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Re: conditional limits

  • To: mathgroup at
  • Subject: Re: conditional limits
  • From: kenward at (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.

Gary Kenward

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>From paul at Wed Sep  8 02:59:43 1993
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Date: Wed, 8 Sep 1993 17:56:19 +0800
To: kenward at (Gary Kenward)
From: paul at (Paul C Abbott)
Subject: Re: conditional limits
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>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)


Power::infy: Infinite expression - encountered.

   Indeterminate expression 0 ComplexInfinity encountered.


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

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


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