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 $n\geq M$.

I'm not quite sure what to do with this one. By definition I know the following:

For $\varepsilon>0$, there exists $k\in\mathbb{N}$ such that $|x_n-x|<\varepsilon$ for all $n\geq k$.

Equivalently I can say: $$-\varepsilon < x_n - x < \varepsilon$$ $$\Rightarrow x-\varepsilon < x_n < x +\varepsilon$$

I'm not quite sure what I can do with the fat that $x>0$, or what I could do to relate $k$ and $M$. This problem makes sense intuitively but I don't quite see what my next step in the proof would be.

  • $\begingroup$ What happens if you pick $\varepsilon = x/2$? $\endgroup$ – Alex Wertheim Oct 2 '15 at 4:48
  • $\begingroup$ I thought I was onto something with that hint but I haven't really gotten anywhere. Perhaps you could give another hint? $\endgroup$ – flubsy Oct 2 '15 at 4:54
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    $\begingroup$ Sure, I'll write up a more complete answer. :) $\endgroup$ – Alex Wertheim Oct 2 '15 at 4:55

You're essentially almost all of the way there. As you correctly noted, that $\lim_{n \to \infty} x_{n} = x$ for some $x > 0$ means that for every $\varepsilon>0$, there exists $k\in\mathbb{N}$ such that $|x_n-x|<\varepsilon$ for all $n\geqslant k$.

Since $x > 0$, $x/2 > 0$, so choose $\varepsilon = x/2$. Then as you pointed out, there exists $k \in \mathbb{N}$ such that for all $n \geqslant k$, $|x_n-x|<\varepsilon$, i.e. $x - \frac{x}{2} < x_{n} < x + \frac{x}{2}$, whence $0 < \frac{x}{2} < x_{n}$.

The idea here is to think about the picture. Since $x$ is the limit of $x_{n}$, we can always go far out in the sequence to get as close to $x$ as we want. Since $x > 0$, we can specifically go far out enough in the sequence so that for large enough $n$, $x_{n}$ in the ball of radius $x/2$ about $x$. This keeps us away from $0$, and specifically greater than $0$.

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  • $\begingroup$ I know this is a while later, but could you explain the reasoning behind choosing $\epsilon = \frac{x}{2}$ $\endgroup$ – Curious Student Nov 7 '17 at 21:18
  • $\begingroup$ @CuriousStudent, it's because $x>0$ and $\frac{x}{2} > 0$ moreover $x-\frac{x}{2} > 0$ (since $\frac{x}{2} < x$) hence $x_n > 0$ for all $n > k$ $\endgroup$ – Florian Suess Sep 8 '18 at 1:10

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