Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > comp.soft-sys.math.mathematica > #1870 > unrolled thread

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

Back to article view | Back to comp.soft-sys.math.mathematica

Contents

  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

#1870 — Re: Limit[f[x], x->a] vs. f[a]. When are they equal?

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

[toc] | [next] | [standalone]

#1875

On Apr 24, 8:25 am, Richard Fateman <fate...@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...@cs.berkeley.edu <mailto:fate...@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.

[toc] | [prev] | [standalone]

Back to top | Article view | comp.soft-sys.math.mathematica

csiph-web