Abbott's Proof $(1/b_n) \to 1/b$

For a certain $N_1\in\Bbb N$,\begin{align}n\geqslant N_1&\implies|b_n-b|<\frac{|b|}2\\&\implies|b_n|>\frac{|b|}2\\&\implies\left|\frac1{b_n}-\frac1b\right|<2\frac{|b_n-b|}{|b|^2}\end{align}and so, if you want to have $\left|\frac1{b_n}-\frac1b\right|<\varepsilon$, it will be enough to have$$|b_n-b|<\frac{\varepsilon|b|^2}2.$$

Abbott wants $|b - b_n|$ to satisfy a bound that, when multiplied by something (namely $\frac{1}{|b| |b_n|}$) that is known to have absolute value less than $\frac{1}{|b| |b|/2}$, will be guaranteed to be less than epsilon. In other words, he wants to conclude from $|b - b_n| < M$ that something known to be less than $M \frac{1}{|b|^2/2}$ is also known to be less than $\epsilon$.

This will certainly happen if $M \frac{1}{|b|^2/2} = \epsilon$, a condition that can be solved for $M$, getting the formula.

This particular arrangement of choices in a limit proof is not the only way to prove the limit. There are generally a number of somewhat arbitrary choices made in proving theorems of this type.