Without additional assumption that x is constant
the limit is not zero (take for example x=(1/2)^(1/(n+1))
(W... alpha:
Assuming[x=const,x<1,x>0];Limit[x^(n+1)/(1-x),n->+Infinity] 0 OK,
Assuming[x<1,x>0];Limit[x^(n+1)/(1-x),n->+Infinity] unevaluated OK)
On 15 Kwi, 06:00, Dan Drake <dr...@kaist.edu> wrote:
> Why doesn't this work?
>
> sage: assume(x > -1)
> sage: assume(x < 1)
> sage: n = var('n')
> sage: limit(x^(n+1)/(1-x), n=infinity)
> -limit(x^(n + 1), n, +Infinity)/(x - 1)
>
> ...when this works:
>
> sage: forget()
> sage: assume(0 < x)
> sage: assume(x < 1)
> sage: limit(x^(n+1)/(1-x), n=infinity)
> 0
>
> and
>
> sage: forget()
> sage: assume(x < 0)
> sage: assume(-1 < x)
> sage: limit(x^(n+1)/(1-x), n=infinity)
> 0
>
> ?
>
> Any ideas? I'm sure this is some Maxima thing, but I don't know how to
> work with Maxima.
>
> Thanks!
>
> Dan
>
> --
> --- Dan Drake
> ----- http://mathsci.kaist.ac.kr/~drake
> -------
>
> signature.asc
> < 1KWyświetlPobierz
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