Prove if lim(x sub n)=x and x>0 then there exists a natural number M such that x sub n > 0 for all n> or equal to M
Any hints on how to start this or what the end result should be would be greatly appreciated, thanks.
if $\displaystyle {\rm{lim}}_{n \to \infty}{x_n}=x,\ \ x>0$ then for all $\displaystyle \varepsilon>0$ there exists an $\displaystyle N_{\varepsilon}$ such that:
$\displaystyle |x_n-x|<\varepsilon$ for all $\displaystyle n\ge N_{\varepsilon}$
Now choose $\displaystyle \varepsilon=x/2$ and you should find the result follows with $\displaystyle M=N_{\varepsilon}.$
CB
You can choose anything you like for $\displaystyle \varepsilon$, that is the definition of convergence for a sequence.
What we do is choose a particular value to ensure that $\displaystyle x_n$ is positive from $\displaystyle N_{\varepsilon}$ onwards. We do this by choosing any value of $\displaystyle \varepsilon \le x/2$ then as $\displaystyle |x_n-x|<\varepsilon$ for all $\displaystyle n \ge N_{\varepsilon}$ we have:
$\displaystyle |x_n-x|<x/2$
so:
$\displaystyle -x/2<x_n-x<x/2$
rearranging:
$\displaystyle x/2<x_n<3x/2$
for all $\displaystyle n \ge N_{\varepsilon}$
But $\displaystyle x>0$, so:
$\displaystyle x_n>0$ for all $\displaystyle n \ge N_{\varepsilon}$
CB