1. Sequence question

I am studying for my final in undergrad analysis, and have a question concerning sequences. Here it is. Let $\{s_n\}$ be a sequence $\backepsilon \{s_n\}>0 \; \forall \; n \in \mathbb{N}.$ Suppose $\lim_{n \to \infty}s_n=s.$ Show that $s \ge 0.$

2. Pf: Assume $s_n > 0 \forall n \in \mathbb{N}$ and $\displaystyle\lim_{n\to\infty} s_n = s$ Now assume $s < 0$ So, By def $\forall \epsilon > 0 \exists N \in \mathbb{N} | \forall n \geq N \: |s_n - s|<\epsilon$ By the triangle ineq $\forall \epsilon > 0 \: |s_n| - |s| < \epsilon$ which means $|s_n| \rightarrow |s| as n \rightarrow \infty$ since $s_n > 0 \forall n$ we have $s_n \rightarrow |s| as n \rightarrow \infty$ since $|s|>0$ this is a contradiction.
3. In other words, if s<0, then for a given $\epsilon$ with $s+\epsilon<0$, you should have elements of $s_n$ existing in the $\epsilon$ neighborhood of s (by limit to s). Since $s+\epsilon<0$, then those elements are also negative which gives a contradiction.
4. I think there's a somewhat easier way. Suppose that $s_{n} > 0$ for all $n\in \mathbb{N}$. Suppose $\lim_{n\to \infty} s_{n} = s < 0$. Then we know there exists $N$ such that $n>N$ implies $|s_{n} - s| < -s$. This is true because such an implication must hold for any $\epsilon > 0$ and if $s < 0$ then $-s > 0$, so this is the particular case where $\epsilon = -s$. So this means $s < s_{n} - s < -s$ for $n>N$. Adding $s$ to each part of the equality, we get for $n>N$, $2s < s_{n} < 0$. That means after some point, $s_{n}$ will be negative! And there arises your contradiction.