How did they get $|\epsilon/(2M)|$ when proving this limit property? I am currently learning calculus by going through this calculus book. The following Theorem and part of its proof are copied from page 31. (I shortened it a bit and left out the rest, since I don't think its relevant to my question)
THEOREM 2.6:
Suppose $\lim_{x\to a}f(x) = L$ and $\lim_{x\to a} g(x) = M$. Then
$\lim_{x\to a}f(x)g(x) = LM$.
Proof:
Given any $\epsilon$ we need to find a $\delta$ so that $0 < |x − a| < \delta$ implies $|f(x)g(x) − LM| < \epsilon$. What do we
have to work with? We know that we can make f(x) close to L and g(x) close to M, and
we have to somehow connect these facts to make f(x)g(x) close to LM.
We use, as is so often the case, a little algebraic trick:
$$|f(x)g(x) − LM| = |f(x)g(x) − f(x)M + f(x)M − LM|\\
= |f(x)(g(x) − M) + (f(x) − L)M|\\
≤ |f(x)(g(x) − M)| + |(f(x) − L)M|\\
= |f(x)||g(x) − M| + |f(x) − L||M| $$
[...] Since $\lim_{x\to a} = L$, there is a value $\delta_1$ so that $0 < |x − a| < \delta_1$ implies $|f(x) − L| <|\epsilon/(2M)|$, This means that $0 < |x − a| < \delta_1$ implies $|f(x) − L||M| < \epsilon/2$. You can see
where this is going: if we can make $|f(x)||g(x) − M| < \epsilon/2$ also, then we’ll be done.
My question:
Can someone explain to me how they got to $\epsilon/(2M)$ in this statement: " Since $\lim_{x\to a} = L$, there is a value $\delta_1$ so that $0 < |x − a| < \delta_1$ implies $|f(x) − L| <|\epsilon/(2M)|$. "

Since $\lim_{x\to a}f(x)=L$, you know that, for every $\varepsilon>0$, there is some $\delta>0$ such that$$|x-a|<\delta\implies\bigl|f(x)-f(a)\bigr|<\varepsilon.$$But $\left|\frac\varepsilon{2M}\right|>0$, and therefore there is some $\delta_1>$ such that$$|x-a|<\delta_1\implies\bigl|f(x)-f(a)\bigr|<\left|\frac\varepsilon{2M}\right|.$$
