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I'm having difficulty understanding the common textbook way of starting from a random small value(often 1) for $\delta$. For example, to prove $f(x)=x^2+x-3 \rightarrow -1 \;as\; x \rightarrow 1$, textbooks often start by setting $\delta <= 1$, then plug in this inequality into the function to get $4\delta < \epsilon$. Isn't doing so only proving for only the $\epsilon$ whose corresponding $\delta$ is less than or equal to 1 intead of for every $\epsilon$ and every corresponding $\delta$?

user917099
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  • Textbooks should start by saying "Let $\epsilon > 0$ be arbitrary" for any $\epsilon-\delta$ proof. You should tell us how your textbook starts a proof, for I have not seen a textbook start with $\delta \leq 1$. Remember, the $\delta$ adjusts itself for the $\epsilon$, so there's no way the $\delta$ comes before the $\epsilon$. – Sarvesh Ravichandran Iyer Sep 10 '19 at 05:20
  • Yes, the textbook I got this example from does start by "Let $\epsilon$ > 0". But it then starts "If $0 < \delta \leq 1,; then \mid x - 1 \mid < \delta;implies ;0<x<2$". And this still seems to me only proves the $\epsilon$ whose $\delta$ is in a certain range, so i don't understand how this way can prove for every $\epsilon$ and its $\delta$ – user917099 Sep 10 '19 at 05:36
  • Is it possible for you to put the whole textbook argument word to word above, if the answers below don't satisfy you? That would be the best couse of action. – Sarvesh Ravichandran Iyer Sep 10 '19 at 05:38
  • Thanks for answering! I think it was my way of looking at this proof made it seems to be an issue from the textbook to you. Please check the below answers and my discussions there then you will see what I mean by the textbook way. – user917099 Sep 10 '19 at 06:40
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    Excellent, I am very happy with the fact that you got this cleared. For your interest in the question and the interaction below, +1. – Sarvesh Ravichandran Iyer Sep 10 '19 at 07:15

3 Answers3

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Your are not prroving anything for every $\delta$. You are required to prove that for every $\epsilon >0$ there exists $\delta >0$ such that something happens. $\delta$ is completely at your choice and it is very often convenient to take $\delta <1$.

Kavi Rama Murthy
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  • So is the relationship between $\epsilon$ and $\delta$ many to one? I thought it is 1 to 1. – user917099 Sep 10 '19 at 05:53
  • $\delta$ is not even uniquely determined by $\epsilon$. If one $\delta$ works then an smaller positive number also works). So the question of $\delta$ being a 1-1 function of $\epsilon$ does not arise. @user917099 – Kavi Rama Murthy Sep 10 '19 at 05:55
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I think the real problem here, faced by many students, is the fact that the proof of the result is written in a specific order BUT we find the information for the proof (eg in this case what we need δ to be) in a different order.

To find the relevant δ we use the fact that our final line needs to be less than ε and work backwards. I teach my students to use the standard proof format but to leave blanks that we return to and complete once we know the information!

In more complicated situations such as this one, look for the |x-a| term and keep it for the |x-a|< δ and then look for a constant value that will give you a bound (any bound) on the other terms. You then choose the minimum of all the values used.

Remember our result has to work for any ε, the "hard" εs are the small ones so we often think of ε as being small, but our result has to work when ε is large as well!

Sally Barton
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  • I'm getting clearer on this definition after the discussions and answers here. I think my problem was that I forgot that essentially the thing that needs to be proved is the smallness/narrowness in the $\delta$-bound and the $\epsilon$-bound. – user917099 Sep 10 '19 at 16:52
  • I agree that what you describe is a problem, but it is merely a symptom and not the underlying problem, which is a lack of grasp of basic logic. It is not hard to teach first-order logic properly to students, and yet it is necessary for them to be able to perform logical reasoning correctly if we want them to be able to handle anything more than simple mathematics. In particular, see here for an explanation of the ε-δ definition, and here for a comparison of various notions of continuity and convergence. – user21820 Sep 11 '19 at 09:54
  • That said, I actually think that after basic logic we should teach asymptotic analysis as the primary tool instead of limit theorems, and definitely not ε-δ definitions. For some examples see here and here. @user917099: You may want to learn asymptotic analysis because it is superior to any other method in almost all cases (and is what computer algebra systems actually use to compute limits). – user21820 Sep 11 '19 at 10:00
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Your proof should start with: "let $\epsilon>0$". Then let $x$ such that $|x-1|<\delta$ for some $\delta>0$ (to be determined as a function of $\epsilon$) and consider the difference $$|f(x)-f(1)|=|x^2+x-2|=|x+2|\cdot |x-1|\leq |x+2|\delta.$$ Now, try to find $\delta>0$ such that $$|x+2|\delta<\epsilon$$ for all $x$ such that $|x-1|<\delta$.

Robert Z
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  • This is exactly what I think it should be!! – user917099 Sep 10 '19 at 05:39
  • But even the teacher in my another real analysis class taught the same method of starting from a random small value for $\delta$. – user917099 Sep 10 '19 at 05:41
  • Fine! So what is your choice for $\delta$ here? – Robert Z Sep 10 '19 at 05:42
  • Sometimes one can guess that the restriction with $\delta$ less than some constant (in your case $1$) will make inequalities a bit easier... – Robert Z Sep 10 '19 at 05:44
  • Actually, what I think should be doesn't have the "to be determined" for $\delta$..... – user917099 Sep 10 '19 at 05:45
  • I mean to be determined as a function of $\epsilon$. Otherwise you have to explain why such $\delta$ exists. – Robert Z Sep 10 '19 at 05:48
  • So I can't end the proof by $\delta = \frac{\epsilon}{\mid x+2 \mid}$....(thinking hard) – user917099 Sep 10 '19 at 05:51
  • No, because such $\delta$ depends on $x$ (which could be also $x=-2$). But if you assume some bound for $\delta$... – Robert Z Sep 10 '19 at 05:52
  • If I assume some bound for $\delta$, will there still be enough $\delta$ mapped to every $\epsilon$? Sorry I might be looking at this definition from some weird pov but I'm trying hard to understand it. – user917099 Sep 10 '19 at 05:56
  • You need a "good" bound. Guess one and verify that it works. For example if $\delta\leq 1$ then $0<x<1+\delta\leq 2$ and therefore $|x+2|<2+2=4$. So we can take any $0<\delta\leq \min(1,\epsilon/4)$. – Robert Z Sep 10 '19 at 05:57
  • But I thought the definition means that for every $\epsilon$/bound-on-y, there is 1 $\delta$/bound-on-x. How can finding a single good bound-on-x satisfy for every such bound-on-x that determines a bound-on-y? – user917099 Sep 10 '19 at 06:21
  • In fact for ANY $\epsilon>0$ the interval $(0,\min(1,\epsilon/4)$ is non-empty and therefore there exists a $\delta>0$ (not unique) which gives you the required inequality. – Robert Z Sep 10 '19 at 06:25
  • Your "(not unique)" hit right on my confusion! Is it that for this example, the interval $(0,\min(1,\epsilon/4))$ the only interval? – user917099 Sep 10 '19 at 06:35
  • No, it is not the only interval because it is only a "good" bound. – user917099 Sep 10 '19 at 06:38
  • @user917099 Exactly. – Robert Z Sep 10 '19 at 09:22