(ε, δ)-definition of limit

[Doom]

For the (ε, δ)-definition of limit, why does the relationship between the two parameters have to be an equation?

For example,


Why can't we choose δ >= ε/4 as |x -3| < ε/4 <= δ will always holds on? I see almost every textbook think they are on an equation relationship but not a inequality by default.

​

 

[Dr.Peterson]

For the (ε, δ)-definition of limit, why does the relationship between the two parameters have to be an equation?

For example,
View attachment 34494

Why can't we choose δ >= ε/4 as |x -3| < ε/4 <= δ will always holds on? I see almost every textbook think they are on an equation relationship but not a inequality by default.

​

You're right that there is not a single delta that must be chosen, and an inequality is appropriate; but the example you quote doesn't say that. What it says is that their work suggests that they should choose a particular value. They are not making a statement about all possible values of delta, but just choosing a particular value to use.

In fact, they aren't even saying that this is the largest possible value of delta; often you can't be sure of that, and you don't need to. All you need is to choose a value.

 

[Dr.Peterson]

Why can't we choose δ >= ε/4 as |x -3| < ε/4 <= δ will always holds on?

I should add that the inequality you wrote would be wrong if you were trying to describe all valid values for delta. It is not "at least ε/4". Do you see why? Is any such value of δ going to work?

 

[JeffM]

I think that your text has simply chosen the wrong modal verb. Rather than “should,” it is better to say “can equal.” If you insist on “should,” then it would better to say “$\text {should be } \le$”

Dr. Peterson, however, has pointed out that the reasoning shown does not prove that a larger value for $\delta$ is impossible. Therefore, as a matter of careful use of English, it is best to say “$\text {can be} \le$”.

 

[Doom]

I should add that the inequality you wrote would be wrong if you were trying to describe all valid values for delta. It is not "at least ε/4". Do you see why? Is any such value of δ going to work?

I'm not sure. Is that because we don't know the particular value of ε/4 or δ cannot be infinity?

 

[Doom]

I think that your text has simply chosen the wrong modal verb. Rather than “should,” it is better to say “can equal.” If you insist on “should,” then it would better to say “$\text {should be } \le$”

Dr. Peterson, however, has pointed out that the reasoning shown does not prove that a larger value for $\delta$ is impossible. Therefore, as a matter of careful use of English, it is best to say “$\text {can be} \le$”.

Yes. Forgive me that I'm not a native speaker. So may I state it as "there exists a δ value such that ε/4 ≤ δ, to make |x -3| < ε/4 <= δ hold on?

 

[Dr.Peterson]

So may I state it as "there exists a δ value such that ε/4 ≤ δ, to make |x -3| < ε/4 <= δ hold on?

What you need to do is to specify a value (or set of values) of $\delta$, given a specified $\epsilon$, so that whenever $|x-3|<\delta$, then $|(4x-5)-7|<\epsilon$. You haven't done that yet.

This will not be true for any $\delta\ge\epsilon/4$, which is what you are still saying.

JeffM has implied the answer: you should have said, for $\delta\le\epsilon/4$.

 

[Doom]

What you need to do is to specify a value (or set of values) of $\delta$, given a specified $\epsilon$, so that whenever $|x-3|<\delta$, then $|(4x-5)-7|<\epsilon$. You haven't done that yet.

This will not be true for any $\delta\ge\epsilon/4$, which is what you are still saying.

JeffM has implied the answer: you should have said, for $\delta\le\epsilon/4$.

Oh right! If ε/4 > δ , whenever |x-3| ≤ δ, |x-3| will surely be less than ε/4, as |x-4| < δ ≤ ε/4 . If |x-3| is smaller than a value, it will definitely be smaller than a larger value. That was what I actually thought of but I got messy when I wrote it

 

[Doom]

Oh right! If ε/4 > δ , whenever |x-3| ≤ δ, |x-3| will surely be less than ε/4, as |x-4| < δ ≤ ε/4 . If |x-3| is smaller than a value, it will definitely be smaller than a larger value. That was what I actually thought of but I got messy when I wrote it

And to prove the limit holds on we don't need to consider the situation δ < ε/4 because ε/4 are somewhat dependent from δ, we are trying to pick a value of x inside the δ interval as small as possible, and it will return a resulting value in the ε interval, that's what the definition is actually saying.

