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

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On 4/23/2011 10:53 AM, Leonid Shifrin wrote:
> Richard,
>
> On Sat, Apr 23, 2011 at 4:49 AM, Richard Fateman 
> <fateman@cs.berkeley.edu <mailto:fateman@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

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Re: Limit[f[x], x->a] vs. f[a]. When are they equal? Richard Fateman <fateman@eecs.berkeley.edu> - 2011-04-24 12:25 +0000
  Re: Limit[f[x], x->a] vs. f[a]. When are they equal? Stefan <wutchamacallit27@gmail.com> - 2011-04-25 11:28 +0000

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