show that if ${a_n} \to a$, and $a>0$, then $\exists N$ such that $a_n>0$ for $n \ge N$.
Solution 1:
We have that $a_n\to a$ and that $a>0$. In particular, since $a/2>0$, there must be some $N$ such that, for $n\ge N$, $a_n$ is within distance $a/2$ of $a$. In symbols, $$ |a_n-a|<\frac{a}{2} $$ for all $n\ge N$. This follows from taking $\varepsilon=\frac{a}{2}$ in the definition of convergence.
Now note that $$ |a_n-a|<\frac{a}{2}\iff -\frac{a}{2}<a_n-a<\frac{a}{2} $$ The left inequality then gives $$ \frac{a}{2}<a_n $$ for all $n\ge N$.