Re: Limit of an expression?
- To: mathgroup at smc.vnet.net
- Subject: [mg67595] Re: Limit of an expression?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sat, 1 Jul 2006 05:12:49 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
On 30 Jun 2006, at 11:49, David W. Cantrell wrote:
> [Message also posted to: comp.soft-sys.math.mathematica]
>
> 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
I agree. (I neglected to pay careful attention to the output when I
posted the above - sometimes I have too much trust in Mathematica I
guess ;-) ). If you look at Trace with the option TraceInternal->True
you can see how Mathematica arrives at this. Essentially, at some
point it evaluates
Simplify[1 - E^(a*t), Assumptions -> {a == 0}]
and then it concludes that the whole expression is 0, without ever
noticing that the denominator is also assumed to be zero.
This, I think, should count as a bug.
Andrzej Kozlowski
Tokyo, Japan