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Re: On Limit[ f[x,y], x->x0 ]

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  • Subject: [mg34792] Re: [mg34780] On Limit[ f[x,y], x->x0 ]
  • From: Andrzej Kozlowski <andrzej at>
  • Date: Fri, 7 Jun 2002 01:08:54 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

I agree that it looks like a possible oversight. The only thing I can 
suggest is something like this:

limit[f_[x_], Rule[a_, b_], opts___] := Limit[f[x], Rule[a, b], opts];
limit[f_[x__], Rule[a_, b_], opts___] /;
       MemberQ[Attributes[f], NumericFunction] :=
     Limit[f[x], Rule[a, b], opts];
limit[f_[x__], Rule[a_, b_], opts___, Analytic -> True, opts1___] :=
     Limit[f[x], Rule[a, b], opts, Analytic -> True, opts1];
limit[f_[x__], Rule[a_, b_], opts___] :=
   HoldForm[Limit[f[x], Rule[a, b], opts]]

Now you get:





while in other cases you ought to get whatever Limit gives (I hope!). Of 
course you should not forget about the HoldForm in Out[45] above.

Andrzej Kozlowski
Toyama International University

On Thursday, June 6, 2002, at 02:55  PM, Jack Goldberg wrote:

> Hi Group,
> If this is a repeat please forgive.  I have not seen my original post.
> There appears to be some unfortunate behavior of  Limit[f[x,y],x->x0].
> Limit[ f[x], x->0 ]  for general f  simply returns the limit unevaluated
> unless the option  Analytic->True is invoked and then
> 	Limit[ f[x], x->0 ]  returns  f[0]
> However, if f  is a function of more that 1 variable or has more than 1
> slot, we get this unexpected behavior:
> 	Limit[ f[x,y,1], x->0 ]  returns f[0,x,1]
> whether or not the option  Analytic->True is invoked.  Likewise for the
> functions   f[x,x],  f[x,y], f[x,y,z] etc.  Of course  x->0 is not the
> essential feature,  x->x0  results in the same "errors".
> I hate to call this a bug, but it sure 'taint a feature!  After all, 
> if  f
> is discontinuous at  x0,  the expression returned could be (and often 
> is)
> false.  A second difficulty with this "feature" is that the conditional
> 	/; "some expression involving Limit"
> will evaluated when such evaluation is not expected.
> Comments appreciated!  As usual -
> Jack

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