Proving the multiplicative property of limits. 
If $\lim_{x\rightarrow a}f(x) =L$ and $\lim_{x\rightarrow a}g(x) =M$ then prove $\lim_{x\rightarrow a}f(x)g(x) =LM$

Attempt
(Should there be any statement that I should write at start of this proof?)
 $$|f(x)g(x) -LM|=|f(x)g(x)+Lg(x)-Lg(x) -LM|=|g(x)[f(x)-L]+L[g(x)-M]<|g(x)|f(x)-L|+|L||g(x)-M|$$
Let $\epsilon>0$ then there exists $\delta_1, \delta_2$ such that 
$$|f(x)-L|<\frac{\epsilon}{2g(x)}~~ whenever~~ 0<|x-a|<\delta_1$$
$$|g(x)-M|<\frac{\epsilon}{2L}~~ whenever~~ 0<|x-a|<\delta_2$$
Choose $0<|x-a|<\delta$ where $\delta = \min(\delta_1,\delta_2)$ then we have
 $$|f(x)g(x) -LM|<\epsilon$$ 
Is this proof correct?
The reason I am concerned is in the second step where I defined the inequalities for the individual expression, I wrote it in terms of $g(x)$  and $L$. Is this step mathematically correct? 
Edit:
Is this continuation correct?
Let $\epsilon>0$ then there exists $\delta_1, \delta_2$ such that 
$$|f(x)-L|<\frac{\epsilon}{2(1+|M|)}~~ whenever~~ 0<|x-a|<\delta_1$$
$$|g(x)-M|<\frac{\epsilon}{2(1+|L|)}~~ whenever~~ 0<|x-a|<\delta_2$$
Choose $0<|x-a|<\delta$ where $\delta = \min(\delta_1,\delta_2)$ then we have
 $$|f(x)g(x) -LM|<|g(x)|\frac{\epsilon}{2(1+|M|)}+|L|\frac{\epsilon}{2(1+|L|)}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ 

Your approach is fine but a few details are not correct.
You should replace 
$$|f(x)-L|<\frac{\epsilon}{2g(x)}$$
with something like
$$|f(x)-L|<\frac{\epsilon}{2M'}.$$
where $M'$ is a positive constant to be decided.
Recall that $\lim_{x\rightarrow a}g(x) =M$ implies that $g$ is bounded in a neighbourhood of $a$.
Moreover to avoid problems with the case when $L=0$, replace
$$|g(x)-M|<\frac{\epsilon}{2L}$$
for example with
$$|g(x)-M|<\frac{\epsilon}{2(1+|L|)}.$$
Are you able to finish to proof correctly now?
P.S. In your edited work add th following line $|g(x)|<{(1+|M|)}$ whenever $0<|x-a|<\delta_3$ and define $\delta = \min(\delta_1,\delta_2,\delta_3)$.
