Sequence Question If $(x_n)$ is a sequence of positive values and $\lim_{n\to\infty} n x_n $ exists, prove that $(x_n) \rightarrow 0$.
Since $\lim_{n\to\infty} n x_n $ exists, we know $(nx_n)$ converges to some positive number; call it $x$. Let $\varepsilon > 0$. Then we can find an $N \in \mathbb{N}$ such that $|nx_n - x| < \varepsilon$. This is implies that $-\varepsilon < nx_n - x < \varepsilon$, which is equivalent to $\frac{x - \varepsilon}{n} < x_n < \frac{x + \varepsilon}{n}$. I am stuck here. How can I show $(x_n) \rightarrow 0$. Thanks. 

Hint. We have that  $\frac{x - \varepsilon}{n}\to 0$ and $\frac{x + \varepsilon}{n}\to 0$ as $n$ goes to infinity. 
Notice that the limit $x$ is greater or equal to zero (it is not necessarily positive).
