Re: Limit with restrictions
- To: mathgroup at smc.vnet.net
- Subject: [mg34288] Re: [mg34269] Limit with restrictions
- From: BobHanlon at aol.com
- Date: Tue, 14 May 2002 04:09:56 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
In a message dated 5/13/02 7:00:15 AM, hannes.egli at uni-greifswald.de writes:
>Is there a possibility to combine the command Limit with parameter
>restrictions?
>
>A simple example reads:
>
>p= m^(-1 + a + b + n)*(a/(a + b + n))^a*((b + n)/(a + b + n))^(b +
>n)*(a + b + n)
>
>The following restrictions are given:
>
>restrictions = {a > 0, b > 0, n > 0, a + b + n < 1}
>
>Since Limit does not allow restrictions, I tried to combine Limit with
>Simplify, since in other examples (e.g. determination of an equation's
>sign) this has worked very well:
>
>Simplify[Limit[p, m -> Infinity],restrictions]
restrictions={a>0,b>0,n>0,a+b+n<1};
p=m^(-1+a+b+n)*(a/(a+b+n))^a*((b+n)/(a+b+n))^(b+n)*(a+b+n);
Needs["Calculus`Limit`"];
Limit[p, m->Infinity]
0
You would generally want to use Simplify (or FullSimplify) with the
restrictions before applying Limit. If the resulting limit is
complicated you might then want to re-use Simplify (or FullSimplify).
Limit[Simplify[p, restrictions], m->Infinity]
0
However, in general Simplify (or FullSimplify) with the restrictions
should be applied when defining p so that all subsequent uses of
p have the simpler form.
p=Simplify[
m^(-1+a+b+n)*(a/(a+b+n))^a*((b+n)/(a+b+n))^(b+n)*(a+b+n),
restrictions];
Bob Hanlon
Chantilly, VA USA