Prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N. Let $(a_n)$ be a convergent sequence. Suppose that $\lim_{n\to\infty} a_n > 0$. Use the definition of a limit to prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N.
Use definition of a limit ($|a_n−L|<ϵ$).
Please check my proof!
My Proof:

Let $\epsilon$ be an arbitrary positive number. Let L = $\lim_{n\to\infty} a_n > 0$. Choose N ∈ (N) (naturals) such that n ≥ N implies that $|a_N−L|<\frac{\epsilon}{2}$ by definition of a limit. Assume n ≥ N. Therefore, $|2a_N−2L|=|2(a_N−L)|=|2||a_N-L|<\frac{\epsilon}{2}*2=\epsilon$. Since L > 0, $a_n$ > 0.

 A: I don't think your proof is quite right. Instead, try expanding the definition of a limit a little bit. The key will be to pick some $\varepsilon >0$ that is strictly less than the limit $L$, say $\varepsilon = \frac{L}{2}$ (we can do that because $L$ is positive). Now note that there exists $N\in \mathbb{N}$ such that for all $n\geq N$ we have  $$\lvert a_n - L\rvert < \varepsilon \iff - \varepsilon < a_n - L < \varepsilon.$$ Adding $L$ results in $$L-\varepsilon < a_n < L + \varepsilon.$$ Since we chose our $\varepsilon = \frac{L}{2}$, we have $$L-\varepsilon = L - \frac{L}{2} = \frac{L}{2} > 0,$$ because $L>0$. Hence $0< a_n$ for all $n\geq N$, as required.
A: I think you nearly have the idea - the only problem is that you never define $\varepsilon$ and then later assume things about it; in particular, you start with
$$|a_N - L | <\varepsilon /2$$
and then later conclude that
$$a_N > 0$$
from this - but this does not follow from what you have written, which implies that $\varepsilon$ could be chosen to be any positive number - and chosen before even inspecting the limit $L$ or the sequence $a_n$. Perhaps you're used to starting analysis proofs this way - as most proofs that a sequence converges do begin with letting $\varepsilon$ be an arbitrary positive number - but it's not appropriate here.
Rather, I would just eliminate $\varepsilon$ and replace its occurrences with, say, $L$. As an example:

Let $L=\lim_{n\rightarrow\infty}a_n$. Using the definition of a limit and the fact that $L>0$, we may choose $N\in\mathbb N$ such that if $n\geq N$ then $|a_n - L| < L$. This implies that $L-a_n < L$ and thus that $a_n>0$.

Where I've simplified some of the algebra as well - though the algebra is not the point. Note that here we are referencing the definition of a limit:

If $L=\lim_{n\rightarrow\infty}a_n$, then for every $\varepsilon>0$ there exists some $N\in\mathbb N$ such that if $n\geq N$ then $|L-a_n|<\varepsilon$.

We then supply some particular value of $\varepsilon$ to use - in this case, we use $L$. Essentially, if you want to prove that a limit exists, you need to write a proof that works for all $\varepsilon > 0$. If you already know that a limit exists, you can use this by choosing some particular $\varepsilon$ that works for your case - and choosing $\varepsilon = L$ gives the existence of some $N$ that works for your proof.
It's possible that your idea was to imagine, in your proof, that $\varepsilon$ will be really small (since we are free to choose it as whatever we want) - and the fact that $a_n$ stays arbitrarily close to $L$ after some time would imply that $a_n$ is positive. However, to capture this intuition formally, it is necessary to choose some particular radius around $L$ where all the values are positive.
