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: Suppose $\lim_{n\rightarrow \infty} nx_n = L$. Fix $\epsilon>0$ then there exists $N$ such that for all $n>N$
\begin{align}
\left||x_n|-\frac{L}{n} \right|\leq \left|x_n - \frac{L}{n} \right| <\frac{\epsilon}{n}.
\end{align}
