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The definition;

$(\forall \epsilon > 0)(\exists \delta > 0)(\forall x$ satisfying $0 <$ $|x-a| < \delta$ also satisfies $|f(x) - L| < \epsilon) \iff \lim_{x\to a} f(x) = L$

To explain my understanding, lets consider the following example;

$$\lim_{x\to 4} 3x^2 = 48.$$

Proof:

Let $\epsilon > 0$ be given such that $|3x^2 - 48| < \epsilon.$Then observe that $$|3x^2 - 48| = 3 \cdot |x-4| \cdot |x+4| < \epsilon.$$

Now, since our aim is to find a $\delta > 0$(probably as a function of $\epsilon$) such that for all $x$ satisfying $0 < |x-4| < \delta$, we have $|3x^2 - 48| < \epsilon$, i.e whatever the value of $\epsilon$ is, we just want to make sure that the condition is satisfied, so to be, sort of, safe and get rid of the factors, we can restrict ourselves by assuming $|x-4| < 1$, so this implies $|x+4| < 9$, so we have $$3 \cdot |x-4| \cdot |x+4| < 27 |x-4|,$$ and we already assumed that $$3 \cdot |x-4| \cdot |x+4| < \epsilon.$$

Now, if $\epsilon < 27 |x-4|$, there will be some $x$ such that $$3 \cdot |x-4| \cdot |x+4| \not < \epsilon,$$ which would contradict with out assumption, so we must have $$27 |x-4| \leq \epsilon,$$ which implies $|x-4| \leq \frac{\epsilon}{27}.$ However, this result is only valid if $|x-4| < 1$, so we must state that $$0 < |x-4| < \min\left\{1, \frac{\epsilon}{27}\right\} = \delta$$ QED.

First of all, is there any flow or misunderstanding in my understanding of the definition of the limit and generally with the proof of the example.

Secondly, in the book that I'm using it says that we don't want to make $|x-4|\cdot |x+4|$ too large, why ? what is wrong with it ? what if we do, then what ?

Edit:

As @md2perpe pointed out in his/her comment, there is logical mistake in the proof, so I'm writing the, so called, corrected version;

Let $\epsilon > 0$ be given. $$|3x^2 - 48| < \epsilon \iff 3 \cdot |x-4| \cdot |x+4| < \epsilon,$$ so let assume $|x-4| < 1$, hence $|x+4| < 9$, so we have $$|3x^2 - 48| = 3 \cdot |x-4| \cdot |x+4| < 27 |x-4|.$$ Since we want to find $\delta(\epsilon) > 0$, we want to relate $\epsilon$ to $\delta$, somehow, so if we say $\epsilon < 27 |x-4| < 27 \delta$, there will be some $x$ such that $0 < |x-4| < \delta$ but $|3x^2 - 48| \not < \epsilon$, so we consider the case $27 |x-4| < \epsilon$.Thus, $$0 < |x-4| < \delta = \min \{ 1, \frac{\epsilon}{27} \}$$

