Proof that $a_n < b_n \implies \lim(a_n) \le \lim(b_n)$ First of all, I'm aware that there are many questions like this on the site, but they all seem to be related to either $\limsup$ or $\liminf$ and I couldn't find anything that would help me with my problem. I've done some Googling and found some great resources, but I'm still not quite sure how to get to some steps and would like your assistance.

The problem is as follows:

Given $a_n &lt; b_n$ prove that $\lim_{n\to \infty}(a_n) \le \lim_{n\to\infty}(b_n)$. The proof is then done by contradiction, assuming that $a = \lim_{n\to \infty}(a_n) &gt; b =\lim_{n\to\infty}(b_n)$.

We take an $\epsilon = \frac{a-b} 2$, so that the $\epsilon$-neighborhoods of $a$ and $b$ are disjoint. From the definition of limits, we now know that there is such a $N$, so that $\forall n &gt; N : |a_n-a|&lt;\frac\epsilon2$ and $|b_n-b|&lt;\frac\epsilon2$.

The next step is absolutely always confusing. Two variants I've found are either:

$a_n&gt;a-\epsilon=a-\left(\frac{a-b} 2\right)=b+\left(\frac{a-b} 2\right)=b+\epsilon&gt;b_n$

In which I do not understand why any two terms of that (in)equality are like that, or it is said that if $a &gt; b$, there must be such an $\epsilon$ so that $a - \epsilon &gt; b + \epsilon$. Then,
$a - \epsilon &gt; b + \epsilon &gt; b_n, a_n &gt; a - \epsilon &gt; b + \epsilon &gt; b_n$, which contradicts $a_n \le b_n$. Here I simply cannot comprehend how we came to the conclusion that $a_n &gt; a - \epsilon$ and $b + \epsilon &gt; b_n$. The definition of the limit uses absolute values everywhere, so surely the values depends on the signs of $a_n$ and $a$, and $b_n$ and $b$. Please help me understand what is it that I'm missing here. Thanks in advance!
$\newcommand{\eps}{\varepsilon}$If $\eps &gt; 0$ is real, then for every real number $x$,
$$
\text{$|x| &lt; \eps$ if and only if $-\eps &lt; x &lt; \eps$.}
$$
Particularly,
\begin{align*}
  |a_{n} - a| &lt; \eps &amp;\quad \text{if and only if}\quad -\eps &lt; a_{n} - a &lt; \eps, \\
  &amp;\quad \text{if and only if}\quad a - \eps &lt; a_{n} &lt; a + \eps.
\end{align*}

Second, if $a$ and $b$ are real numbers such that $b &lt; a$, then $b &lt; \frac{1}{2}(b + a) &lt; a$ (the mean/midpoint lies between). A bit of algebra shows that if $\eps = \frac{1}{2}(a - b)$, then $\eps &gt; 0$, and
$$
b + \eps = \tfrac{1}{2}(b + a) = a - \eps.
$$

In each case, it's a pleasant and instructive exercise to sketch a number line and see the obvious geometric fact these inequalities express.
$a_n=0$ for all $n$ and $b_n=\frac{1}{n}$