In my textbook (S. Abbott. Understanding Analysis 1 ed. pp 47 Theorem 2.3.3.iv), the author proves $b_n \to b$ implies $\frac{1}{b_n} \to \frac{1}{b}$ the following way:


therefore, since we can make the numerator as small as we like by definition, we just need to find a "worst-case estimate" for $\frac{1}{|b||b_n|}$ by considering $|b_n| \geq \lambda > 0$. He then argues that we need to look far into the sequence so that the terms are closer to $b$ than to $0$. So, considering $\epsilon_0$=$|b|/2$, we can use $|b_n-b|<|b|/2$ for all $n \geq N_1$. Thus, $|b_n| \geq |b|/2$. Then we choose $N_2$ s.t. $n \geq N_2$ implies $|b_n-b|<\frac{\epsilon |b|^{2}}{2}$. Finally, if $N$ = max {$N_1, N_2$}, then $n \geq N$ implies $$\frac{|b-b_n|}{|b||b_n|}<\frac{\epsilon |b|^{2}}{2} \frac{1}{|b|\frac{|b|}{2}}=\epsilon.$$ What confuses me is why we need to pick elements closer to $b$ than to 0. What makes picking $\epsilon_0$=$|b|/3$ and $|b_n-b|<\frac{\epsilon |b|^{2}}{3}$ wrong for the proof? Thanks.

  • $\begingroup$ Your choice $\epsilon_0=|b|/3$ works even better and gets you the upper bound $3\epsilon/4$ (instead of $\epsilon$) at the end. $\endgroup$ – Did Jul 31 '11 at 13:54
  • $\begingroup$ @DidierPiau I think the OP meant $\epsilon_0 = 2|b|/3$. Note that he is making the numerator $|b_n-b|$ even smaller than before, presumably in order to compensate for the shrinking denominator $|b| |b_n|$. $\endgroup$ – Srivatsan Jul 31 '11 at 13:57
  • $\begingroup$ In this case, say you set $\varepsilon$ as a function/fraction of b, i.e., $\varepsilon:=tb$ (so that your expression is all in terms of variables that can cancel each other out), for some $t\neq 0$. Then your expression becomes (assuming $b \neq 0$):$\frac{|b_n-b|}{|b||b_n|}>=\frac{tb}{b(b-tb)}=\frac{t}{b-tb}=\frac{t}{b(1-t)}$, where t is defined as in $\varepsilon:=tb$. So any non-zero choice of t will do. $\endgroup$ – gary Jul 31 '11 at 15:52
  • $\begingroup$ @gary Don't we want an upper bound on $|\frac{1}{b_n} - \frac{1}{b}|$? How did you end up with a lower bound? $\endgroup$ – Srivatsan Jul 31 '11 at 16:09
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    $\begingroup$ @Geoff: Indeed. And the other way round, one is happy with a proof getting $|f(x)-f(a)|$ less than $18\epsilon$, or something similar... $\endgroup$ – Did Jul 31 '11 at 17:06
  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.

  • $\begingroup$ This makes sense, thank you. You are right, picking a smaller epsilon actually leads me closer to the sequence, I missed that at first. Thanks again! $\endgroup$ – confused Jul 31 '11 at 18:42

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.

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    $\begingroup$ I upvoted for the "math written backwards" part :-) . Your explanation demystifies picking $\epsilon$s and $\delta$s very nicely, thanks! $\endgroup$ – Srivatsan Aug 1 '11 at 1:18
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    $\begingroup$ We share pet-peeves here! In fact, I remember vividly that I was absolutely amazed that my teacher could see those weird choices of $\delta$ from the beginning of a proof. I was only convinced that he did in fact "cheat" when he messed up one of his brilliantly clairvoyant choices and did the whole thing backwards in front of my eyes... $\endgroup$ – t.b. Aug 1 '11 at 6:10
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    $\begingroup$ @D Lim: Ha, serves her right! In this particular case one can avoid the problem with a slick trick by breaking up the problem in three steps: Step 1: Prove that $s_n \to s$ converges then $s_{n}^2 \to s^2$. Step 2: Prove that $s_n \to s$ and $t_n \to t$ implies $s_n + t_n \to s + t$. Step 3: Write $st = \frac{1}{2} [(s + t)^2 - (s^2 + t^2)]$. $\endgroup$ – t.b. Aug 21 '11 at 8:28
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    $\begingroup$ @DJC Upvote for "magical way" I always hated this as it always gave the impression these people had received some divine knowledge in the beginning in picking these epsilons and deltas... $\endgroup$ – user38268 Aug 21 '11 at 8:28
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    $\begingroup$ @Theo Buehler I was just going through baby Rudin and he too comes up magically with his choices of epsilons and deltas.....hmmm now I remember vividly my supervisor telling me how he was "lying" to me when he came up with these epsilons out of the hat... $\endgroup$ – user38268 Aug 21 '11 at 11:34

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