Proving that $(b_n) \to b$ implies $\left(\frac{1}{b_n}\right) \to \frac{1}{b}$

Solution 1:

This question has already been answered, but I believe that this scenario is a case of math being written backwards. By this I mean that most $\epsilon$-$\delta$ proofs are written in a "magical" way in which constants are chosen such that one gets $\epsilon$ in the end. This process is analogous to writing a first draft of an essay and then later making it shiny. Let's try and prove this and then see what happens.

(It is also of note that if $f(x)$ is continuous on some interval containing $b$, then $f(b_n) \to f(b)$. It was for this reason we avoided $b \neq 0$ in your proof above. Now, onto the proof at hand).

We aim to show that if $b_n \to b$, then $\frac{1}{b_n} \to \frac{1}{b}$, whenever $b \neq 0$. Since $b_n \to b$, then for all $\epsilon_0 > 0$, there exists a positive integer $N_0$ such that

$$ n \geq N_0 \quad \Longrightarrow \quad |b_n - b| < \epsilon_0. $$

Furthermore, since $b \neq 0$, for any $\lambda \in (0, |b|)$, there exists an integer $N_1$ such that $$ n \geq N_1 \quad \Longrightarrow \quad |b_n| > \lambda. $$
Let $\epsilon > 0$. As you point out, for $n \geq \max\{ N_0, N_1\}$, we have $$ \left| \frac{1}{b_n} - \frac{1}{b}\right| = \frac{|b-b_n|}{|b||b_n|} < \frac{\epsilon_0}{|b| |b_n|} < \frac{\epsilon_0}{|b| \lambda}, $$

so it would be sufficient to choose $\epsilon_0$ so that $\frac{\epsilon_0}{|b|\lambda} = \epsilon$.

There has been nothing magical about this proof as of yet, but now let's cover our work, and run the argument again:

Let $\epsilon > 0$ and $\lambda \in (0, |b|)$. There exists an integer $N$ such that for $n \geq N$, we have $$ |b_n| > \lambda $$ and $$ |b_n - b| < \epsilon |b| \lambda. $$

Therefore, for $n \geq N$, we have $$ \left| \frac{1}{b_n} - \frac{1}{b} \right| = \frac{|b_n - b|}{|b| |b_n|} < \frac{\epsilon |b| \lambda}{|b| \lambda} = \epsilon. $$

Note that any fixed $\lambda \in (0, |b|)$ would work for the proof. The author of your text choose $\lambda = \frac{|b|}{2}$. The mysterious choice of constants chosen by the author is again nothing more than running the argument twice. The first time you run the argument, you need not be careful with the exact constants. The second time through, you can cover your work. While it seems that the previous answers certainly answered your questions, I hope this answer is also helpful.

Solution 2:

1. It does not matter that the elements are closer to $b$ than to $0$. The main point to recognize is that beyond a stage $N_1$, all terms in the sequence are bounded away from $0$. So, you will be in good shape as long as there is a $N_1 \in \mathbb N$ and constant $\lambda > 0$ such that all terms beyond the $N_1$th term are at least $\lambda$ in magnitude. The choice $\lambda = |b|/2$ is just a convenient one, but $|b|/3$ or $2|b|/3$ is equally good.

2. I think you meant to ask about picking $\epsilon_0$ to be $\mathbf{2} |b|/3$. Picking $\epsilon_0$ to be smaller (e.g.: $\epsilon_0 = |b|/3$) only makes the sequence closer to $b$. As @DidierPiau points out, by triangle inequality, we now have $\frac{1}{|b_n|} \leq \frac{3}{2|b|}$ which is smaller than before.

   Finally, the elements will also continue to be closer to $b$ than to $0$, although this point has no real significance in the proof IMO.