Alternative Definitions of the Epsilon-Delta

[Calculuser]

Formal definition (epsilon-delta definition) of limit is symbolically as follows: $$\lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0,\ \exists \delta > 0,\ \forall x \in I,\ (0 < |x - c| < \delta \implies |f(x) - L| < \epsilon)]$$

Now I want to create alternative definitions out of this by replacing some of the terms.

$$D[1]:\ \lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0,\ \exists \delta > 0,\ \forall x \in I,\ (|f(x) - L| < \epsilon \implies 0 < |x - c| < \delta)]$$

$0 < |x - c| < \delta$ was replaced with $|f(x) - L| < \epsilon$ in the original definition to get $D[1]$.

$$D[2]:\ \lim_{x \to c}f(x) = L \iff [\forall \delta > 0,\ \exists \epsilon > 0,\ \forall x \in I,\ (0 < |x - c| < \delta \implies |f(x) - L| < \epsilon)]$$

$\epsilon$ was replaced with $\delta$ in the original definition to get $D[2]$.

Of course we can generate many other alternatives as we did above. What I want to know or expect a detailed, step by step answer because I could not figure it out myself is why $D[1]$ and $D[2]$ fail in that context so that we stick with the original definition. In other words, what counterexample satisfies the original definition, does not satisfy $D[1]$ and $D[2]$? Can anyone explain this clearly for me?

By the way, I am familiar with quantifiers, logical connectives, etc. and propositional logic in general.

 

[BvU]

D[1] fails if $f(x)$ has value $f(c)$ more than once in $I$.
D[2] fails near a vertical asymptote

 

[fresh_42]

You will learn more, if you try to figure out yourself why these are completely different concepts!
Sketch a bell curve for D1 and a step function for D2 and investigate which of your conditions are satisfied and which are not.

 

[PeroK]

Formal definition (epsilon-delta definition) of limit is symbolically as follows: $$\lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0,\ \exists \delta > 0,\ \forall x \in I,\ (0 < |x - c| < \delta \implies |f(x) - L| < \epsilon)]$$

Now I want to create alternative definitions out of this by replacing some of the terms.

$$D[1]:\ \lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0,\ \exists \delta > 0,\ \forall x \in I,\ (|f(x) - L| < \epsilon \implies 0 < |x - c| < \delta)]$$

$0 < |x - c| < \delta$ was replaced with $|f(x) - L| < \epsilon$ in the original definition to get $D[1]$.

$$D[2]:\ \lim_{x \to c}f(x) = L \iff [\forall \delta > 0,\ \exists \epsilon > 0,\ \forall x \in I,\ (0 < |x - c| < \delta \implies |f(x) - L| < \epsilon)]$$

$\epsilon$ was replaced with $\delta$ in the original definition to get $D[2]$.

Of course we can generate many other alternatives as we did above. What I want to know or expect a detailed, step by step answer because I could not figure it out myself is why $D[1]$ and $D[2]$ fail in that context so that we stick with the original definition. In other words, what counterexample satisfies the original definition, does not satisfy $D[1]$ and $D[2]$? Can anyone explain this clearly for me?

By the way, I am familiar with quantifiers, logical connectives, etc. and propositional logic in general.

D1 fails because it says that the only points where $f(x)$ is close to the limit $L$ are close to $c$. There may be points elsewhere that are close to $c$. Counterexamples include constant and periodic functions.

D2 fails because there is no requirement for $\epsilon$ to be small. The limit $L$, therefore, will not be unique. And, indeed, in many cases you could have $L$ to be any number as long as you choose $\epsilon$ large enough.

 

[jedishrfu]

An aside:

In my undergraduate days, I was bored with Differential Calculus (DC) and asked to skip ahead to Integral Calculus having learned the rudiments over summer break between HS and College.

The prof quizzed me on various aspects of DC and I did well except for the definition of a limit. He asked me to explain the definition in my own words and then said close but that's not quite right. He gave me permission to move on provided I come back and try again.

I did so a second time and again not quite right. Finally, in a move of student desperation, I memorized the definition and recited it exactly and he smiled and said I think you've got it now.

