Suppose $\lim \limits_{n \to \infty} \frac{a_n}{b_n} = L$, and $a_n > 0$, $b_n > 0$. Then, for all $\varepsilon$ there is an $N$ such that for all $n \geq N$:
$$\left \vert \frac{a_n}{b_n} - L \right \vert < \varepsilon$$ $$-\varepsilon + L < \frac{a_n}{b_n} < \varepsilon + L$$ $$b_n < \frac{1}{L-\varepsilon} a_n$$
Am I making a mistake here? I ask because, say I choose $\varepsilon$ such that it is greater than $L$. Wouldn't this last inequality be false, since the term would be negative and $b_n$ is positive?
If $L-\varepsilon < 0$, then you would need to reverse the inequality when you divide through by it, so you would actually have $$b_n > \frac{1}{L-\varepsilon} a_n.$$