Proving that the limit of product of functions is equal to product of limits with epsilon delta definition

Question

If $\lim_{x\to a} {f(x)}=L$ and $\lim_{x\to a} {g(x)}=M$
Then I wanted to prove that $\lim_{x\to a} {f(x).g(x)}=L.M$
In a book I saw the proof as follows
They initially proved that |f(x)|< 1+|L| by taking ε=1------(1)
Then the proof goes as follows
Let ε>0 be given.
Then from (1) there exists a δ1>0 such that such that if 0<|x-a|<δ1 then, |f(x)|< 1+|L|
since $\lim_{x\to a} {g(x)}=M$ there exists δ2 >0 such that if 0<|x-a|<δ2 then,
|g(x)-M|< ε/2(1+|L|) ----(3) And using these conditions then they prove the existence of the required limit using the epsilon-delta definition of limit.
But I can't understand how they reached at (3) because (1) is true for f(x) when ε=1, then how can they use this condition for g(x) for any ε and how did they get |g(x)-M|< ε/2(1+|L|)
I have just started this topic, and I am confused in this part though I understood the rest of the proof. It would be helpful if someone could help me with this.

Answer 1

If you know that the limit for $f$ and $g$ converges, then you know that $$0\lt|x-a|\lt\delta_0\implies|f(x)-L|\lt\epsilon_0$$ and $$0\lt|x-a|\lt\delta_1\implies|g(x)-M|\lt\epsilon_1.$$ What you are trying to prove now is that $$0\lt|x-a|\lt\delta\implies|f(x)g(x)-LM|\lt\epsilon.$$ Here is the hint: $$|f(x)g(x)-LM|=|f(x)g(x)-f(x)M+f(x)M-LM|=|f(x)[g(x)-M]+[f(x)-L]M|=|[f(x)-L][g(x)-M]+L[g(x)-M]+[f(x)-L]M|\leq|[f(x)-L][g(x)-M]|+|L[g(x)-M]|+|[f(x)-L]M|=|f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L|.$$ Can you take it from here?