I now defer to Brilliant for explaining it "better":

https://brilliant.org/wiki/epsilon-delta-definition-of-a-limit/

 

[PeroK]

A good alternative is the sequence-based definition:

$$\lim_{x \to c}f(x) = L \iff$$

For any sequence $x_n \in I - \{c\}$ that converges to $c$, the sequence of function values $f(x_n)$ converges to $L$.

This gives a nice intuitive negation:

There exists a sequence $x_n \in I - \{c\}$, which converges to $c$, but the sequence of function values $f(x_n)$ does not converge to $L$.

 

[Infrared]

Since you asked for concrete examples:

The constant function $f(x)=1$ is not continuous according to definition 1.

The step function $f(x)=\begin{cases} 0, & x\leq 0\\ 1, & x>0\end{cases}$ is continuous according to definition $2$ (take $\varepsilon>1$).

Your definition 2 is equivalent to the function $f$ being locally bounded. This is much weaker than being continuous.

 

[FactChecker]

In D1, you just changed the direction of the "implies". That should raise red flags in your mind.
In D2, what prevents you from just making $\epsilon= 10^{999}$? Then you would say that practically all functions are continuous.
You should study the details of the official definition to see how it works better than your definitions. If you continue your studies of mathematics, you will see other definitions that work as well and are more general than the one you know now.

 

[Calculuser]

Sketch a bell curve for D1 and a step function for D2 and investigate which of your conditions are satisfied and which are not.

Correct me if I am wrong in my thinking:

$D[1]$ fails for a bell curve, or in general non-bijective functions, in the application of $\implies$ connective since "if $|f(x) - L|< \epsilon$, then $0 < |x - c| < \delta$". That is, we first take $|f(x) - L|< \epsilon$ interval along the y-axis inside the bell curve and find two $0 < |x - c| < \delta$, which means two different delta values. So this contradicts "for all epsilon there exists a delta". For $D[2]$ we cannot find an epsilon for all delta values where the jump occurs.

 

[fresh_42]

Correct me if I am wrong in my thinking:

$D[1]$ fails for a bell curve, or in general non-bijective functions, in the application of $\implies$ connective since "if $|f(x) - L|< \epsilon$, then $0 < |x - c| < \delta$". That is, we first take $|f(x) - L|< \epsilon$ interval along the y-axis inside the bell curve and find two $0 < |x - c| < \delta$, which means two different delta values. So this contradicts "for all epsilon there exists a delta". For $D[2]$ we cannot find an epsilon for all delta values where the jump occurs.

No. You should really draw some figures.

Assume $f(c)=L$ and $f(x)$ approaches $L$ if $x$ gets bigger and bigger. Then your condition $D1$ reads: For all $c>0$ there is an $N>0$ such that $|f(x)-L|<c$, which is given for some big values of $x$, we have $|x-c|<N$.

Now $|x-c|$ gets bigger and bigger, but there is always an $N$ which is even greater! Hence the entire condition is worthless. Although the function values are closing down, the gap on the $x-$axis increases and increases. Nevertheless, we will always find a natural number $N=\delta$ which exceeds this gap.

Congratulations: You defined the Archimedean order of the real numbers.

Now try something similar in the second case. A step function should do.

The clue is the following insight: Continuity is a local property of the behavior of a function. That is: A property which describes what $f$ does around a certain point $x=c$. The example uses the fact that $D1$ resolves locality. We are no longer restricted to a neighborhood of $x=c$. This contradicts our goal. It is something completely different.

 

[PeroK]

Correct me if I am wrong in my thinking:

$D[1]$ fails for a bell curve, or in general non-bijective functions, in the application of $\implies$ connective since "if $|f(x) - L|< \epsilon$, then $0 < |x - c| < \delta$". That is, we first take $|f(x) - L|< \epsilon$ interval along the y-axis inside the bell curve and find two $0 < |x - c| < \delta$, which means two different delta values. So this contradicts "for all epsilon there exists a delta". For $D[2]$ we cannot find an epsilon for all delta values where the jump occurs.

D1 fails fundamentally because the implication is the wrong way round. Your approach is completely on the wrong track altogether.

