0**0 follows Python conventions. The rest are likely equally useful.

By the way, 0*oo does give nan as it should.

Aaron Meurer

On Apr 7, 2013, at 4:59 PM, "hacm...@gmail.com" <hacm...@gmail.com> wrote:

Why are the indeterminate forms oo/oo, 0/0, and oo-oo recognized as nan,
but not 1**oo, 0**0, 0*oo, oo**0?

On Sunday, April 7, 2013 2:38:14 PM UTC-4, Aaron Meurer wrote:
>
> That's because subs just works dumbly, so it produces (1 + 1/oo)**oo,
> which gives 1**oo, which gives 1. A limit procedure doesn't fall into
> this trap because it knows that 1**oo is an indeterminate form (for
> exactly this reason).  We could probably "fix" this by making 1**oo
> give nan, but that would no doubt break a ton of other stuff.
>
> I really think that we should scrap these limit heursitics. They have
> lead to dozens of wrong results like this one, where limit() is wrong,
> but gruntz() is right. The whole point of limit() is that you can't
> trust subs, so the fact that limit() uses subs is contradictory. And
> the ironic thing is that limit() doesn't always return the result when
> subs does (for example, limit() on Piecewise gives
> NotImplementedError), hence workarounds like the one in
> Expr._eval_interval.
>
> Aaron Meurer
>
>
> On Sun, Apr 7, 2013 at 12:21 PM, Ankit Agrawal 
> <aaaag...@gmail.com<javascript:>>
> wrote:
> >
> >
> >
> >
> > On Sunday, April 7, 2013 11:25:11 PM UTC+5:30, Sergey Kirpichev wrote:
> >>
> >> Yes.  This is a bug in the limit() heuristic.
> >>
> >>
> >> gruntz() works well:
> >> In [5]: from sympy.series import gruntz
> >> In [6]: gruntz(1+(1+1/x)**x,x,oo)
> >> Out[6]: 1 + ℯ
> >>
> > Yes. I tried to narrow down the root of the bug and it is in _sift() at
> line
> > 188 of series.limits in current master.
> > It is actually the .subs that generates value of 1 for on (1 + 1/a)**a
> >
> >>>> ((1 + 1/a)**a).subs(a,oo)
> > 1
> >
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