lim_{x \to c}{f(x)} = f(c).
>> If the point c in the domain of f is not a limit point of the domain,
>> then this condition is vacuously true, since x cannot approach c through
>> values not equal c. Thus, for example, every function whose domain is
>> the set of all integers is continuous.
>>
>> ........
>> There is now a kind of semantic gap, it seems to me. There are, perhaps,
>> several gaps.
>
>
> Here is another point worth (perhaps) commenting upon. The function 1/x defined on the real line excluding the point 0 is continuous everywhere because it is continuous wherever it is defined. Sometimes however mathematicians do say that this function "is not continuous at 0" and when they do so they mean that there is no way to extend its domain of definition so as to include 0 and make the new function continuous. This is unlike the function Sin[x]/x which is also continuous everywhere on the real line minus 0, but can be continuously extended to include 0 (and then it is nowadays called Sinc).
>
> It is correct to say that Sin[1/z] is not continuous at zero because there is no way to give it a value there that would make it continuous. In the same sense Sin[z] is not continuous at ComplexInfinity: it cannot be continuously extended to the Riemann sphere. On the other hand the function 1/x can be and hence:
>
> 1/ComplexInfinity
>
> 0
>
> or, if you prefer
>
> In[34]:= Limit[1/x, x -> ComplexInfinity, Direction -> 1]
>
> Out[34]= 0
>
> or
>
> In[35]:= Limit[1/x, x -> ComplexInfinity, Direction -> I]
>
> Out[35]= 0
>
> etc.
>
> The functions x and  1/x are "continuous at Infinity" as is every meromorphic function. They define continuous mappings from the Riemann sphere to itself.  The function Sin[x] is not meromorphic and cannot be continuously extended to the Riemann sphere.
>
> Andrzej Kozlowski


Perhaps (just to avoid any further pointless quibbles) I should add that whenever I speak of "extending" a function continuously I am always assuming that continuity is meant with respect to the given natural subspace topology. Any function can be extended continuously if we add an isolated point to it's domain.
Similarly, when speaking of "meromorphic functions" I mean "meromorphic on the Riemann sphere" (thus while Sin[z] is meromorphic on the complex plane it is not meromorphic on the Riemann sphere however one defines the value at ComplexInfinity).

Finally -  ComplexInfinity and DirectedInfinitiy[z] and belong really to two quite distinct compactifications of the complex plane and should not be considered together. ComplexInfinity results from the one point compactification of the complex plane. A consistent interpretation of DirectedInfinity[z] requires a quite  different compactification, which is not compatible with the complex structure (the compactified space is no longer a complex manifold) but nevertheless can be useful when studying certain phenomena involving complex functions. I have discussed this in more detail the past and anyone really interested can look up the list archives.

Andrzej Kozlowski


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