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Definition of $\lim_{x \to a} f(x) = L$:

$\forall \epsilon > 0, \exists \delta > 0 s.t. |f(x) - L| < \epsilon$

$ if \ 0 < |x-a| < \delta$

Question: Why can't we weaken the assumption to

$\exists N > 0$ s.t.

$\forall \epsilon \in (0, N), \exists \delta > 0 s.t. |f(x) - L| < \epsilon$

$ if \ 0 < |x-a| < \delta$

?

I think they are not equivalent. If they are, please explain how the latter proves the former and why we still need to have case 1 below.


Consider proving

$\forall \epsilon > 0, \exists \delta > 0 s.t. |x^2 - 25| < \epsilon$

$if \ 0 < |x-5| < \delta$


We first try to find some $\delta$.

$|x^2 - 25|$

$ = |x - 5| |x + 5| < \epsilon$ if we maybe choose $\delta$ s.t. ...:

Let $M > 0$ (further restrictions may be needed).

If $|x-5| < M$, then we have

$$- M < x-5 < M$$

$$\to 5 - M < x < 5 + M$$

$$\to 10 - M < x + 5 < 10 + M$$

$$\to (-10 - M) < 10 - M < x + 5 < 10 + M$$

$$|x + 5| < 10 + M$$

So we might choose $\delta = \min\{M, \frac{\epsilon}{10+M} \}$ for the two cases in the proof (it seems no further restrictions on M are needed).


Proof: Let $\epsilon > 0$.

Case 1: $$\epsilon > M(10+M)$$

$$\delta = M$$

$$\to |x - 5| |x + 5| < M |x+5| < \frac{\epsilon}{10+M} (10+M) = \epsilon$$

Case 2: $$0 < \epsilon < M(10+M)$$

$$\delta = \frac{\epsilon}{10+M}$$

$$\to |x - 5| |x + 5| < \frac{\epsilon}{10+M} (10+M) = \epsilon$$

Case 3: $$\epsilon = M(10+M)$$

Pick either value of $\delta$.

QED


Question in the case of this example: Cases 1 and 3 refer to tolerance levels $\ge M(10+M)$. Why do we care about those? Why isn't enough that we have proved case 2?

I'm thinking that we could just find $\delta$'s that work for $\epsilon \in (0,N)$ for some $N > 0$. Why do we care about all $\epsilon$ ie $\epsilon \ge N$?

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    $\begingroup$ I wouldnt say the two are equivalent. I'd say they are identical. They both so for all $\epsilon$ such that $N > \epsilon> 0$. What I can't figure out is why you think have an utterly unnecessary N helps in any way. $\endgroup$ – fleablood Feb 6 '16 at 17:44
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    $\begingroup$ The equivalence is precisely what justifies formulations often found in proofs such as: "Let $\epsilon>0$ be given. We may assume without loss of generality that $\epsilon<\frac12$. ..." $\endgroup$ – Hagen von Eitzen Feb 6 '16 at 17:45
  • $\begingroup$ @fleablood I'm just wondering why anyone cares about cases 1 or 3 in the proof above. We already proved the case for positive numbers up to M(10+M) so to me that should be enough. I think we should be concerned about SMALLER not LARGER tolerance levels $\endgroup$ – BCLC Feb 6 '16 at 17:54
  • $\begingroup$ @HagenvonEitzen Cool so I can just show Case 2 then add 'WOLOG' beforehand? $\endgroup$ – BCLC Feb 7 '16 at 5:40
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    $\begingroup$ Who says we care about all $\epsilon>0$? It is simply easier to write the definition that way! That definition is also clearly equivalent to limiting the choices of $\epsilon$ to some sequence of positive numbers converging to $0$. Say, $\epsilon=1/n$, or $\epsilon=10^{-n}$ for all naturals $n>0$. IIRC some textbooks do it using such a sequence. But defining it the usual way avoids drawing undue attention to any specific sequence, so it is more elegant. $\endgroup$ – Jyrki Lahtonen Feb 7 '16 at 6:33
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You can "weaken" it that way (it turns out to not be a weakening at all, except in apparence). Even take $N=1$ (or any fixed positive number) if you want.

Clearly, the first definition implies the second. Now, for the converse: assume we have $\exists N>0 \text{ s.t. }\forall \varepsilon \in (0, N), \exists \delta > 0 s.t. |f(x) - L| < \varepsilon \ if \ 0 < |x-a| < \delta$.

Then, fix any $\varepsilon > 0$, and choose $\varepsilon^\prime = \min(\varepsilon, N/2)$. Then use the above assumption: there exists one $\delta$ for $\varepsilon^\prime$, and this $\delta$ will then also work for $\varepsilon$ as $\varepsilon^\prime \leq \varepsilon$.

