# Proof

We have to prove the proposition only for the special case of a constant complex sequence $$(a_n)_{n\in\mathbb N}$$ with $$a_n=1$$ for all $$n\in\mathbb N$$, because the general case follows then immediately from the corresponding proposition about the product of convergent complex sequences and from the identity $\frac {a_n}{b_n}=a_n\cdot\frac 1{b_n}.$

By hypothesis, the limit of the convergent complex sequences $$(b_n)_{n\in\mathbb N}$$ does not equal the complex zero ($$b\neq 0$$). Thus, we can find some index $$N(b)$$ such that

$|b_n - b| < \frac{|b|}{2}\quad\quad\forall n \ge N(b).$

Therefore, it follows1 that $$b_n\neq 0$$ for all $$n \ge N(b)$$, because $$|b_n|\ge |b|/2$$ for all $$n \ge N(b)$$, or $$1/|b_n|\le 2/|b|$$. We will need this estimation a little bit later.

Now, it follows from the convergence of $$(b_n)_{n\in\mathbb N}$$ that for a given $$\epsilon > 0$$, there exists an index $$N(\epsilon)\in\mathbb N$$ such that

$|b_n - b |< \frac{\epsilon|b|^2}{2}\quad\quad\forall n\ge N(\epsilon).$

It follows for all $$n\ge M(\epsilon,b):=\max(N(b),N(\epsilon))$$ that

$\left|\frac 1{b_n} - \frac 1b \right| = \left|\frac{ b - b_n}{b_n b}\right|=\frac{1}{|b_n||b|}\cdot |b_n - b| < \frac{2}{|b|^2}\cdot\frac{\epsilon|b|^2}{2}=\epsilon\quad\quad\forall n\ge M(\epsilon,b).$

Thus, we have proven that :

$\lim_{n\rightarrow\infty,~n\ge M(\epsilon,b)} \frac{a_n}{b_n}=\frac{\lim_{n\rightarrow\infty,~n\ge M(\epsilon,b)} a_n}{\lim_{n\rightarrow\infty,~n\ge M(\epsilon,b)} b_n}.$

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983

#### Footnotes

1. This is just because each sequence member $$b_n$$ is "closer" to $$b$$ then it is to $$0$$, interpreting $$|b|/2$$ as half the distance from $$b$$ to $$0$$.