On Sunday, January 22, 2017 at 4:13:53 PM UTC, William wrote:
>
> On Sun, Jan 22, 2017 at 7:14 AM, Peleg Michaeli <free...@gmail.com 
> <javascript:>> wrote: 
> > Hi, 
> > 
> > sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity) 
> > -Infinity 
> > 
> > This is a wrong answer. It should be 2. 
> > 
> > Replacing 3/2 in the power by 1, 2, or 3 (at least) gives correct 
> answers 
> > (2, inf, inf).  Replacing it by 5/2 given a wrong answer again. 
> > 
> > Is this related to a known bug? 
>
> I don't know if it is known - - you can try searching 
> https://trac.sagemath.org/, which lists all bugs we know of. 
>
> This is actually a bug in **MAXIMA**, which Sage calls to do this 
> limit: http://maxima.sourceforge.net/ 
>

no, it's not in Maxima 5.39.0 (see https://trac.sagemath.org/ticket/18920 
and give it a positive review :-))
called with its default settings.

$ sage --maxima
;;; Loading #P"/home/dima/Sage/sage-dev/local/lib/ecl/sb-bsd-sockets.fas"
;;; Loading #P"/home/dima/Sage/sage-dev/local/lib/ecl/sockets.fas"
;;; Loading #P"/home/dima/Sage/sage-dev/local/lib/ecl/defsystem.fas"
;;; Loading #P"/home/dima/Sage/sage-dev/local/lib/ecl/cmp.fas"
Maxima 5.39.0 http://maxima.sourceforge.net
using Lisp ECL 16.1.2
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) limit((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x),x,inf);
(%o1)                                  2

Although calling this version of Maxima as  
sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity)

gives -Infinity, still, as it's not called with default settings, but with 
something
that breaks it sometimes.


> If you call sympy instead it gets the right answer: 
>
> sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity, 
> algorithm="sympy") 
> 2 
>
> sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity, 
> algorithm="maxima_taylor") 
> -Infinity 
>
> sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity, 
> algorithm="maxima") 
> -Infinity 
>
>
>
> > 
> > 
> > Thanks, 
> > Peleg. 
> > 
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>
>
>
> -- 
> William (http://wstein.org) 
>

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