Alex Ortiz
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Our
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    You may not make any assumptions on $\epsilon$ other than its $>0$. Especially you may not assume $|3x^2-48|<\epsilon $. – md2perpe Jul 30 '17 at 16:52
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    @md2perpe if we are not going to assume that $|3x^2-48|<\epsilon$, then how can you conclude anything about the limit ? I mean then what is the purpose of $\epsilon$ ? – Our Jul 30 '17 at 16:53
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    You shall show that $|3x^2-48|<\epsilon $ if $|x-4|<\delta $ for some $\delta>0$. – md2perpe Jul 30 '17 at 16:57
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    @md2perpe Yeah, you are right, but still we do something like "Let $\epsilon > 0$ be given, then $|3x^2 - 48| < \epsilon \Longleftrightarrow ...$ " as if we are going backwards. – Our Jul 30 '17 at 16:59
  • @md2perpe Thanks for your comment by the way. – Our Jul 30 '17 at 17:05
  • "Let $\epsilon >0$ be given. Then let $\delta = \ldots $. If $|x-4|<\delta $ we then have $|3x^2-48| = \cdots <\epsilon $." – md2perpe Jul 30 '17 at 17:06
  • When the author says "we dont want to make $|x-4||x+4| $ too large", he means the following. Here we computing the limit as $x\to 4$, so what matters is the behavior of the function in the vicinity of $4$. When we assume $|x-4|<1$, we mean to investigate the behavior of the function in some small neighborhood of $4$. You may even take $|x-4|<1/2$, it does not matter. – Naive Jul 30 '17 at 17:07
  • You have a food beginning in factoring $3x^2-48$. Also the restriction $|x-4|<1$ is fine; this implies taking $\delta = \min (1, \ldots) $. – md2perpe Jul 30 '17 at 17:10
  • @Naive small relative to what ? I mean, what if chose 100 instead of 1, would this make proof invalid ? If not, why shouldn't I choose it ? – Our Jul 30 '17 at 17:16
  • A $small$ neighborhood around $4$. – Naive Jul 30 '17 at 17:17
  • @md2perpe About the proof template you have given, I generally would like to write these kind of proofs as if they were scratch work, but still rigorous, while writing my notes. – Our Jul 30 '17 at 17:19
  • @Naive What is neighbourhood around 4 ? can you define it ? – Our Jul 30 '17 at 17:20
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    You could restrict $\delta < 100$ if you want. Then you should take $\delta \leq \min (100, \epsilon/324) $. – md2perpe Jul 30 '17 at 17:21
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    A neighborhood of a point $a $ is a set having an open subset containing $a $. Often, however, a neighborhood is taken to be open itself. – md2perpe Jul 30 '17 at 17:23
  • When you assume $|x-4|<1$, you are considering all $x$'s in the neighborhood $(3,5)$. – Naive Jul 30 '17 at 17:23
  • @Naive I mean it is clear that OP, I'm, trying to eliminate any vagueness in the process, so it is not a good time to post vague comments :) – Our Jul 30 '17 at 18:05
  • Given a set $X$ and any $x\in X$ we say $U$ is a nbd of $x$ if : $x\in U$ and $U$ is open in $X$. I thought at this point of your studies you would have been familiar with what a nbd is. Hence I did not bother to give a precise definition. – Naive Jul 30 '17 at 18:11
  • @Naive of course, that is quite possible, thanks. – Our Jul 30 '17 at 18:14
  • You are still apparently making assumptions on $\epsilon$, which is a no-go. Try looking at a couple of $\epsilon$-$\delta$ proofs of limits to get the flavor for the argument. – Alex Ortiz Jul 30 '17 at 21:02
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    I still believe that using lot of logical symbols here is the main issue. Writing in plain English (substitute your natural language) is much easier to understand and far more unambiguous than believe think. After all who understands legalese apart from lawyers? Also these concepts of analysis are not supposed to be thought of just an exercise in mathematical logic, but rather they should be thought of as concepts built on the structure of an ordered field and hence the focus should be on understanding the basic inequalities and their consequences involved in the definition of a limit. – Paramanand Singh Jul 31 '17 at 05:44
  • You may have a look at this answer https://math.stackexchange.com/a/2349196/72031 where the $\epsilon, \delta$ is handled without using symbols from mathematical logic. – Paramanand Singh Jul 31 '17 at 05:51
  • @AOrtiz No, I'm not, as far as I see. I have checked lots of proofs, and what I'm trying to do is not learn how to imitate the other proofs, I'm trying to understand the logic, so it would be nice if you explain what is wrong in my proof. – Our Jul 31 '17 at 10:10
  • @AOrtiz Note that, in the beginning of the proof, I'm saying that the expressions $|3x^2 - 48| < \epsilon$ and $3 \cdot |x-4| \cdot |x+4| < \epsilon$ are equivalent. – Our Jul 31 '17 at 10:13
  • If you check the linked answer in my previous answer you will at once notice the issue in both your proofs. You are supposed to ensure that inequality $|3x^{2}-48|<\epsilon$ holds if values of $x$ are constrained in a certain manner. Your approach to assume $0<|x-4|<1$ is fine. This allows us to have the inequality $|3x^{2}-48|<27|x-4|$. Here is the crux of the main argument. The inequality $|3x^{2}-48|<\epsilon$ will now hold if we can ensure $27|x-4|<\epsilon$ and thus we get the desired constraint $0<|x-4|<\min(1,\epsilon/27)$. – Paramanand Singh Jul 31 '17 at 10:41
  • You are trying to use some sort of double implication like $$|3x^{2}-48|<\epsilon \Leftrightarrow 27|x-4|<\epsilon$$ this is wrong. The correct implication in above equation is $\Leftarrow$ and not $\Leftrightarrow $ and most importantly the $\Leftrightarrow $ implication is not needed. – Paramanand Singh Jul 31 '17 at 10:45
  • @ParamanandSingh I'm going backward in there, so I need $\Longleftrightarrow$. – Our Jul 31 '17 at 12:35
  • That is the fundamental mistake you are making. The definition of limit requires the implication $0<|x-a|<\delta\Rightarrow |f(x) - L|<\epsilon $ and this is just one way. It is not supposed to be two way in general. And the right approach is by going sort of backwards but not via chain of equivalent inequalities. That has to be avoided seriously because limit definition is not an exercise in algebraic manipulation to solve inequalities. Please see the linked answer for more details. – Paramanand Singh Jul 31 '17 at 13:50

