Confusion on proof of limit laws epsilon delta at https://www.youtube.com/watch?v=9tYUmwvLyIA from 35:33 to 39:33, Herb Gross says:
\begin{align}
f(x) &= L+ [f(x)-L] \\
g(x) &= M+ [g(x)-M]
\end{align}
Multiplying these 2, we get:
\begin{align}
f(x)g(x) &= LM + L[g(x) -M] + M[f(x) -L] +[f(x) -L][g(x)-M] \\
f(x)g(x) - LM &= L[g(x) -M] + M[f(x) -L] +[f(x) -L][g(x)-M]
\end{align}
From here we can say:
\begin{align*}
|f(x)g(x) - LM| &= |L[g(x) -M] + M[f(x) -L] +[f(x) -L][g(x)-M]| \\
&\leq |L[g(x) -M]| + |M[f(x) -L]| +|[f(x) -L][g(x)-M]|
\end{align*}
From here we can impose:
\begin{align}
|L[g(x) -M]| &< \frac{\epsilon}{3} \\ 
|M[f(x) -L]| &<\frac{\epsilon}{3} \\
|f(x) -L] &< \sqrt{\frac{\epsilon}{3}} \\
|g(x)-M| &< \sqrt{\frac{\epsilon}{3}}
\end{align}
My problem is:
\begin{align}
|L[g(x) -M]| &< \frac{\epsilon}{3} \implies|g(x) -M| &< \frac{\epsilon}{3|L|} \\ 
|M[f(x) -L]| &<\frac{\epsilon}{3} \implies |f(x) -L| <\frac{\epsilon}{3|M|}\\
|f(x) -L] &< \sqrt{\frac{\epsilon}{3}} \\
|g(x)-M| &< \sqrt{\frac{\epsilon}{3}}
\end{align}
So does that mean:
$$|f(x) -L] < \min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|M|}  \}$$
$$|g(x) -M] < \min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|L|}  \}$$
Is this how it would look like in a proof:
Given $\epsilon > 0$
Let $\epsilon_1 =\min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|M|}  \} $
then $\exists \delta_1 >0$ such that:
$$0<|x-a|<\delta_1 \implies |f(x) - L| < \epsilon_1$$
Let $\epsilon_2 =\min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|L|}  \} $
then $\exists \delta_1 >0$ such that:
$$0<|x-a|<\delta_2 \implies |g(x) - M| < \epsilon_2$$
then let $\delta \leq \min\{\delta_1,\delta_2\}$
\begin{align*}
0<|x-a|<\delta &\implies |L||g(x) -M| + |M||f(x) -L| +|f(x) -L||g(x)-M| \\
&< |L|\times\min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|L|}  \} + |M|\times\min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|M|}  \} + \min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|L|}  \} \times \min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|M|}  \}\\
&< \epsilon
\end{align*}
This last part is not mentioned in the video is it all correct?
 A: That is needlessly complicated.
(1). Prove that if $A$ is a constant and $\lim_{x\to y}u(x)=V$ then $\lim_{x\to y}A+u(x)=A+V.$
(2). Prove that if $A$ is a constant and  $\lim_{x\to y}u(x)=V$ then $\lim_{x\to y}Au(x)=AV.$
(3). Prove that if $\lim_{x\to y}u_1(x)=V_1$ and $\lim_{x\to y}u_2(x) =V_2$ then $\lim_{x\to y}u_1(x)+u_2(x)=V_1+V_2.$
(4). Prove that if $\lim_{x\to y}u_1(x)=0=\lim_{x\to y}u_2(x)$ then $\lim_{x\to y}u_1(x)u_2(x)=0$.... See (**) below.
Suppose $\lim_{x\to y}f(x)=L$ and $\lim_{x\to y}g(x)=M.$ Let $f(x)=L+h(x)$ and $g(x)=M+i(x).$
By (1) with $A=-L$ and $u=f$ and $V=L$ we have $\lim_{x\to y}h(x)=0.$ Similarly we have $\lim_{x\to y}i(x)=0. $
Now $f(x)g(x)-LM=h(x)M+i(x)L+h(x)i(x).$ By (2) we have $\lim_{x\to y}h(x)M=0\cdot M=0.$ And similarly  $ \lim_{x\to y}i(x)L=0.$ So by (3) with $u_1(x)=h(x)M$ and $u_2(x)=i(x)L$ and $V_1=V_2=0,$ we have $\lim_{x\to y}h(x)M+i(x)L=0.$
By (4) we have $\lim_{x\to y}h(x)i(x)=0.$ So by (3) with $u_1(x)=h(x)M+i(x)L$ and $u_2(x)=h(x)i(x)$ we have $\lim_{x\to y}f(x)g(x)-LM=0.$
Finally by (1) with $u(x)=f(x)g(x)-LM$ and $A=LM$ we have $\lim_{x\to y}f(x)g(x)=LM.$
(**). To prove (4): Given $e>0$ let $e'=\min (1,e).$ Note that $0<(e')^2\le e'\le e.$ For $j\in \{0,1\}$ take $d_j>0$ such that $0<|x-y|<d_j\implies |u_j(x)|<e'.$ Let $d=\min(d_1,d_2).$ Then $d>0$ and $0<|x-y|<d\implies |u_1(x)u_2(x)|<(e')^2\le e.$
