Limit of convergent sequences Suppose that $\{a_n\}$ and $\{b_n\}$ are convergent sequences. Show that if $\lim a_n &lt; \lim b_n$, then there exists a natural number $N$ such that $a_n &lt; b_n$ holds for all $n\geq N$. 


I don't know how to solve this problem. Could someone help, thanks in advance.
We have $$\lim_na_n=A&lt;B=\lim_nb_n.$$ By definition

$$\forall \epsilon&gt;0\exists N_1\in\mathbb{N}: n\ge N_1\implies A-\epsilon&lt;a_n&lt;A+\epsilon$$ and 

$$\forall \epsilon&gt;0\exists N_2\in\mathbb{N}: n\ge N_2\implies B-\epsilon&lt;b_n&lt;B+\epsilon.$$ In particular, for $\epsilon=\frac{B-A}{2}$ we have that

$$\exists N_1\in\mathbb{N}: n\ge N_1\implies a_n&lt;A+\frac{B-A}{2}=\frac{A+B}{2}$$ and 

$$\exists N_2\in\mathbb{N}: n\ge N_2\implies \frac{A+B}{2}=B-\frac{B-A}{2}&lt;b_n.$$ Thus, we have that

$$n\ge \max\{N_1,N_2\}\implies a_n&lt;\frac{A+B}{2}&lt;b_n.$$
Hint: Study the sequence $c_n = b_n - a_n$. What is the sign of $\lim_{n \to \infty}c_n$? What does that say about the sign of $c_n$ for large $n$? What does that say about the relationship between $a_n$ and $b_n$ for correspondingly large $n$?