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    $\begingroup$ No, it is equivalent. It looks like a weakening, but it turns out to be exactly equivalent -- see the answer, which shows it. $\endgroup$ – Clement C. Feb 4 '16 at 17:22
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    $\begingroup$ To be honest, I have some trouble parsing what you do in the second part of your question. What is this $M$ supposed to be? You are given $\varepsilon$ and aim at finding a $\delta$ -- the introduction of an extra parameter $M$ looks weird, especially given the way it is introduced. $\endgroup$ – Clement C. Feb 4 '16 at 17:29
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    $\begingroup$ The thing is, here you write "If $\exists M$ such that $\lvert x-5\rvert < M$." This is not very helpful in defining $M$ -- there is always such an $M$, depending on $x$: take $M=\lvert x-5\rvert +1$. What you should do is to choose yourself a fixed $M>0$ (e.g., 1 as you say) independent of $x$, and then say "for all $x$ s.t. $\lvert x-5\rvert < M$, we get [...]." $\endgroup$ – Clement C. Feb 4 '16 at 17:34
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    $\begingroup$ Look at the sentence you wrote: $M$ is defined by the statement $\exists M > 0 \text{ s.t. } \lvert x-5\rvert < M$. This, mathematically, means that $M$ is allowed to depend on $x$. I am basically pointing out that the way you wrote the proof (not its spirit, but what is effectively written) is not rigorous: you probably did not write what you meant to write. $\endgroup$ – Clement C. Feb 4 '16 at 18:10
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    $\begingroup$ @BCLC $\delta$ can depend on $a$, and $\varepsilon$ (and, I reckon, possibly on $L$) -- not on $x$. $\endgroup$ – Clement C. Feb 5 '16 at 20:27
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The definitions are equivalent. Assume that there exists $N > 0$ such that for any $\varepsilon \in (0, N)$ there exists $\delta > 0$ for which $0 < |x - a| < \delta$ implies that $|f(x) - L| < \varepsilon$.

We want to show that given $\varepsilon' > 0$ there exists $\delta' > 0$ for which $0 < |x - a| < \delta'$ implies that $|f(x) - L| < \varepsilon'$. Consider two cases

  1. If $\varepsilon' \geq N$, take $\varepsilon = \frac{N}{2}$ and obtain $\delta > 0$ for which $0 < |x - a| < \delta$ implies that $|f(x) - L| < \varepsilon = \frac{N}{2} < \varepsilon'$ so taking $\delta' = \delta$ works.
  2. If $0 < \varepsilon' < N$, take $\varepsilon = \varepsilon'$ and obtain $\delta > 0$ for which $0 < |x - a| < \delta$ implies that $|f(x) - L| < \varepsilon = \varepsilon'$ so taking $\delta' = \delta$ works.

Thus, once you have proven the equivalence, cases $1$ and $3$ become redundant.

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  • $\begingroup$ Thanks levap. does this proof that they are equivalent answer 'why we still need to have case 1 below' ? great if it does $\endgroup$ – BCLC Feb 4 '16 at 17:21
  • $\begingroup$ re your edit, aha! so the proof is correct if we omit cases 1 and 3? $\endgroup$ – BCLC Feb 4 '16 at 17:23
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    $\begingroup$ Yeah, I've added a line that makes this explicit. $\endgroup$ – levap Feb 4 '16 at 17:23
  • $\begingroup$ @BCLC What the proof means is that to show that the limit exists, we need to find a $\delta$ for sufficiently small $\varepsilon$ and not necessarily all $\varepsilon$ where $N$ is our "smallness" threshold. We have the freedom to find some $N$ for which our proof works, given that $\varepsilon < N$ but we still need to find such $N$. $\endgroup$ – levap Feb 4 '16 at 17:25
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    $\begingroup$ @BCLC Well, if you work with the standard definition, then no but if you prove the equivalence and then quote it, then you can omit the rendurant cases. $\endgroup$ – levap Feb 4 '16 at 19:16
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Your intuition is pretty good. You actually claimed that it doesn't matter what the function does "far away" from $f(x_0)$ (More precisely, outside some neighborhood of $f(x_0)$).

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  • $\begingroup$ Oh thanks. Ummm is the claim correct? XD $\endgroup$ – BCLC Feb 4 '16 at 17:54
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    $\begingroup$ Yes, it's correct $\endgroup$ – Elimination Feb 4 '16 at 21:53
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    $\begingroup$ Sooooo why? Thanks Elimination, but your answer seems more to be suitable for a comment. No offense $\endgroup$ – BCLC Feb 5 '16 at 2:12

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