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:
$$ \lim_{n \to \infty} n x^{n} = 0$$

We have $nx^n=\exp(\log n)\exp(n\log x)=\exp(n\log x+\log n)$, so it's enough to show that $n\log x+\log n\to -\infty$ as $n\to +\infty$. We use the fact that $\log n\leq \sqrt n$ for $n$ large enough to see that 
$$n\log x+\log n\leq n\log x+\sqrt n=n\left(\log x+\frac 1{\sqrt n}\right).$$
As $\log x<0$, $\log x+\frac 1{\sqrt n}<\frac{\log x}2$ for $n$ large enough hence 
$$n\log x+\log n\leq n\frac{\log x}2,$$
which gives the result.
