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 -----