My question is about the following proof:
Lemma: If $\lim_{n \to \infty} a_n = l$ and $l > 0$, then there is an integer $N$ such that $a_n > \frac{1}{2}l$ for all $n > N$.
Quotient Rule for convergent sequences
If $\lim_{n \to \infty} a_n = l$ and $\lim_{n \to \infty} b_n = m$, then $\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{l}{m}$, for $m \ne 0$.
Proof: For $m > 0$.
- Write the required expression in terms of the null sequences $a_n - l$ and $b_n - m$.
\begin{equation*} \frac{a_n}{b_n} - \frac{l}{m} = \frac{m(a_n -l) - l (b_n - m)}{b_n m} \end{equation*}
Note that $m(a_n -l) - l (b_n - m)$ is a null sequence, but the denominator $b_n m$ can take the value of $0$. However by the lemma, we know that for some integer $N$ we have $b_n > \frac{1}{2}m$, for all $n > N$, so the terms of $(b_n)$ are eventually positive.
Thus, for all $n > N$, we find
\begin{align*} \bigg| \frac{a_n}{b_n} - \frac{l}{m} \bigg| = \frac{|m(a_n -l) - l (b_n - m)|}{b_n m} \le \frac{|m(a_n -l) - l (b_n - m)|}{ \frac{1}{2} m^2 } \end{align*}
Since the RHS defines a null sequence, it follows by the Squeeze Theorem for null sequences that $\big( \frac{a_n}{b_n} - \frac{l}{m} \big)$ is null, as required.
In general I follow how the proof is structured, but I get lost with the last inequality.
Using the lemma, we know there must exist some $N$ such that the denominator $b_n m$ will be eventually positive. So we don't have to worry about division by $0$ for all $n > N$.
What I don't get: But why do they introduce absolute value signs in the last inequality? Couldn't we just continue with $\frac{a_n}{b_n} - \frac{l}{m} = \frac{m(a_n -l) - l (b_n - m)}{b_n m}$, as for all $n > N$, we avoided division by $0$.
First of all, I really wouldn't call this result a Theorem, but that's not really the point of your question. The reason why the absolute value of the aforementioned quantity appears in the proof of your proposition is precisely the way convergence is set up in $\mathbb{R}.$ To elaborate on that, recall that convergence is a topological concept and, given $(X, \tau)$ a topological space, we say that a sequence $(x_n)_{n \in \mathbb{N}} \subset X$ is convergent to $\lambda \in X$ if for every open set $U \in \tau$ such that $\lambda \in U,$ there is $n_U \in \mathbb{N}$ so that $x_n \in U$ for all $n \geq n_U.$
In the particular case of a metric space $(X, d)$, we have a very natural way to topologise $X,$ namely we consider the topology generated by the family of balls $B(x, r) := \{ y \in x: d(x, y) < r \},$ where $x \in X$ and $r > 0.$ On $\mathbb{R}$ we have exactly the topology generated by the standard metric, so putting all of these facts together explains why we need to consider the absolute value of the quantity you mentioned, not just the quantity itself.
I hope I understood your question correctly and that you will find this helpful.