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[Î±*t]], t -> Infinity,
Assumptions -> {Î± > 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[{Î± = 1}, Limit[Erf[1/Sqrt[Î±*t]],
t -> Infinity]]

Out[2]=
0

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

Out[3]=
0

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

Out[4]=
0

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

Out[5]=
Infinity

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

Out[6]=
Infinity

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

Out[7]=
Infinity

Regards,
Jean-Marc

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