From: SourceForge.net <noreply@so...>  20071125 15:49:50

Bugs item #1483121, was opened at 20060506 16:47 Message generated for change (Comment added) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1483121&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core  Limit Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: limit(1/x, x, infinity) Initial Comment: (%i15) limit(1/x, x, infinity); (%o15) 0 (%i16) limit(1/x, x, infinity); (%o16) 1/infinity (should it be 0!) Please include the following build information with your bug report:  Maxima version: 5.9.3 Maxima build date: 0:52 3/20/2006 host type: i686pcmingw32 lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.7   >Comment By: Dan Gildea (dgildea) Date: 20071125 10:49 Message: Logged In: YES user_id=1797506 Originator: NO limit(1/x, x, infinity) probably indicates user misunderstanding, but limit(1/x, x, inf) seems natural  see bug 1594977. For this reason, I added a call to infsimp on the third arg of limit. So as of limit.lisp rev 1.47: (%i51) limit(1/x, x, infinity); (%o51) 0 If we don't call infsimp, I think we should generate an error in this case.  Comment By: Stavros Macrakis (macrakis) Date: 20060829 11:16 Message: Logged In: YES user_id=588346 It's very simple. The third argument to limit can only be a finite number, inf, minf, inf, or infinity. Limit doesn't accept other values. So, for example, limit(2*inf) => inf, but limit(x,x,2*inf) gives 2*inf, because limit interprets that as a finite value. I suppose as a matter of userfriendliness, we could have limit first apply the singleargument version of limit to its third argument. But that might encourage misunderstandings.  Comment By: Robert Dodier (robert_dodier) Date: 20060515 20:08 Message: Logged In: YES user_id=501686 Given that limit (1/infinity) => 0, the result limit (1/x, x, infinity) => 1/infinity certainly looks like a bug to me (i.e., can't be explained away by invoking the inf/infinity distinction).  Comment By: Barton Willis (willisbl) Date: 20060513 17:30 Message: Logged In: YES user_id=895922 infinity is the complex infinity. Maybe you wanted to do limit(1/x,x,minf). Notice that limit(infinity) > infinity and limit(1/infinity) > 0. Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1483121&group_id=4933 