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 $\delta$?

  • $\begingroup$ 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 $\epsilon$. $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 at 5:20
  • $\begingroup$ 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 $\delta$ $\endgroup$ – user917099 Sep 10 at 5:36
  • $\begingroup$ 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. $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 at 5:38
  • $\begingroup$ 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. $\endgroup$ – user917099 Sep 10 at 6:40
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    $\begingroup$ Excellent, I am very happy with the fact that you got this cleared. For your interest in the question and the interaction below, +1. $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 at 7:15

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 $|x-1|<\delta$.

  • $\begingroup$ This is exactly what I think it should be!! $\endgroup$ – user917099 Sep 10 at 5:39
  • $\begingroup$ But even the teacher in my another real analysis class taught the same method of starting from a random small value for $\delta$. $\endgroup$ – user917099 Sep 10 at 5:41
  • $\begingroup$ Fine! So what is your choice for $\delta$ here? $\endgroup$ – Robert Z Sep 10 at 5:42
  • $\begingroup$ Sometimes one can guess that the restriction with $\delta$ less than some constant (in your case $1$) will make inequalities a bit easier... $\endgroup$ – Robert Z Sep 10 at 5:44
  • $\begingroup$ Actually, what I think should be doesn't have the "to be determined" for $\delta$..... $\endgroup$ – user917099 Sep 10 at 5:45

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 $\delta <1$.

  • $\begingroup$ So is the relationship between $\epsilon$ and $\delta$ many to one? I thought it is 1 to 1. $\endgroup$ – user917099 Sep 10 at 5:53
  • $\begingroup$ $\delta$ 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 $\endgroup$ – Kabo Murphy Sep 10 at 5:55

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!

  • $\begingroup$ 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. $\endgroup$ – user917099 Sep 10 at 16:52
  • $\begingroup$ 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. $\endgroup$ – user21820 Sep 11 at 9:54
  • $\begingroup$ 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). $\endgroup$ – user21820 Sep 11 at 10:00

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