real analysis - Alternative proofs that if $a_n leqslant b_n$ then $lim a_n leqslant lim b_n$

A well-known limit property asserts that if $a_n$ and $b_n$ are convergent sequences and $a_n \leqslant b_n$ for all $n$, then $\lim a_n \leqslant \lim b_n$.The most common means of proof is by contrapositive, but are there any other nice ways of proving this without using contrapositive?

Answer

Assume $a_n \to A$ and $b_n \to B$. We use the fact that $A \le B$ if and only if $A < B + \epsilon$ for every $\epsilon > 0$.

Let $\epsilon > 0$ be given.

There exists an index $n$ with the property that $|a_n - A| < \epsilon/2$ and $|b_n - B| < \epsilon/2$. Thus $$A < a_n + \epsilon/2 \le b_n + \epsilon/2 < B + \epsilon.$$

Thus $A \le B$.