Let's take $f(x)= 0$ to be the constant function, equal to zero, on the interval $[0, 1]$. And let's take $c = 1/2$. Note that $\forall x \in [0, 1]: \ |x - 1/2| < 1$. So, $1$ becomes a universal $\delta$ for all $L$ and all $\epsilon$.

For example, we can show that by D1 we have $$\lim_{x \rightarrow 1/2} f(x) = 2$$

Proof:

Let $\epsilon > 0$. Choose $\delta = 1$.

Note that $\forall x \in [a, b]: \ |f(x) - L| = | 0 - 2| = 2$.

If $\epsilon \le 2$, then there are no values of $x$ for which $|f(x) - L| < \epsilon$, hence $0 < |x - 1/2| < 1$ is vacuously true.

And, if $\epsilon > 2$, then for all $x \in [a, b]$ we have $|f(x) - L| < \epsilon$. But, $0 < |x - 1/2| < 1$ is again true, as it is universally true.

And, indeed by the same argument we can show that the limit is any number $L$. And, indeed this will be the case for almost any function defined on a finite interval or bounded set.

You can see from this that by having the implication the wrong way round, the whole logic of the process breaks down. It's not just wrong for certain functions at certain points, it's logically the wrong way round.

 

[FactChecker]

In D(1), consider the constant function $f(x)\equiv0$. Clearly it is continuous everywhere. Consider $c= 0$ and $\epsilon = 1$. D(1) would say that there is a $\delta$ such that $|x-0|\lt \delta, \forall x$.

In D(2), consider $f(x)=0, x \lt 0; f(x)=1, x \ge 0$. Clearly it is not continuous at $x=0$. For $\epsilon = 100$, D(2) would say that the function is continuous at $x=0$.

 

[PeroK]

... in summary, this is the precise definition for good reason:

Formal definition (epsilon-delta definition) of limit is symbolically as follows: $$\lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0,\ \exists \delta > 0,\ \forall x \in I,\ (0 < |x - c| < \delta \implies |f(x) - L| < \epsilon)]$$

 

[Calculuser]

In D(1), consider the constant function $f(x)\equiv0$. Clearly it is continuous everywhere. Consider $c= 0$ and $\epsilon = 1$. D(1) would say that there is a $\delta$ such that $|x-0|\lt \delta, \forall x$.

That explanation is quite vague to me. If it says there is such a delta, so we can pick one, say $\delta = 1$. Why not? What prevents us from such a selection?

In D(2), consider $f(x)=0, x \lt 0$; $f(x)=1, x \ge 0$. Clearly it is not continuous at $x=0$. For $\epsilon = 100$, D(2) would say that the function is continuous at $x=0$.

What if we used that $\epsilon =100$ value in the original definition, would that say it is discontinuous if it would? If so, how come?

Let's look at the original definition and $D[1]$ with a single example. Imagine you want to prove a function's limit by two definitions. One makes sense and the other is nonsense. For simplicity imagine there is a constant function $f(x) = 3$ and I want to find its limit as $x$ goes to $1$ so that $c=1$ and $L=3$. Basically it asks to prove $\lim_{x \to 1}f(x) = 3$.

We have the boxed relations shown below for both the original definition and $D[1]$ to begin with, $$0<|x-c|< \delta \rightarrow \fbox{$0<|x-1|< \delta$}$$ $$|f(x)-L|< \epsilon \rightarrow |f(x)-3|< \epsilon \rightarrow |3-3|< \epsilon \rightarrow \fbox{$0 < \epsilon$}$$

I am stuck right here in both definitions. How do we combine these two relations in the boxes to interpret them in both of the definitions so that the original makes sense, but $D[1]$ does not?

 

[Stephen Tashi]

$$D[1]:\ \lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0,\ \exists \delta > 0,\ \forall x \in I,\ (|f(x) - L| < \epsilon \implies 0 < |x - c| < \delta)]$$

What is "$I$"? The set of real numbers?

By the way, I am familiar with quantifiers, logical connectives, etc. and propositional logic in general.

Then do you understand how to write the negation of the statement $\lim_{x \rightarrow c} = L$, according to $D[1]$?

