$\infty\cdot 0$ Indetermination without L'Hopital? Evaluate $\lim_{n\to\infty}n\cdot r^n$ , being $0<r<1$.
I dont know if I took the proper steps, but I get to this point:
$$\lim_{n\to\infty}n\cdot r^n=\lim_{n\to\infty}n \lim_{n\to\infty}r^n = \infty \cdot  0 $$ 
I dont know how to solve this indetermination without L'hopital's rule.

$0<r<1.$
Set $r=\dfrac{1}{1+x}$, where $x >0$.
$(1+x)^n =$
$ 1+ nx + (1/2)n(n-1)x^2+......$
$\dfrac{n}{(1+x)^n} =$
$\dfrac{n}{1+nx +(1/2)n(n-1)x^2+...} < $
$\dfrac{2n}{n(n-1)x^2}<(2/x^2) \dfrac{n}{(n-1)^2} <$
$(2/x^2)\dfrac{n}{(n-n/2)^2}= (8/x^2)\dfrac {n}{n^2}=$
$(8/x^2)\dfrac{1}{n}.$
The limit is?
