In Calculus, Apostol writes the following proof:
Proof of (iv). Since the quotient $f(x) / g(x)$ is the product of $f(x) / B$ with $B / g(x)$, it suffices to prove that $B / g(x) \rightarrow 1$ as $x \rightarrow p$ and then appeal to (iii [the product rule]). Let $h(x)=g(x) / B$. Then $h(x) \rightarrow 1$ as $x \rightarrow p$, and we wish to prove that $1 / h(x) \rightarrow 1$ as $x \rightarrow p$. Let $\epsilon>0$ be given. We must show that there is a $\delta>0$ such that $$ (3.14) \left|\frac{1}{h(x)}-1\right|<\epsilon \quad \text { whenever } 0<|x-p|<\delta . $$ The difference to be estimated may be written as follows. $$ (3.15) \left|\frac{1}{h(x)}-1\right|=\frac{|h(x)-1|}{|h(x)|} $$ Since $h(x) \rightarrow 1$ as $x \rightarrow p$, we can choose a $\delta>0$ such that both inequalities $$ (3.16) |h(x)-1|<\frac{\epsilon}{2} \quad \text { and } \quad|h(x)-1|<\frac{1}{2} $$ are satisfied whenever $0<|x-p|<\delta$. The second of these inequalities implies $h(x)>\frac{1}{2}$ so $1 /|h(x)|=1 / h(x)<2$ for such $x$. Using this in (3.15) along with the first inequality in (3.16), we obtain (3.14). This completes the proof of (iv).
Please forgive the numbering not being left aligned. I am not sure how to do that with the math exchange formatting.
I understand how he did everything, but I do not understand how this proves the quotient rule. I suppose my point of confusion is the first paragraph. How exactly does it "it suffice[] to prove that $B / g(x) \rightarrow 1$ as $x \rightarrow p$ and then appeal to (iii [the product rule])"?