Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- To: mathgroup at smc.vnet.net
- Subject: [mg118396] Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Tue, 26 Apr 2011 06:51:20 -0400 (EDT)
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