In the last post one can replace 0.99 by 1
but I wanted to exclude the following situation:
sage: n=var('n')
sage: x=var('x')
sage: x=(1/2)^(1/(n+1))
sage: limit(x^(n+1)/(1-x),n=+oo)
+Infinity # OK
On 16 Kwi, 15:36, achrzesz <achrz...@wp.pl> wrote:
> 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
>
>
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