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MathGroup Archive 2001

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Re: Limit and Abs

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31219] Re: [mg31194] Limit and Abs
  • From: Tomas Garza <tgarza01 at prodigy.net.mx>
  • Date: Fri, 19 Oct 2001 03:12:12 -0400 (EDT)
  • References: <200110170935.FAA19657@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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, 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.

Tomas Garza
Mexico City
----- Original Message -----
From: "Oliver Friedrich" <oliver.friedrich at tz-mikroelektronik.de>
To: mathgroup at smc.vnet.net
Subject: [mg31219] [mg31194] 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. Help states, that Limit will return
> expressions unevaluated, if there are functions with unknown behaviour
(Abs
> unknown ??) so I'm not too surprised or disappointed.
> But anyway, how can I workaround or bypass this problem, maybe an option
or
> another function in the extra packages ?
>
> Thanks
>
> Oliver Friedrich
>
>



  • References:
    • Limit and Abs
      • From: "Oliver Friedrich" <oliver.friedrich@tz-mikroelektronik.de>
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