$lim_{x \rightarrow c} f(x) \ne L$ would be written:

$\exists \epsilon > 0, \lnot (\exists \delta > 0, \forall x \in I, |f(x) - L| < \epsilon \implies 0 < |x -c | < \delta)$

This is equivalent to

$\exists \epsilon > 0, \forall \delta > 0, \lnot ( \forall x \in I, |f(x) - L| < \epsilon \implies 0 < |x -c | < \delta)$

Which is equivalent to

$\exists \epsilon > 0, \forall \delta > 0, \exists x \in I, \lnot (|f(x) - L| < \epsilon \implies 0 < |x-c| < \delta)$
Which is equivalent to
$\exists \epsilon > 0, \forall \delta > 0, \exists x \in I, |f(x) - L| < \epsilon) \land \lnot ( 0 < |x-c| < \delta)$

Which is equivalent to
$\exists \epsilon > 0, \forall \delta > 0, \exists x \in I, |f(x) - L| < \epsilon) \land ( \lnot (0 < |x - c|) \lor \lnot ( |x-c| < \delta))$

Which is equivalent to
$\exists \epsilon > 0, \forall \delta > 0, \exists x \in I, (|f(x) - L| < \epsilon) \land ( |x -c| \le 0 \lor |x -c| \ge \delta)$

Using that form, consider the example $f(x) = x$, $c = 3$, $L = 3$. Prove $\lim_{x \rightarrow 3} x \ne 3$.

Proof: Let $\epsilon = 6$. For an arbitrary $\delta > 0$ select $x = 3 + 100 \delta$. This establishes the existence of an $\epsilon > 0$ and the existence of an $x \in I$ such that $|f(x) -L| = |3 - 3| = 0 < 6 = \epsilon$ and $|x - c| = |3 + 100 \delta - 3| = 100 \delta \ge \delta$.

 

[FactChecker]

That explanation is quite vague to me. If it says there is such a delta, so we can pick one, say $\delta = 1$. Why not? What prevents us from such a selection?

Because the conclusion is ridiculous. How can you conclude that $|x|<1$ for all real values of x? You should think about what the D(1) definition implies when it is used in a simple case like this.

.
What if we used that $\epsilon =100$ value in the original definition, would that say it is discontinuous if it would? If so, how come?

That value for $\epsilon$ would not fit the requirements of the standard definition, but it does fit your D(2) requirements. For the standard definition, there is no value for $\epsilon$ that will work, so it does not say that the function is continuous, whereas your D(2) does.

 

[Calculuser]

I think my confusion results from the concept of "domain of discourse", was not specified by me above in the formal definition. What I have eventually come to realize, yet I am still not sure of it, that if we define $x \in \mathbb{R}$ and $\epsilon,\ \delta \in \mathbb{R}^{+}$, what you all explained above makes sense to me. The right-hand side of $D[1]$ after $\iff$ becomes false since we know $0 < \epsilon$ is automatically true for any constant function because of lack of $x$ and what we are left to test with is $0 < |x-1| < \delta$. If we recall that x is defined on $\mathbb{R}$ so we can find an $x$ for a specific $\delta$ to make $0 < |x-1| < \delta$ false because $x$ is not strictly defined to take values less than for any $\delta$. Therefore from $0 < \epsilon \implies 0 < |x-1| < \delta$ becomes false. After all, I think if we had a condition $x < \delta$ in the original definition, swapping $|f(x) - L| < \epsilon$ and $0 < |x - c| < \delta$ would not matter.

 

[FactChecker]

After all, I think if we had a condition $x < \delta$ in the original definition, swapping $|f(x) - L| < \epsilon$ and $0 < |x - c| < \delta$ would not matter.

I'm not sure that I followed your logic, but I will say that the use of $\epsilon$ and $\delta$ are not interchangable. $\epsilon$ is used to define the requirement and $\delta$ is used to show that the requirement, based on that $\epsilon$ is met. $\delta$ depends on the value of $\epsilon$, and it has to work no matter how small a positive number $\epsilon$ is.

 

[Stephen Tashi]

After all, I think if we had a condition $x < \delta$ in the original definition, swapping $|f(x) - L| < \epsilon$ and $0 < |x - c| < \delta$ would not matter.

What you mean by that is unclear.

If we modify the definition of $lim_{x \rightarrow a} f(x) = L$ in some manner so it no longer defines the same thing, then why would we be interested in whether swapping $\epsilon$ and $\delta$ matters in the definition of this new thing?

