# Prove that if $\displaystyle \lim_{n\to \infty} x_n = x$ and if $x > 0$, there exists a natural number $M$ such that $x_n >0$ for all $n > M$.

Prove that if $\displaystyle \lim_{n\to \infty} x_n = x$ and if $x > 0$, then there exists a natural number $M$ such that $x_n >0$ for all $n > M$.

Is this not just a proof of the Archemedian property?

• Thanks for the edit ^^. I'm not too familiar with MathJax yet haha – Jesse Liu Feb 22 '15 at 3:02

Let $\epsilon = \dfrac{x}{2} > 0 \to \exists M > 0 : |x_n-x| < \dfrac{x}{2} \to -\dfrac{x}{2} < x_n - x < \dfrac{x}{2} \to x_n > \dfrac{x}{2} > 0$ for $n > M$.