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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