Which is to say, why would we be interested vis-a-vis the topic of this thread which is alternative definitions of the limit of a function - as opposed to definitions of some different concept ?

 

[Calculuser]

We discussed above why we cannot use alternative definition(s) of the formal limit definition. What I mean by,

After all, I think if we had a condition $x < \delta$ in the original definition, swapping $|f(x) - L| < \epsilon$ and $0 < |x - c| < \delta$ would not matter.

is if we had extra $x < \delta$ condition in the original definition, $$\lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0\ \exists \delta > 0\ \forall x \in I\ (x < \delta \land (0 < |x - c| < \delta \implies |f(x) - L| < \epsilon))]$$
we could use an alternative definition formed swapping $|f(x) − L| < \epsilon$ and $0 < |x − c| < \delta$ to get, $$\lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0\ \exists \delta > 0\ \forall x \in I\ (x < \delta \land (|f(x) - L| < \epsilon \implies 0 < |x - c| < \delta))]$$
I know these are not logically equivalent statements or any other alternatives that can be generated that way. The point I am more or less argue here is the uniqueness of that definition as it seemed to me that this original definition can be change a little to give the same concept, or in other words the existence of two definitions, from which one is made up by changing the other a little as I said, to encapsulate the same concept.

 

[Stephen Tashi]

if we had extra $x < \delta$ condition in the original definition, $$\lim_{x \to c}f(x) = L \iff [\forall \epsilon > 0\ \exists \delta > 0\ \forall x \in I\ (x < \delta \land (0 < |x - c| < \delta \implies |f(x) - L| < \epsilon))]$$

According to that definition, limits don't exist. A statement of the form $...\forall x ( (x < c) \land ...)$ is False since there exists values of $x$ for which $x < c$ is not true.

as it seemed to me that this original definition can be change a little to give the same concept,

The change did not result in the same concept.

 

[FactChecker]

Adding a constraint on x to keep it near zero has nothing to do with the concept of limits at other places where x is large. I don't think that you are thinking about your ideas as skeptically as you should.

 

[Calculuser]

According to that definition, limits don't exist. A statement of the form $\forall x ( (x < c) \land ...)$ is False since there exists values of $x$ for which $x < c$ is not true.

I think there is a typo above. You meant $\delta$ instead of $c$, I guess.

What is $I$, then? Is it the domain that function $f(x)$ is defined on, or is it a set that is tacitly assumed to be defined within $(0,\ \delta)$ like $0 < x < \delta$? Did you get my point? If the latter is the case, then it would be impossible $|f(x) - L| < \epsilon \implies 0 < |x - c| < \delta$ to be false in the new definition $D[1]$ above.

 

[PeroK]

I think there is a typo above. You meant $\delta$ instead of $c$, I guess.

What is $I$, then? Is it the domain that function $f(x)$ is defined on, or is it a set that is tacitly assumed to be defined within $(0,\ \delta)$ like $0 < x < \delta$? Did you get my point? If the latter is the case, then it would be impossible $|f(x) - L| < \epsilon \implies 0 < |x - c| < \delta$ to be false in the new definition $D[1]$ above.

Seriously, you need to forget this whole idea and try to learn Real Analysis, the genuine article.

Kicking these ideas around is okay for a while, but it soon becomes a complete waste of time.

 

[member 587159]

I also suggest to learn about limits of sequences first. They are easier to grasp imo, the structure is somewhat clearer.

 

[Stephen Tashi]

What is $I$, then? Is it the domain that function $f(x)$ is defined on, or is it a set that is tacitly assumed to be defined within $(0,\ \delta)$ like $0 < x < \delta$?

You introduced the notation "$I$", so you'll have to answer that question. In the (standard) definition of the limit of a real valued fuction, the domain of the function is a subset of the real numbers.

Did you get my point?

No.

If the latter is the case,

It isn't. A given function has only one domain. If the variable $\delta$ was used to define the domain of a function, then a statement of the form $\forall \epsilon > 0, \exists \delta > 0 (...)$ would not allow us to choose a $\delta$ that could vary with the value of $\epsilon$.