Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- To: mathgroup at smc.vnet.net
- Subject: [mg118380] Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- From: Stefan <wutchamacallit27 at gmail.com>
- Date: Mon, 25 Apr 2011 07:28:32 -0400 (EDT)
- References: <ip14ro$gfa$1@smc.vnet.net>
On Apr 24, 8:25 am, Richard Fateman <fate... at eecs.berkeley.edu> wrote: > On 4/23/2011 10:53 AM, Leonid Shifrin wrote: > > > > > > > > > > > Richard, > > > On Sat, Apr 23, 2011 at 4:49 AM, Richard Fateman > > <fate... at cs.berkeley.edu <mailto:fate... 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 With regards to Limit[1/(1 - Exp[I x]), x -> 0], this does indeed yield i*infinity as Mathematica says, a short bit of calculation will verify that. ComplexInfinity is a more general case where the result has some infinite real and some infinite imaginary piece, but with some arbitrary phase (ie. Infinity*Exp[I*theta] for some arbitrary theta). When Mathematica can discern more information about your limit, it does so. In this case, it is not some arbitrary phase, but in fact the phase is exactly Pi/2, and so the result is much more specific than ComplexInfinity, it is indeed I*Infinity. This is the same reason that Limit[1/(I - Exp[I x]), x -> Pi/2] yields Infinity. If the direction of the infinity is known but not simply that it is real or imaginary, Mathematica uses DirectedInfinity[z_Complex], which is exactly what it sounds like. You can try things like In[3]:= Limit[1/(Exp[I Pi/4] - Exp[I x]), x -> Pi/4] Out[3]:= DirectedInfinity[(1 + I)/Sqrt[2]] I would recommend reading the documentation for ComplexInfinity, DirectedInfinity, and Indeterminate. They are all linked together in the "See Also" section, as they are closely related.