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MathGroup Archive 2007

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Re: Limit of Error function Erf

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73565] Re: Limit of Error function Erf
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Thu, 22 Feb 2007 04:27:07 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <ergpqn$hbq$1@smc.vnet.net>

ben wrote:
> Dear group
> 
> I dont understand the following behaviour of Mathematica,
> I would say the first result is simply wrong
> 
> 1.)
> 
> Limit[Erf[1\/Sqrt[\[Alpha]\ t]], t -> \[Infinity],
>     Assumptions -> {\[Alpha] > 0}]
> 
> gives infinity
> 
> 2.)
> 
> Limit[Erf[1\/Sqrt[\[Alpha]\ t]], t -> \[Infinity],
>     Assumptions -> {\[Alpha] == 1}]
> 
> gives zero
> 
> Any suggestions? Is this a bug?
> Bye
> Ben
> 
> 
Same behavior with Mathematica 5.2 for Microsoft Windows (June 20, 2005).

It seems that the issue is related to the combination parameter and 
fractional powers.

In[1]:=
Limit[Erf[1/Sqrt[α*t]], t -> Infinity,
   Assumptions -> {α > 0}]

Out[1]=
Infinity

(* It is usually better to use a clause With to temporarily set the 
value of a parameter rather than using Assumption -> param == value. *)

In[2]:=
With[{α = 1}, Limit[Erf[1/Sqrt[α*t]],
    t -> Infinity]]

Out[2]=
0

In[3]:=
Limit[Erf[1/Sqrt[t]], t -> Infinity]

Out[3]=
0

In[4]:=
Limit[Erf[1/(α*t)], t -> Infinity,
   Assumptions -> {α > 0}]

Out[4]=
0

In[5]:=
Limit[Erf[1/(α*t)^(1/2)], t -> Infinity,
   Assumptions -> {α > 0}]

Out[5]=
Infinity

In[6]:=
Limit[Erf[1/(α*t)^(1/4)], t -> Infinity,
   Assumptions -> {α > 0}]

Out[6]=
Infinity

In[7]:=
Limit[Erf[(α*t)^(-(1/2))], t -> Infinity,
   Assumptions -> {α > 0}]

Out[7]=
Infinity

Regards,
Jean-Marc


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