On Apr 29, 3:08 pm, Ronan Lamy <[email protected]> wrote:
> Le vendredi 29 avril 2011 à 11:53 -0700, Julien Rioux a écrit :
>
>
>
> > On Apr 29, 2:17 pm, Ronan Lamy <[email protected]> wrote:
> > > I would argue that the correct results are limit(x - oo, x, oo) == -oo
> > > and limit(oo - x, x, oo) == oo.
>
> > > The expression whose limit is taken belongs to the extended real line,
> > > so we need to consider the topology of the extended real line. Unless
> > > otherwise specified, limits are always evaluated for values of the
> > > variable close to, but different from, the "destination". In this case,
> > > this means arbitrarily large, but finite, reals. For any real x, x - oo
> > > = -oo, so the function Lambda(x, x - oo) is constant over the reals, but
> > > discontinuous, undefined actually, at x = +oo. However, the latter
> > > doesn't matter for the limit, and the result is the constant value, -oo.
> > > All this is completely parallel to limit(abs(x)/x, x, 0, '+'), for
> > > instance.
>
> > Take f(x,y)=x-y. By your account lim(lim(f(x,y),y,oo),x,oo) == -oo but
> > lim(lim(f(x,y),x,oo),y,oo) == oo. This can't be right.
>
> Why can't it be right?
> Limits that don't commute at a singular point are
> a common occurrence.
> For instance:
> lim(lim(exp(-b/a), b, 0+), a, 0+) == 1
> lim(lim(exp(-b/a), a, 0+), b, 0+) == 0
>
> Intuitively, the inner variable goes to infinity first, i.e. much faster
> than the outer one.
Intuitively?! Oh, I see how I must have been misleading. I suppose I
should have used a different syntax than what I used. But
mathematically what is lim_{x,y->oo} x-y ? My impression is that if
the order of the limit matters I would expect "undefined" as an
answer. My intuition says that there isn't one variable going to
infinity faster. If you assume one does, then you assume a particular
answer.
Also, f(x,y)=x-y isn't singular.
--
Julien
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