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MathGroup Archive 2006

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Re: Limit Question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg66465] Re: Limit Question
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Sun, 14 May 2006 02:58:01 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <e43vun$90e$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Blimbaum, Jerry CIV NSWC PC wrote:
> Why do these give no output?
> 
> Simplify[Limit[Exp[-a t],t->Infinity],a>0]
> 
> and  same for
> 
> Simplify[Limit[Exp[a t],t->Infinity],a<0]

Hi Jerry,

First, *Simplify* evaluates its arguments. The assumption 'a > 0' is not 
passed on to the *Limit* function. Since the limit statement returned 
unevaluated -- the result depends on the value of the parameter 'a' and 
*Limit* does nothing about 'a' at this stage -- *Simplify* has nothing 
to simplify. You can see this process using the *Trace* function (see 
In[2]).

To pass the assumption to the limit statement, wrap it inside an 
*Assuming* function [1], as in In[3] and In[4].

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

Out[1]=
        -a t
Limit[E    , t -> Infinity]

In[2]:=
Trace[Simplify[Limit[Exp[(-a)*t], t -> Infinity],
    a > 0]]

In[3]:=
Assuming[a > 0, Limit[Exp[(-a)*t], t -> Infinity]]

Out[3]=
0

In[4]:=
Assuming[a < 0, Limit[Exp[(-a)*t], t -> Infinity]]

Out[4]=
Infinity

In[5]:=
$Version

Out[5]=
5.2 for Microsoft Windows (June 20, 2005)

Best regards,
Jean-Marc

[1] http://documents.wolfram.com/mathematica/functions/Assuming


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