# 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).
• But since $(x_n)$ is a sequence of positive values, doesn't that imply that $x_n > 0$ for all $n \in \mathbb{N}$ and so $x > 0$? – user378513 Oct 14 '16 at 7:42
• @user378513 No, take for example $x_n=1/n\to 0.$ – Robert Z Oct 14 '16 at 7:48
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}