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I've been reading Spivak's chapter on limits and something that I don't feel I understand entirely is how the epsilon is decided upon. It makes sense to me in the context of $\,|f(x)-L|<\epsilon$ where it appears just on it's own (representing any positive number) but Spivak seems to invoke an arbitrary value of epsilon in some of the given proofs. For example in proving that: $$ \lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}[f(x)]+\lim_{x\to a}[g(x)]$$ he says if: $$\lim_{x\to a}f(x)=l \;\; \text{and} \;\;\lim_{x\to a}g(x)=m$$ then for any $\epsilon>0$ then there are $\delta_1,\delta_2>0$ such that for all $x$: $$ 0<|x-a|<\delta_1 \implies |f(x)-l|<\frac{\epsilon}{2} \\ 0<|x-a|<\delta_2 \implies |g(x)-m|<\frac{\epsilon}{2} $$ He then carries on to show that $|(f+g)(x)-(l+m)|<\epsilon$. I feel like I understand the proof he gives but I just wanted to clarify whether it matters how he defined what the epsilon was. Since for example if he had started with expressions without the $\frac{\epsilon}{2}$ but rather: $$0<|x-a|<\delta_1 \implies |f(x)-l|<\epsilon$$ wouldn't the outcome be $|(f+g)(x)-(l+m)|<2\epsilon$; which I assume still says that it's bounded, and since $\epsilon$ was any positive number it shouldn't matter. Though I'm not certain, so I suspect there's an error in my understanding.

Another example is if you tried proving that:

$$\lim_{x\to 1} 2x-2=0$$ then instead of writing just epsilon, one wrote some positive constant $c$ times epsilon so: $$ |(2x-2)-0|<c\times\epsilon\;\; \rightarrow \;\; |x-1|<\frac{c\times \epsilon}{2}$$ therefore let $\delta =\frac{c\times\epsilon}{2}$ which would still seem to prove it even if we hadn't had the $c$ that is with $\delta=\frac{\epsilon}{2}$ . I know it's not a proper proof but hopefully you see what I'm trying to say (or indeed what I'm doing incorrectly).

Hopefully my question isn't too badly written or that there isn't one asking the same thing, I have found this which is from the same part of the book but I still don't feel sure whether or not the choice of epsilon changes the validity proof.

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    $\begingroup$ The proof must hold for all positive numbers $\varepsilon$. If you can choose an arbitrary $\varepsilon > 0$ and prove the result using it, the result must hold for all positive numbers. In the first example you give, you explicitly state "any $\varepsilon > 0$", which is where $\varepsilon$ is being "chosen". And if the result holds for all $c\varepsilon$ then it holds for all positive numbers because any positive number can be written as $c\varepsilon$ for some $\varepsilon$. $\endgroup$ – Reinstate Monica Nov 20 '14 at 21:23
  • $\begingroup$ @jef That's what I suspected was the case. So for example in the linked question (math.stackexchange.com/questions/864538/…) it's the case there too that it doesn't matter in what form epsilon is written, so long as we can logically find a $\delta$ (for that particular form of epsilon) then the proof would be valid? Also would it be valid for example if: $|f(x)-L|<\epsilon_1<\epsilon_2$ that we proved $|f(x)-L|<\epsilon_1$ then it would automatically be true for $\epsilon_2$? Thanks for the help! It's starting to make more sense now. $\endgroup$ – Jay Nov 20 '14 at 21:33
  • $\begingroup$ It does matter in what form the "epsilon" is written. In particular, that "epsilon" must be able to take on all positive values (or more precisely, all sufficiently small positive values). So the "epsilon" in the linked question (the entire $\min(\ldots)$ expression) can take on any value $< 1$ by choosing the appropriate $\varepsilon$. But if the "epsilon" Is given in the form $1+\varepsilon$ for $\varepsilon > 0$ then the proof is invalid. Your latter statement about $\varepsilon_1$ and $\varepsilon_2$ is true but I'm not sure how it connects to your original question. $\endgroup$ – Reinstate Monica Nov 20 '14 at 21:37
  • $\begingroup$ @jef That makes sense, especially the example of $1+\epsilon$. The last part was about if one were to prove it true for one epsilon value then it's true for all (including larger values). It's stemmed from when completing a problem going from: $|(1/x)(x-1)|<\epsilon$ (given that $|1/x|<2$) to: $|(1/x)(x-1)|<2(x-1)$ and then to say that $2(x-1)<\epsilon$ since it's not stated that $2(x-1)$ is less than epsilon (and technically it's larger than the thing that was less than epsilon). But I'm guessing that since epsilon is any number it doesn't matter in that situation. Thanks for the Help! $\endgroup$ – Jay Nov 20 '14 at 22:00
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    $\begingroup$ @jef How about writing that as an answer? :) $\endgroup$ – Mark Fantini Nov 20 '14 at 22:05
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$\varepsilon$-$\delta$ proofs seem to be the most confusing concept of first-year Calculus (or pre-Calculus, depending on how it's taught).

The $\varepsilon$-$\delta$ definition of a limit is as follows. $\lim_{x\rightarrow a} f(x) = l$ means for all $\varepsilon > 0$, there is a $\delta > 0$ such that if $|x - a| < \delta$ then $|f(x) - l| < \varepsilon$. All your answers are contained within this definition so let's break it down and see how all the proofs you reference satisfy the definition.

