Proof verification for limit quotient rule Please help me check the mathematical rigor of my following proof of the quotient rule for limits of functions, thanks a lot in advance. 
Statement: Suppose $\;lim_{x\to c}f(x)=L\;$ and $\;lim_{x\to c} g(x)=M.\;$ Then $\;\lim_{x\to c}\dfrac{f(x)}{g(x)}=\dfrac{L}{M}$, provided $M\ne 0$.
Proof:
 $$\forall{\varepsilon \gt 0}, \exists{\delta_1 \gt 0, \delta_2 \gt 0}$$ such that $|x-c|\lt \delta_1 \;\text{implies} \;|f(x)-L|\lt \dfrac{\varepsilon}{|M| + 1},\;|x-c|\lt \delta_2 \;\text{implies} \;|g(x)-M|\lt \dfrac{\varepsilon}{|L| + 1}$
Now, 
$$|\frac{f(x)}{g(x)}-\frac{L}{M}|=\frac{|Mf(x)-Lg(x)|}{|M||g(x)|}\le \frac{|M||f(x)|+|L||g(x)|}{|M||g(x)|}\lt \frac{(|M|+1)|f(x)|+(|L|+1)|g(x)|}{|M||g(x)|}$$
By the definitions above, we have 
$$\frac{(|M|+1)|f(x)|+(|L|+1)|g(x)|}{|M||g(x)|}\lt  \frac{\varepsilon +(|M|+1)|L|+\varepsilon+(|L|+1)|M|}{|M||g(x)|}$$
(*)
Let $\varepsilon +(|M|+1)|L|+\varepsilon+(|L|+1)|M|=\epsilon,$ then $$\frac{\varepsilon +(|M|+1)|L|+\varepsilon+(|L|+1)|M|}{|M||g(x)|}\lt \frac{\epsilon}{|M||g(x)-M|}\lt \frac{\epsilon}{\dfrac{|M|\varepsilon}{|L|+1}}$$
Let $$\dfrac{|M|\varepsilon}{|L|+1}= \varepsilon_1$$
Then $$|\frac{f(x)}{g(x)}-\frac{L}{M}|\lt \frac{\epsilon}{\varepsilon_1}$$
Let  $\dfrac{\epsilon}{\varepsilon_1}=\epsilon_1, \delta = min\{\delta_1,\delta_2\}$, we have 
$$\forall{\epsilon_1 \gt 0}, \exists \delta \gt 0 \;\text{such that}\; |x-c|\lt \delta \;\text{implies}\; |\frac{f(x)}{g(x)}-\frac{L}{M}|\lt \epsilon_1$$
Hence finishing the proof. 
EDIT: An attempt to fix the flaws in the proof. Rewriting from (*) on:
$\forall{g(x) \ne 0}, \exists{Q \gt 0}$ such that $Q \lt |g(x)|.$
Let $\varepsilon +(|M|+1)|L|+\varepsilon+(|L|+1)|M|=\epsilon,$ then $$\frac{\varepsilon +(|M|+1)|L|+\varepsilon+(|L|+1)|M|}{|M||g(x)|}\lt \frac{\epsilon}{|M|Q}$$
Fix one such Q and let $\dfrac{\epsilon}{|M|Q}=\varepsilon_1.$ Then $$|\frac{f(x)}{g(x)}-\frac{L}{M}|\lt \varepsilon_1$$
which concludes the proof.
 A: First of all, this question has nothing to do with the quotient rule (it's for derivatives). This is just about proving a basic limit axiom. Secondly, I personally would prefer to use $\epsilon_1,\epsilon_2,$ etc. to avoid confusion and you incorrectly assumed $|g(x)|<|g(x)-M|,$ which is not true for negative $M.$ 
I would prefer the proof below.
Suppose $\lim\limits_{x\to c} f(x) = L$ and $\lim\limits_{x\to c} g(x) = M.$
We want to estimate $\left|\dfrac{f(x)}{g(x)}-\dfrac{L}{M} \right|$ when $x\to c.$ Rewriting, like you did, gives
$\left|\dfrac{f(x)M-Lg(x)}{Mg(x)}\right|.$ Now I would prefer adding and subtracting $ML$ to the numerator to yield
$\left|\dfrac{M(f(x)-L)-L(g(x)-M)}{Mg(x)}\right|.$ Let $\epsilon = \dfrac{|M|}{2}.$ Then since $\lim\limits_{x\to c} g(x)= M,$ $\exists \delta_1$ such that $|x-c|<\delta_1 \Rightarrow |M-g(x)|<\epsilon=\dfrac{|M|}{2}\\
 \Rightarrow|g(x)|=|M-(M-g(x))|\geq |M| -|M-g(x)|>\dfrac{|M|}{2}.$ 
So we have that when $|x-c|<\delta_1,$
$$\left|\dfrac{f(x)M-Lg(x)}{Mg(x)}\right|<\dfrac{2}{|M|}|f(x)-L| +\dfrac{2|L|}{M^2}|g(x)-M|.$$ Now let $\epsilon_2 = \dfrac{|M|\epsilon}{4}$ and $\epsilon_3 = \dfrac{M^2\epsilon}{4|L|}.$ By convergence of $f(x)$ and $g(x),$ $\exists \delta_2 >0$ such that $|x-c|<\delta_2 \Rightarrow |f(x)-L|<\epsilon_2\\
\Rightarrow \dfrac{2}{|M|}|f(x)-L| < \dfrac{\epsilon}{2}.$ 
Similarly, $\exists \delta_3 >0$ such that $|x-c|<\delta_3 \Rightarrow |g(x)-M| < \epsilon_3 = \dfrac{M^2\epsilon}{4|L|}\\
\Rightarrow \dfrac{2|L|}{M^2}|g(x)-M|<\dfrac{\epsilon}{2}.$ 
Therefore, when $|x-c|<\delta = \min\{\delta_1,\delta_2,\delta_3\},$
we have
$$\left|\dfrac{f(x)M-Lg(x)}{Mg(x)}\right|<\dfrac{2}{|M|}|f(x)-L| +\dfrac{2|L|}{M^2}|g(x)-M|\\
<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon.$$
