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
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>From paul at earwax.pd.uwa.oz.au Wed Sep 8 02:59:43 1993
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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
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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)
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
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