Prove that if $M \neq 0$, then $\frac{a_n}{b_n} \Rightarrow \frac{L}{M}$ Question: Let $\{a_n\}$ and $\{b_n\}$ be convergent sequences with $a_n \Rightarrow L$ and $b_n \Rightarrow M$ as $n \Rightarrow \infty$. 
Prove that if $M \neq 0$, then $\frac{a_n}{b_n} \Rightarrow \frac{L}{M}$ 
My solution:
WTS:
Let $B := max(|M - \epsilon|, |M + \epsilon|, |b_n|, \text{for } n > N)$
(1) $\exists L \in R, \forall \epsilon > 0, \exists N_1 > 0$, such that for all $n \in N$, if $n > N_1$, then $|a_n - L| < B\frac{\epsilon}{2}$
(2) $\exists M \in R, \forall \epsilon > 0, \exists N_2 > 0$, such that for all $n \in N$, if $n > N_2$, then $|b_n - M| < \frac{\epsilon |M|B}{2(|L|+1)}$
Let $\epsilon > 0$ be arbitrary
Choose N = $max(N_1, N_2) > 0$
Suppose $n > N$, then
$$|\frac{a_n}{b_n} - \frac{L}{M}| = |\frac{Ma_n  - Lb_n}{b_nM}|$$
$$= |\frac{Ma_n +  LM - LM - Lb_n}{b_nM}| = |\frac{M(a_n - L) + L(M-b_n)}{b_nM}|$$
$$\leq \frac{|M(a_n - L)| + |L(b_n - M)|}{|b_nM|} \text{ By triangle inequality}$$
$$< \frac{|(a_n - L)|}{|b_n|} +  \frac{|(|L|+1)(b_n - M)|}{|b_nM|}$$
$$< B\frac{\epsilon}{2B} + \frac{\epsilon|M|(|L|+1)B}{2(|L|+1)B|M|}$$
$$= \epsilon$$
 A: If $L=0$ and $b_n \to M$ so $\tfrac{|M|}{2} \le \left| b_n \right| \le \tfrac{3|M|}{2}$ for $n$ sufficiently large $(*)$, then:
$$\left| \frac{a_n}{b_n} - \frac{L}{M} \right| = \left| \frac{a_n}{b_n}\right| \le 2\frac{a_n}{|M|}$$
Now since $a_n \to 0$, pick $N$ to get $|a_n| < \tfrac{|M|\varepsilon}{2}$ and such that $(*)$ holds to finish.

Usually splitting into two cases is avoided and you could use ΘΣΦGenSan's suggestion from the comments. Since for $L \ne 0$, $\tfrac{1}{|L|}<\tfrac{1}{|L|+1}$, you can write for any $L$:
$$|b_n - M| < \frac{\epsilon |M|B}{2\left(|L|+1\right)}$$
Note that the right-hand side works as an upper bound (compare it to yours) and the denominator can never become $0$, not even when $L=0$.
A: It is easier this way. First show that if the sequence $a_n\rightarrow a$ and the sequence $b_n\rightarrow b$ then the sequence $a_nb_n\rightarrow ab$. 
Now show that if $c_n\rightarrow c$ and $c\neq 0$ then $\frac{1}{c_n}\rightarrow \frac{1}{c}$. Then use both these facts to conclude your answer.
A: Wow, no offense, but your proof seems to me so cumbersome! We actually can do it in one stroke neatly.
To begin with, I suppose you mean there is some $N_{1}$ such that $b_{n} \neq 0$ for all $n \geq N_{1}$ (to ensure that $a_{n}/b_{n}$  is at least eventually meaningful.).
Note that
$$
|\frac{a_{n}}{b_{n}} - \frac{L}{M}| = | \frac{Ma_{n} - Lb_{n}}{Mb_{n}}| = \frac{|Ma_{n} - Lb_{n}|}{|M||b_{n}|} = \frac{|M(a_{n} - L) - L(b_{n} - M)|}{|M||b_{n}|}
$$
for all $n \geq N_{1}$.
Since $b_{n} \to M$, there is some $N_{2}$ such that $||b_{n}| - |M|| \leq |b_{n}-M| < |M|/2$ for all $n \geq N_{2}$; so $|b_{n}| > |M|/2$ for all $n \geq N_{2}$.
Then
$$
|\frac{a_{n}}{b_{n}} - \frac{L}{M}| < \frac{2|M(a_{n} - L) - L(b_{n} - M)|}{M^{2}} \leq \frac{2|a_{n}-L|}{|M|} + \frac{2|L||b_{n}-M|}{M^{2}}
$$
for all $n \geq \max \{N_{1},N_{2} \}$.
Now again by the convergence assumption, given any $\varepsilon > 0$, there are some $N_{3},N_{4}$ such that $|a_{n}-L| < |M|\varepsilon/4$ for all $n \geq N_{3}$ and $|b_{n}-M| < M^{2}\varepsilon/4|L|$ for all $n \geq N_{4}$ (if $L = 0$ only $N_{3}$ is needed to bound away $|a_{n}-L|$). All in all, if $L = 0$ , taking $N := \max \{N_{1}, N_{2},N_{3} \}$ suffices; if $L \neq 0$, taking $N := \max \{N_{1},N_{2},N_{3},N_{4} \}$ suffices.
