Prove the following: If $\lim(x_n) = x$ and if $x > 0$, then there exists a natural number $M$ such that $x_n > 0$ for all $n\ge M$.

Question

I recently encountered the following question:

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

My solution manual suggests that I should choose $\epsilon=\frac{x}{2}$. I tried working out the same procedure with $\epsilon=\frac{x}{3}$. I could still obtain $x_n>0$ for $n\geq M$ ($x_n > \frac{2x}{3} $ to be precise).

Is it correct to choose $\epsilon=\frac{x}{3}$? To extend the principle, is it correct to choose $\epsilon=\frac{x}{a}$ where a is a natural number and $a<x $?

Can anyone please help?

Thanks in advance!

Answer 1

Yes, it is correct. More generally, you can take $\varepsilon$ equal to any number smaller than or equal to $x$ (and greater than $0$). As far as I am concerned, the more natural choice is to take $\varepsilon=x$, but that's a matter of taste.