For the limit to be $l$, we require that every $\epsilon > 0$ has a corresponding $\delta > 0$. A standard way to prove that something holds for all values of a variable in a certain domain is to choose an arbitrary (unspecified) value for that variable in its domain and make a proof that works for that arbitrary choice. Spivak's proof does exactly that: he says "for any $\varepsilon > 0$" and then proceeds to show that there is a $\delta$ corresponding to this specific but arbitrary $\varepsilon$. The unstated conclusion is that there is a $\delta$ for every $\varepsilon$.

Note that the proof of a limit requires that there is a $\delta$ corresponding to every positive "epsilon", but there is no requirement to denote the "epsilon" value as the symbol $\varepsilon$. Instead, we may use an expression such as $c\varepsilon$ for constant $c>0$ so long as this expression can take on all values in the domain of the "epsilon" (namely, all positive numbers). Clearly we may write any positive number as $c\varepsilon$ for some choice of $\varepsilon$. So if we prove that a $\delta$ exists for all $c\varepsilon$ in place of $\varepsilon$, the proof holds for every value that $c\varepsilon$ can take, which means it holds for every positive number, so we are good.

However, if the expression we choose can't take on all possible "epsilon" values, such as $1+\varepsilon$ for $\varepsilon > 0$, then if we prove that a $\delta$ exists for all $1+\varepsilon$ then we haven't proven that a $\delta$ exists for all values for which the definition of limit requires it exist. Such a proof would be invalid.

Finally, you posted a link to another question where a complex expression $\min\left(1,\frac{\epsilon}{2(|m|+1)}\right)$ stands in for the "epsilon". Note that by letting $\epsilon$ vary, this expression can take on all positive values up to but not exceeding 1. Therefore it doesn't take on all values that the definition of limit requires. However, if a $\delta$ works for some "epsilon" then it automatically works for any larger "epsilon" - you can immediately verify this statement in the definition of a limit. So if our expression can take on all positive numbers up to 1, that's actually good enough. (In the most generality, we can prove that there is a $\delta$ corresponding to all "epsilon" in a set of positive numbers containing a sequence going to 0.)

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  • $\begingroup$ Thanks for taking the time to write a great answer. I feel like I understand it now. Just to check if I do: If you had: $0<|x-1|<\delta$ with $|1/x|<2$ so that $|(1/x)(x-1)|<2|(x-1)|<\epsilon$; then proving that $2|(x-1)|<\epsilon$ (so $\delta=\epsilon/2$) would automatically be equivalent to proving that $|(1/x)(x-1)|<\epsilon$ since all that's required is a larger value of epsilon? $\endgroup$ – Jay Nov 20 '14 at 22:48
  • $\begingroup$ I'm not exactly sure what the argument that you are presenting is. What are we proving the continuity of? Are we choosing $\varepsilon$ before the $\delta$ or vice versa? $\endgroup$ – Reinstate Monica Nov 20 '14 at 23:30
  • $\begingroup$ Sorry for not being clear. The question was to prove that $\lim_{x\to1}1/x=1$. Starting from $|f(x)-L|<\epsilon$ gives: $|1/x-1|<\epsilon$. So $|1/x(1-x)|=|1/x|\times|x-1|<\epsilon$. I then let $0<|x-1|<\delta\leq 1/2$ so that we can get $|1/x|<2$. Thus we can show that: $|1/x|\times|x-1|<2 |x-1|$. This is where I was confused. Since it's easy to find a delta if: $2|x-1|<\epsilon$ since $\delta=\epsilon /2$ but I was confused making the step from $|1/x|\times|x-1|<\epsilon$ to $2|x-1|<\epsilon$. Since $2|x-1|$ is bigger than $|1/x|\times|x-1|$. $\endgroup$ – Jay Nov 20 '14 at 23:42
  • $\begingroup$ I think it would be clearer if you reordered your argument to make it clear when you are choosing your $\varepsilon$ and $\delta$. Before choosing either, you know that $|f(x)-L| = |\frac{1}{x}| \times |x-1|$. You want to show that for all $\varepsilon > 0$ there is a $\delta > 0$ such that if $|x-1| < \delta$ then $|\frac{1}{x}| \times |x-1| < \varepsilon$. At this point we choose an arbitrary $\varepsilon < 1$ and let $\delta = \frac{\varepsilon}{2}$. Since $|x-1| < \delta < \frac{1}{2}$, $|\frac{1}{x}| \times |x-1| < 2 |x-1| < 2 \delta = \varepsilon$. This completes the proof. $\endgroup$ – Reinstate Monica Nov 21 '14 at 1:07
  • $\begingroup$ Yeah, that order does make more sense since it highlights that if you pick any $\epsilon$ then there's a $\delta$ which works. I think I was getting hung up on having a specific $\epsilon$ in the inequality $|\frac{1}{x}|\times|x-1|<\epsilon$ when it's not, but rather, it just represents any positive number. Again thanks for the help! :) $\endgroup$ – Jay Nov 21 '14 at 5:59
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This is not so much a stand-alone answer as an appendix to jef's excellent answer above.

I think of $\varepsilon$-$\delta$ proofs as "challenge-and-response." The challenge is $\varepsilon$: "Can you find a $\delta$ so that ...?" The response is $\delta$: "Yes, I can!"

The proof is a strategy that will guarantee you an answer to the challenge, no matter what (positive) number you're handed.

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What Spivak calls an "${\epsilon}/{2}$ argument" is presented in some other books as a "$2\epsilon$ argument" where the sum of two things less than $\epsilon$ is bounded by $2\epsilon$.

It might seem nicer to have the final bound be exactly $\epsilon$, but the second approach is more flexible when you get to more complicated bounds like $\epsilon^5 + 7\sqrt{\epsilon}$ that do not have a simple algebraically described $\delta$ that can be chosen at the start to get $\epsilon$ at the end.

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