Advanced Calculus - Limit Proof

Prove that if $\displaystyle lim(x_n)=x$ and if $\displaystyle x>0$, then there exists a natural number $\displaystyle M$ such that $\displaystyle x_n>0$ for all $\displaystyle n \geq M$.

**My thoughts**

Intuitively, this makes sense. If as $\displaystyle n -> \infty, x_n=x>0$ then this means that at some point $\displaystyle n \geq M$ where $\displaystyle x_n>0$.

So far I've gotten:

Given $\displaystyle \epsilon > 0$ then,

$\displaystyle |x_n-x|<\epsilon$

$\displaystyle -\epsilon<x_n-x<\epsilon$

$\displaystyle -\epsilon + x <x_n<\epsilon + x$