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MathGroup Archive 2006

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Re: Limit of an expression?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67557] Re: Limit of an expression?
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.org>
  • Date: Fri, 30 Jun 2006 04:15:25 -0400 (EDT)
  • References: <200606280751.DAA03399@smc.vnet.net> <e7vkut$smg$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> On 28 Jun 2006, at 16:51, Virgil Stokes wrote:
>
> > In the following expression, s is an integer (>= 1), Lambda, Mu, and t
> > are real numbers and all > 0.
> > What is the limit of the following as t goes to infinity?
> >
> > \!\(\(1 - \[ExponentialE]\^\(\(-\[Mu]\)\ t\ \((s - 1 - \
> > \[Lambda]\/\[Mu])\)\)\)\/\(s - 1 - \[Lambda]\/\[Mu]\)\)
> >
> > --V. Stokes
> >
>
> Unless you made a mistake in the formula you posted,  the answer
> depends  on the sign of s - 1 - λ/μ. Mathematica can deal with all
> three possible cases (it is also pretty obvious when done by hand):
>
> (Limit[(1 - E^((-μ)*t*
>          (s - 1 - λ/μ)))/
>       (s - 1 - λ/μ),
>      t -> Infinity,
>      Assumptions ->
>       {μ > 0 && #1[s,
>          1 + λ/μ]}] & ) /@
>    {Greater, Equal, Less}
>
> {-(μ/(λ - s*μ + μ)),
>    0, Infinity}

Much of the above is illegible to me, but I'm guessing that the middle case
is equivalent to

In[1]:= Assuming[a==0, Limit[(1 - Exp[a t])/a, t->Infinity]]

Out[1]= 0

which does not seem to be reasonable in Mathematica. I would have expected
Indeterminate instead.

David


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