Need of $1$ in this proof?

Question

What is the need of a $1$ inside the $\mathbb{max}$ in this proof?

Answer 1

I don't think it's actually needed. The only thing I could of is to ensure that no division by zero takes place, but the condition $-m'<a(n)<M'$ means that not both $m'$ and $M'$ can be zero, so it's actually unnecessary.

Answer 2

The $1$ is there so that we have $$\frac{\varepsilon}{2M}\leq\frac\varepsilon2.$$ This way, we have each sequence being with $\frac\varepsilon2$ of its limit.

Answer 3

It ensures that $M$ is not zero, which is problematic as you divide by $M$.

Edit: However, by strictness in $m' \lt a(n) \lt M'$ we have $m' \neq M'$ and else $M$ could not be zero even without a $1$ there. I guess the $1$ makes it clearer that we do not divide by zero?

Answer 4

The $1$ is redundant. We know that $M$ cannot equal zero as $-m'<a(n)<M'$ are strict inequalities.

Writing $M=\{|m'|, |m''|, |M'|, |M''|, |L_1|, |L_2|\}$ would be sufficient for the proof.