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| From | Richard Fateman <fateman@eecs.berkeley.edu> |
| Newsgroups | comp.soft-sys.math.mathematica |
| Subject | Re: Limit[f[x], x->a] vs. f[a]. When are they equal? |
| Date | Sun, 24 Apr 2011 12:25:28 +0000 (UTC) |
| Organization | Steven M. Christensen and Associates, Inc and MathTensor, Inc. |
| Sender | steve@smc.vnet.net |
| Approved | Steven M. Christensen <steve@smc.vnet.net>, Moderator |
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| Lines | 43 |
| NNTP-Posting-Date | 24 Apr 2011 11:26:09 GMT |
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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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