Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- To: mathgroup at smc.vnet.net
- Subject: [mg118359] Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- From: Richard Fateman <fateman at eecs.berkeley.edu>
- Date: Sun, 24 Apr 2011 08:25:35 -0400 (EDT)
On 4/23/2011 10:53 AM, Leonid Shifrin wrote: > Richard, > > On Sat, Apr 23, 2011 at 4:49 AM, Richard Fateman > <fateman at cs.berkeley.edu <mailto:fateman at 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