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MathGroup Archive 2001

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Re: Limit and Abs

  • To: mathgroup at
  • Subject: [mg31230] Re: Limit and Abs
  • From: "David W. Cantrell" <DWCantrell at>
  • Date: Sat, 20 Oct 2001 04:27:03 -0400 (EDT)
  • References: <> <9qokm5$mt3$>
  • Sender: owner-wri-mathgroup at

Tomas Garza <tgarza01 at> wrote:
> And what, may I ask, is the "correct answer"? Your function is periodic
> with period Pi. The only answer is
> In[1]:=
> Limit[Cot[a]/(2 + Cot[a]), a -> Infinity]
> Out[1]=
> Limit[Cot[a]/(2 + Cot[a]), a -> Infinity]
> And, unless there is a misprint in your second expression below, the
> limit with *n* -> Infinity has to remain unevaluated, because a is not
> defined. But, then, if there is a misprint and you meant a -> Infinity
> instead of n->Infinity,

As we now know, he didn't mean either! But taking a -> Infinity does
raise an interesting point.

> the function is again periodic with period Pi
> (look at the graphs). The limit doesn't exist, and Mathematica has no
> alternative but to leave the expression unevaluated.

False! As others have pointed out, Mathematica does have other
alternatives, and the one it chooses, returning a limit of 1, is
incorrect. In my opinion, it should have returned Interval[{0,1}]
as the answer instead. Mathematica has a more generalized notion of
limit than is often used. For example, although most of us would
normally say simply that Limit[Sin[a], a -> Infinity] does not exist,
Mathematica gives Interval[{-1,1}]. This is correct in a sense. I have
no objection to it; indeed, it provides more useful information than
merely saying "does not exist".

  David Cantrell

> From: "Oliver Friedrich" <oliver.friedrich at>
To: mathgroup at
> Subject: [mg31230]  Limit and Abs
> > Hallo,
> >
> > if I evaluate
> > Limit[Cot[a]/(Cot[a]+2),a->Infinity]
> > i get the correct answer.
> >
> > But I want to evaluate
> >
> > Limit[Abs[Cot[a]]/(Abs[Cot[a]]+2),n->Infinity]
> >
> > and that's being returned unevaluated.

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