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MathGroup Archive 2006

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Re: Limit of an expression?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67518] Re: Limit of an expression?
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Thu, 29 Jun 2006 00:09:50 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <e7td7i$3lg$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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
> 
Hi Virgil,

Contrary to some functions such as Integrate, Limit does not returned 
conditional answers; that is, Limit will return a solution only if it 
can establish the convergence or divergence of the expression without 
taking in account the values of parameters. For instance, compare the 
following answers to a simple limit

In[1]:=
Limit[E^(-t), t -> Infinity]

Out[1]=
0

In[2]:=
Limit[E^((-a)*t), t -> Infinity]

Out[2]=
        -a t
Limit[E    , t -> Infinity]

Of course, in the latter case you can add an assumption on the parameter 
'a' and get the answer

In[3]:=
Limit[E^((-a)*t), t -> Infinity, Assumptions -> a > 0]

Out[3]=
0

Now, let's go back to your original question. In Mathematica form, we have

In[4]:=
Limit[(1 - E^((-μ)*t*(s - 1 - λ/μ)))/
    (s - 1 - λ/μ), t -> Infinity,
   Assumptions -> {s â?? Integers, s > 0, λ > 0,
     μ > 0, t > 0}]

Out[4]=
            -t (-1 + s - λ/μ) μ
       1 - E
Limit[------------------------, t -> Infinity,
                       λ
              -1 + s - -
                       μ

   Assumptions ->

    {s â?? Integers, s > 0, λ > 0, μ > 0, t > 0}]

As stated above, the expression is returned unevaluated since there is 
indeed three possible answers depending on the respective value of s, 
mu, and lambda. We can see by inspection that the limit of the 
exponential function depends on the value of

In[5]:=
Simplify[E^((-μ)*t*(s - 1 - λ/μ)),
   Assumptions -> {s â?? Integers, s > 0, λ > 0,
     μ > 0, t > 0}]

Out[5]=
  t (λ + μ - s μ)
E

Therefore, we compute the different values in three steps

In[6]:=
Limit[(1 - E^((-μ)*t*(s - 1 - λ/μ)))/
    (s - 1 - λ/μ), t -> Infinity,
   Assumptions -> {s â?? Integers, s > 0, λ > 0,
     μ > 0, t > 0, λ + μ - s*μ > 0}]

Out[6]=
Infinity

In[7]:=
Limit[(1 - E^((-μ)*t*(s - 1 - λ/μ)))/
    (s - 1 - λ/μ), t -> Infinity,
   Assumptions -> {s â?? Integers, s > 0, λ > 0,
     μ > 0, t > 0, λ + μ - s*μ == 0}]

Out[7]=
0

In[8]:=
Limit[(1 - E^((-μ)*t*(s - 1 - λ/μ)))/
    (s - 1 - λ/μ), t -> Infinity,
   Assumptions -> {s â?? Integers, s > 0, λ > 0,
     μ > 0, t > 0, λ + μ - s*μ < 0}]

Out[8]=
        μ
-(-----------)
   λ + μ - s μ

Best regards,
Jean-Marc


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