I must correct myself
W... alpha:
Assuming[x>-0.99,Assuming[x<0.99,Limit[x^(n+1)/(1-x),n->+Infinity]]
gives correct answer 0
On 16 Kwi, 09:27, achrzesz <achrz...@wp.pl> wrote:
> In W... alpha
> Assuming[x>-0.99,x<0.99];Limit[x^(n+1)/(1-x),n->+Infinity]
> remains unevaluated, so Maxima, Sage are nol alone
>
> On 16 Kwi, 08:18, achrzesz <achrz...@wp.pl> wrote:
>
> > One can discuss if in limits of f(x,n) as n-->oo
> > x may depend on n or not but in the following version:
>
> > sage: assume(x>-0.99,x<0.99)
> > sage: n=var('n')
> > sage: sage: limit(x^(n+1)/(1-x), n=infinity)
> > -limit(x^(n + 1), n, +Infinity)/(x - 1)
>
> > sage: assume(x>0)
> > sage: sage: limit(x^(n+1)/(1-x), n=infinity)
> > 0
>
> > the dependence of x on n is not essential
> > The limit does not depend on the sign of x
> > so the Maxima behaviour inherited by Sage is inconsistent.
>
> > On 16 Kwi, 05:56, ancienthart <joalheag...@gmail.com> wrote:
>
> > > (Respectfully) Then why does splitting the range of values for x into
> > > positive, zero and negative ranges work?
>
> > > Joal Heagney
>
>
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options, visit this group at
http://groups.google.com/group/sage-support
URL: http://www.sagemath.org
