How to prove that $\lim_{n \to \infty} n x^{n} = 0 $ when $0Intuitively it's easy, but hard to prove by the epsilon-delta method:
$$ \lim_{n \to \infty} n x^{n} = 0$$

Let $a_n=nx^n$. You have
$$\frac{a_{n+1}}{a_n}= \frac{n+1}n x = \left(1+\frac1n\right) x.$$
Let $q=\frac{1+x}2$. (Or we could choose any $q$ such that $x<q<1$.)
There exists $n_0$ such that $\left(1+\frac1n\right) x \le q$ for $n\ge n_0$. (Since $\lim\limits_{n\to\infty} \left(1+\frac1n\right) x = x < q$.)
Thus we have 
$$0 \le a_n \le a_{n_0} \cdot q^{n-n_0}$$
for $n\ge n_0$. (You can show this easily by induction.)
Since $q^n\to 0$ for $n\to\infty$ we get
$$\lim_{n\to\infty} a_n=0.$$