4 Answers4

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Example of proof-writing: Let $\epsilon >0.$ Let $\delta =\min (1,\epsilon /27).$ $$\text { Then }\;\; \delta >0, \;\;\text { and } \; 0<|x-4|<\delta \implies |3x^2-48|<\epsilon.$$ Proof: (i). That $\delta >0:$ Obvious.

$$\text {(ii). }\; 0<|x-4|<\delta \implies |3x^2-48|=3\cdot |x-4|\cdot |x+4|=$$ $$=3\cdot |x-4|\cdot |(x-4)+8|\leq$$ $$\leq 3\cdot |x-4|\cdot (|x-4|+8)\leq$$ $$\leq 3\cdot |x-4| \cdot (\delta +8)\leq $$ $$\leq 3\cdot |x-4|\cdot 9=27 |x-4|<$$ $$<27 \delta \leq 27 (\epsilon /27)=\epsilon.$$

In your work in your edited version, you did some successful "exploratory surgery" to find that $\delta =\min (1.\epsilon /27)$ would suffice. Essentially the implications in your work are going from right to left, so when you reach $\delta =\min (1,\epsilon /27)$ you should then present a re-arrangement into the form I have shown, so that you present a clear and complete picture of the "logical flow".... Or,if you prefer, just present the finished proof with no explanation of how you found it, in the style of Carl F. Gauss.

DanielWainfleet
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Analisys is the art of bounding expressions.

In this case we have to bound $|3x^2 - 48|$ by $\epsilon$ when $x$ is not too far from $4$. Here is how

$$ |3x^2 - 48| = 3|x^2 - 16| = 3|x - 4||x + 4| \le 3|x - 4|c $$

provided that $|x + 4| \le c$ if $x$ is close enough to $4$.

How close? That's up to us. For instance, we can impose $3 \le x \le 5$, which means $|x - 4| \le 1$.

So, $7 \le x + 4 \le 9$ and therefore $|x + 4| = x + 4 \le 9$. Hence, we can take $c = 9$. Replacing

$$ |3x^2 - 48| \le 3\cdot9|x - 4| $$

and if we put $3\cdot9|x - 4| < \epsilon$ we get $|x - 4| <\epsilon/27$.