 

[Dr.Peterson]

Oh right! If ε/4 > δ , whenever |x-3| ≤ δ, |x-3| will surely be less than ε/4, as |x-4| < δ ≤ ε/4 . If |x-3| is smaller than a value, it will definitely be smaller than a larger value. That was what I actually thought of but I got messy when I wrote it

Correct. You had your implications, and your inequality, in the wrong direction. That's why you can't be "messy" in proofs.

And to prove the limit holds on we don't need to consider the situation δ < ε/4 because ε/4 are somewhat dependent from δ, we are trying to pick a value of x inside the δ interval as small as possible, and it will return a resulting value in the ε interval, that's what the definition is actually saying.

I presume you don't mean what you said there; $\epsilon$ is not dependent on $\delta$; the latter is dependent on the former.

 

[Doom]

Correct. You had your implications, and your inequality, in the wrong direction. That's why you can't be "messy" in proofs.


I presume you don't mean what you said there; $\epsilon$ is not dependent on $\delta$; the latter is dependent on the former.

Yes, we get the δ value by plugging into ε, δ = ε/4 can be regarded as y = x/4. Or it's like δ(ε) = ε/4, δ is like a function value or dependent variable here and ε is an independent variable

I was thinking ε interval is on y-axis and δ interval is on x-axis so ε was like a "y" and δ is like a "x", I take it for granted, that is lack of thinking, my bad.

 

[JeffM]

$\epsilon$ is an arbitary positive number.

 

[Doom]

$\epsilon$ is an arbitary positive number.

So can I understand it as we express δ by substitution of ϵ? Do δ and ϵ actually have a linear relationship like x-value and y-value which are independent and dependent variables?

 

[JeffM]

So can I understand it as we express δ by substitution of ϵ? Do δ and ϵ actually have a linear relationship like x-value and y-value which are independent and dependent variables?

It is dangerous to try to explain mathematical statements in natural languages because natural languages must be able to express concisely wide ranges of meaning whereas mathematics involves narrowly defined concepts.

The relationship between $\delta$ and $\epsilon$ is similar in one respect and dissimilar in another respect to the relationship between independent variable x and dependent variable y. It is similar in that $\delta$ will usually be expressed in terms of a function of $\epsilon$. It is dissimilar in that the relationship between x and y is a unique equality, $y = f(x)$ whereas the relationship between $\epsilon$ and $\delta$ is not. There are an infinite number of numbers that could be substituted for the $\delta$ described in the function $\delta = g( \epsilon )$. In particular, every positive number less than $\delta$ will work (I am assuming here that $\delta$ is specified as a positive number). The point being made is that there is at least one number, namely the number specified by $g( \epsilon)$, for ANY arbitrary positive $\epsilon$.

And the relationship between $\delta$ and $\epsilon$ need not be, and frequently is not, linear.

I’ll let Dr. Peterson clear up any infelicities in the above. He is a real mathematician.

 

[pka]

If you want a more intuitive approach then see Jerome Keisler's
Elementary Calculus: An Infinitesimal Approach
The chapters or the whole book is a free down-load at HERE.
Do you understand that $[-3,7]=\{x:|x-2|\le 5\}~?$
The midpoint of the interval $[-3,7]$ is $2$ its radius is $5$
Thus in the limit notation, $\bf{\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L}$
means that if $x\approx x_0$ then $f \left( x_0\right)\approx L$.
OR for $\bf x\approx x_0$ then $\bf f \left( x\right)\approx L$
What does it mean for one number to to be near another?
Well $\{x\in (a-10^{-100},a+10^{-100})\}$ is a set of all numbers very very near to $\bf a$.
Thus if $\epsilon>0$ such that $\exists \delta>0$ and if $\{t\in (x_0-\delta,x_0+\delta\}$ then $f(t)\in (L-\epsilon,L+\epsilon)$
i.e. if $\bf t$ is near $\bf x_0$ then $\bf f(t)$ is near $\bf L$
In my view you in post #1 were correct if only you had stopped after: this suggests that $\bf\large\delta=\dfrac{\epsilon}{4}$

 

[Doom]

Thank you all!