Question

lim xr^x as x approsches infinity and abs r <1

Answer 1

\[\left| r \right| <1, \lim_{x \rightarrow \infty} xr^x \]

Answer 2

Limit will limit to 0. x is simply a number, integer mind you. Taking any r < 1 and raising it to an infinite amount will be a smaller number than x will be. Think of it as a convergence test if you know how MacLauren Series work. While it obviously won't prove the limit of it, you can think that 10000000000(.1)^10000000000 will be a very small number.

Answer 3

I'm not able to calculate / prove the limit either, but logically the exponential (r^x) function goes to zero many times faster than the linear (x) function goes to infinity, and therefor the limit is 0.

Not sure how this can be proved.

Answer 4

Why no use L'Hospital ?
Encode |r|<1 as
a) r=1/u
and
b) r=-1/u
then you get by substituting
lim x/(u^x) as x goes to inf
this is an indeterminate case inf/inf
use L'Hospital for both cases a and b
and you get 1/( x * (u^(x-1))) and 1/(x*(u^(x-1))), that is 1/inf
that is 0