In sum, we have two restrictions for $|x - 4|$. Firstly, it must be $\le1$ and secondly $<\epsilon/27$. In other words, we have proven that

$$ |x - 4| < \min(1, \epsilon/27) \implies |3x^2 - 48| < \epsilon $$

Leandro Caniglia
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Here comes another discussion (and limit proof):

Let $\epsilon>0$ be given. We want to find $\delta>0$ such that $|3x^2-48|<\epsilon$ whenever $|x-4|<\delta$.

We have $|3x^2-48| = 3 |x+4| |x-4| < 3 |x+4| \delta$ when $|x-4|<\delta$. If we were to set $\delta = \epsilon/(3 |x+4|)$ then we get $|3x^2-48| < \epsilon$, but we can not do this since $\delta$ may not depend on $x.$

The factor $|x+4|$ thus is problematic. But using the triangle inequality we can eliminate it: $|x+4| = |(x-4)+8| \leq |x-4| + |8| < \delta + 8.$ Thus we have $|3x^2-48| < 3 (\delta+8) \delta.$

If we set $\delta$ so that $3(\delta+8)\delta = \epsilon$, i.e. $\delta = -4 + \sqrt{16 + \epsilon/3}$ (note: taking a minus sign makes $\delta<0$) we get $|3x^2-48| < \epsilon$. We can do this, but it's ugly having to solve a quadric, and in other cases we can not solve for $\delta$. We therefore want to eliminate the factor $\delta+8$ and only keep the linear factors.

By always making sure that $\delta \leq 1$ we get $\delta+8 \leq 9$ so that we have $|3x^2-48| < 27 \delta.$ We now can take $\delta = \epsilon/27$ or at least $\delta \leq \epsilon/27.$

Thus we take $\delta \leq 1$ and $\delta \leq \epsilon/27,$ i.e. $\delta \leq \min(1, \epsilon/27),$ and then we have $|x+4| \leq 9$ so that $$|3x^2-48| = 3 |x+4| |x-4| < 3 \cdot 9 \cdot \delta = 27 \delta \leq \epsilon$$ whenever $|x-4| < \delta.$

md2perpe
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I'll try to answer your questions one at a time:

First of all, is there any flow or misunderstanding in my understanding of the definition of the limit?

No, you have stated the definition of the limit accurately. I would have dispensed with the logical symbols and simply said that $\lim_{x\to a}f(x) = L$ if corresponding to each $\epsilon>0$ is a $\delta>0$ such that $|f(x)-L| < \epsilon$ whenever $0<|x-a|<\delta$, however.

Is there any flow or misunderstanding in my understanding of the proof of the example?

Yes, there is at least one point where you are tripped up. You have the following:

Since we want to find $\delta(\epsilon) > 0$, we want to relate $\epsilon$ to $\delta$, somehow, so if we say $\epsilon < 27 |x-4| < 27 \delta$, there will be some $x$ such that $0 < |x-4| < \delta$ but $|3x^2 - 48| \not < \epsilon$, so we consider the case $27 |x-4| < \epsilon$.

At this point in your proof, you are already done. This entire paragraph can be eliminated in favor of saying

If $|x-4| < 1$, then $|x+4|<9$, hence $|3x^2 - 48| = 3 |x-4| \cdot |x+4| < 27 |x-4|$. Consequently, if we set $\delta=\min(1,\epsilon/27)$, the claim follows. $\qquad \square$

It's not just about style. When you are considering separately the cases $\epsilon>\text{something}$ and $\epsilon<\text{something}$, you are opening up opportunities for innacuracy. In each $\epsilon$-$\delta$ proof, we never have to make assumptions on $\epsilon$ since $\epsilon$ is handed to us, and we have to work around it.

Secondly, in the book that I'm using it says that we don't want to make $|x−4|\cdot|x+4|$ too large. Why? What is wrong with it? If we do, then what?

The idea is that we want the quantity $|x−4|\cdot|x+4|$ to be small, and to do this, we really want to get some control on the quantity $|x+4|$, since if we had a bound like $|x+4|<M$ for some $M$, we could do something like set $\delta < \epsilon/M$ to finish up.

Alex Ortiz
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