I am confused about Abbott's proof that $(1/b_n)\to (1/b)$. Specifically, I cannot figure out where in the world the term $\frac{\epsilon|b|^2}{2}$ comes from.
(iv) This final statement will follow from (iii), if we can prove that $$(b_n) \to b_n \qquad \text{implies}\qquad \left(\frac{1}{b_n}\right) \to \left(\frac{1}{b}\right)$$ whenever $b \neq 0$. We begin by observing that $$\left|\frac{1}{b_n} – \frac{1}{b}\right| = \frac{|b – b_n|}{|b||b_n|}.$$
Because $(b_n) \to b$, we can make the preceding numerator as small as we like by choosing $n$ large. The problem comes in that we need a worst-case estimate on the size of $1/(|b||b_n|)$. Because the $b_n$ terms are in the denominator, we are no longer interested in an upper bound on $|bn|$ but rather in an inequality of the form $|b_n| \geq \delta > 0$. This will then lead to a bound on the size of $1/(|b||b_n|)$.The trick is to look far enough out into the sequence $(b_n)$ so that the terms are closer to $b$ than they are to $0$. Consider the particular value $\epsilon_0 = |b|/2$. Because $(b_n) \to b$, there exists an $N_1$ such that $|b_n − b| < |b|/2$ for all $n \geq N_1$. This implies $|b_n| > |b|/2$.
Next, choose $N_2$ so that $n \geq N_2$ implies $$|b_n – b| < \frac{\epsilon |b|^2}{2}.$$ Finally, if we let $N = \max\{N_1, N_2\}$, then $n \geq N$ implies $$\left|\frac{1}{b_n} – \frac{1}{b}\right| = |b – b_n|\frac{1}{|b||b_n|} < \frac{\epsilon |b|^2}{2}\frac{1}{|b|\frac{|b|}{2}} = \epsilon. \qquad \square$$
Best Answer
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$$|b_n-b|<\frac{\varepsilon|b|^2}2.$$