For $0I have problems with computing the following limit:

Given a sequence $0<q_n<1$ such that $\lim\limits_{n\to\infty} q_n =q < 1$, prove that for a fixed $k \in \mathbb N$, $\lim\limits_{n\to\infty} n^k q_n^n= 0$.

I know how to prove this, but I can't do it without using L'Hôpital's Rule. Does someone have an elementary proof?

Take the positive series
$$\sum_{n=1}^\infty n^kq^n$$
apply now, say Cauchy's $\,n$-th root test:
$$\sqrt[n]{n^kq_n^n}=\left(\sqrt[n]n\right)^kq_n\xrightarrow[n\to\infty]{}1^k\cdot q=q<1$$
thus the series converges and from here the general term's sequence converges to zero.
