Groups > comp.soft-sys.math.mathematica > #1886

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

Path	csiph.com!x330-a1.tempe.blueboxinc.net!newsfeed.hal-mli.net!feeder1.hal-mli.net!nx02.iad01.newshosting.com!209.197.12.242.MISMATCH!nx01.iad01.newshosting.com!newshosting.com!novia!news-out.readnews.com!transit4.readnews.com!newspump.sol.net!posts.news.twtelecom.net!nnrp2.twtelecom.net!not-for-mail
From	Andrzej Kozlowski <akoz@mimuw.edu.pl>
Newsgroups	comp.soft-sys.math.mathematica
Subject	Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
Date	Tue, 26 Apr 2011 10:51:16 +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
Message-ID	<ip6834$bmt$1@smc.vnet.net> (permalink)
Lines	42
NNTP-Posting-Date	26 Apr 2011 09:51:59 GMT
NNTP-Posting-Host	ce91abd9.news.twtelecom.net
X-Trace	DXC=?\8WY;eC?mcE0G1fKX\WG`C_A=>8kQj6m;[h;PUXBgbdU3bdY^;K09nEFiONJ7[Gof@L@aGO<P]@h
X-Complaints-To	abuse@twtelecom.net
Xref	x330-a1.tempe.blueboxinc.net comp.soft-sys.math.mathematica:1886

Show key headers only | View raw

On 26 Apr 2011, at 10:43, Richard Fateman wrote:

> Back to Andrzej's note.
>
>  Sin(1/z) as z->0 is not the same as Sin[z] as z->ComplexInfinity.
>
> 1/z as z->0 has a neighborhood. There is no neighborhood around
> ComplexInfinity.
>
> (I am not, incidentally, proposing that I have a complete solution to
> these issues.)

I have skipped the rest of the post since I have written a great deal about this on this very forum in the past and, unlike some, I do not have the habit of repeating myself ad nauseam. However, the above point involves basic confusion. Mathematica computes only directional limits.

Hence
 Limit[Sin[1/z], z -> 0]
Interval[{-1,1}]
is  a real interval (since the default direction is -1) and

In[31]:= Limit[Sin[1/z], z -> 0, Direction -> I]

Out[31]= DirectedInfinity[I]

and so on. In Mathematica is no sense in taking limits as z ->ComplexInfinity without specifying a direction as there is no natural direction. If you specify the same direction, you will get the same answer:

In[25]:= Limit[Sin[1/z], z -> 0, Direction -> I]

Out[25]= DirectedInfinity[I]

In[27]:= Limit[Sin[z], z -> ComplexInfinity, Direction -> I]

Out[27]= DirectedInfinity[I]

Does ComplexINfinity not have neighbourhoods? It depends, of course, how you interpret it. It certainly has neighborhoods  if you interpret it as the the opposite pole to zero on the Riemann sphere, which is, in fact, what most people expect.

Andrzej Kozlowski

Back to comp.soft-sys.math.mathematica | Previous | Next — Next in thread | Find similar | Unroll thread

Thread

Re: Limit[f[x], x->a] vs. f[a]. When are they equal? Andrzej Kozlowski <akoz@mimuw.edu.pl> - 2011-04-26 10:51 +0000
  Re: Limit[f[x], x->a] vs. f[a]. When are they equal? Noqsi <noqsiaerospace@gmail.com> - 2011-04-30 09:51 +0000
    Re: Limit[f[x], x->a] vs. f[a]. When are they equal? Richard Fateman <fateman@cs.berkeley.edu> - 2011-05-02 10:51 +0000
  Re: Limit[f[x], x->a] vs. f[a]. When are they equal? Richard Fateman <fateman@cs.berkeley.edu> - 2011-04-28 10:35 +0000
  Re: Limit[f[x], x->a] vs. f[a]. When are they equal? Andrzej Kozlowski <akoz@mimuw.edu.pl> - 2011-04-28 10:36 +0000
  Re: Limit[f[x], x->a] vs. f[a]. When are they equal? Richard Fateman <fateman@eecs.berkeley.edu> - 2011-04-29 11:29 +0000

csiph-web