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.
Best Answer
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 $0$.