Let $s_n$ be a convergent sequence, and suppose that $\lim s_n>a$. Prove that there exists a number $N\in\mathbb{N}$ such that for every $n>N$ we have that $s_n>a$
Proof: Let $s_n$ be a convergent sequence with $\lim s_n=s$, and $s>a$. Since $s-a>0$ and $s_n$ is a convergent sequence, we can choose $\epsilon=s-a$ in the definition of convergence. So there exists an $N\in\mathbb{N}$, such that for every $n\in\mathbb{N}$, with $n>N$, we have, \begin{align*} |s_n-s|<& s-a\\ -s+a< s_n-&s< s-a\\ a< s_n<& 2s-a. \end{align*} Hence there exists an $N\in\mathbb{N}$ such that for all $n>N$ we have $s_n>a$. QED