$\infty\cdot 0$ 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.

You want $\lim_{n\to\infty}n\exp -cn$ with $c:=-\ln r>0$. Since $\int_0^\infty n\exp -cn\operatorname{d}n=c^{-2}$ is finite, the limit is $0$.
