Did I do the proof of the product of limits is the limit of the products correctly If $\displaystyle\lim_{x\to a}f(x) = L$ and $\displaystyle\lim_{x\to a}g(x) = M$, then $\displaystyle\lim_{x \to a}(f(x)g(x)) = L \cdot M$.
Proof:
$0 < |x-a| < \delta_{1} \Rightarrow |f(x) - L| < 1$
$0 < |x-a| < \delta_{2} \Rightarrow |f(x) - L| < \displaystyle\frac{\epsilon}{2(|M|+1)}$
$0 < |x-a| < \delta_{3} \Rightarrow |g(x) - M| < \displaystyle\frac{\epsilon}{2(|L|+1)}$
For the first line of reasoning, I wanted the condition that $|f(x) - L|$ must  be less then one. It will be useful later. To simplify all of this information, just let $\delta = \min(\delta_{1},\delta_{2},\delta_{3})$ and $|f(x) - L| < \min(1,\displaystyle\frac{\epsilon}{2(|M|+1)})$.
$|f(x)g(x) - LM| = |f(x)(g(x)-M)+M(f(x)-L)|$
$\leq |f(x)(g(x)-M)| + |M(f(x)-L)|$
$= |f(x)||g(x)-M| + |M||f(x)-L|$
Note that $|x| - |y| \leq |x-y|$ so $|f(x)| - |L| \leq |f(x) - L| < 1$
Here we used the fact that we wanted $|f(x) - L| < 1$.
So $|f(x)| - |L| < 1 \Rightarrow |f(x)| < |L| + 1$ 
Also, $|M| < |M|+1$ (Here I am unsure of the reasoning other then its just an obvious observation. Is that really the reasoning here or do we use something from the definitions formulated earlier)?
Then,
$|f(x)||g(x)-M| + |M||f(x)-L| < (|L| + 1)|g(x) - M| + (|M|+1)|f(x)-L|$
$< (|L| + 1) \cdot \displaystyle\frac{\epsilon}{2(|L|+1)} + (|M|+1) \cdot \displaystyle\frac{\epsilon}{2(|M|+1)}$
$= \displaystyle\frac{\epsilon}{2} + \frac{\epsilon}{2}$
$= \epsilon$.
 A: It looks good, although a little confused, for my taste.
You could reorder it by first noting that
\begin{align}
|f(x)g(x)-LM|
&=|f(x)g(x)-f(x)M+f(x)M-LM| \\[4px]
&\le|f(x)|\,|g(x)-M|+|M|\,|f(x)-L| \\[4px]
&\le|f(x)|\,|g(x)-M|+|M|\,|f(x)-L|+|f(x)-L| \\[4px]
&=|f(x)|\,|g(x)-M|+(|M|+1)|f(x)-L|
\end{align}
If $\lim_{x\to a}f(x)=L$, then there is $\delta_1>0$ such that, for $0<|x-a|<\delta_1$,
$$
|f(x)-L|<1
$$
which implies $L-1<f(x)<L+1$. If $K=\max(|L-1|,|L+1|)$ we can say that, for $0<|x-a|<\delta_1$, $|f(x)|<K$.
Fix $\varepsilon>0$.
We can find $\delta_2>0$ such that, for $0<|x-a|<\delta_2$,
$|g(x)-M|<\varepsilon/(2K)$.
We can find $\delta_3>0$ such that, for $0<|x-a|<\delta_3$,
$|f(x)-L|<\varepsilon/(2(|M|+1))$
If $\delta=\min(\delta_1,\delta_2,\delta_3)$, then, for $0<|x-a|<\delta$,
\begin{align}
|f(x)g(x)-LM|
&\le|f(x)|\,|g(x)-M|+(|M|+1)|f(x)-L| \\[4px]
&<K\frac{\varepsilon}{2K}+(|M|+1)\frac{\varepsilon}{2(|M|+1)} \\[4px]
&=\frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\[4px]
&=\varepsilon
\end{align}
