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MathGroup Archive 1993

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Re: conditional limits

  • To: mathgroup at yoda.physics.unc.edu
  • Subject: Re: conditional limits
  • From: kenward at mdd.comm.mot.com (Gary Kenward)
  • Date: Wed, 8 Sep 93 08:23:52 PDT

Thanks. The best suggestion I've had so far.
So simple, and it works well. Not as obvious as
it seems, for you were the first of a half dozen or 
so responders to suggest this solution.

I have forwarded your suggestion to the mathgroup.

Regards,
Gary Kenward

----- Begin Included Message -----

>From paul at earwax.pd.uwa.oz.au Wed Sep  8 02:59:43 1993
Mime-Version: 1.0
Content-Type: text/plain; charset="us-ascii"
Date: Wed, 8 Sep 1993 17:56:19 +0800
To: kenward at mdd.comm.mot.com (Gary Kenward)
From: paul at earwax.pd.uwa.oz.au (Paul C Abbott)
Subject: Re: conditional limits
Content-Length: 937
X-Lines: 36

>I have a function of two variables, f[x, y], which when evaluated for values of
>x=y results in:
>                                               1
>              Power::infy: Infinite expression -- encountered.
>                                               0.              
>
>As there is an (x-y) term in the denominator of the function.  However,
>the Limit[f[x,y], x-y] does exist, and evaluates to a finite number.
>
>
>The question is, how do I construct a rule that conditionally evaluates
>the limit of f[x,y] rather than f[x,y] whenever x=y?

How about the following:

f[x_, y_] := Sin[x-y]/(x-y)

f[1,1]

                                 1
Power::infy: Infinite expression - encountered.
                                 0

Infinity::indet: 
   Indeterminate expression 0 ComplexInfinity encountered.

Indeterminate

All you need to do is enter an extra rule like:

f[x_,x_] = Limit[f[x,y], y -> x];

Regards,
        Paul




----- End Included Message -----







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