# Calculus, Definition of limit, Concept

## Main Question or Discussion Point

According to spivak, the defition of limit goes as follows:

" For every ε > 0, there is some δ > 0, such that, for every x, if 0 < |x-a| < δ,
then |f(x) - l |< ε. "

After some exercices, I came across with a doubt.

Say that I could prove that | f(x) - l |< 5ε, for some δ$_{1}$ such that 0 < |x-a| < δ$_{1}$.
Since ε > 0, and thus 5ε > 0, could I say that lim$_{x→a}$f(x) = l based on this proof?

Regards,

HallsofIvy
Homework Helper
Yes, if $|f(x)- L|< 5\epsilon$, for any $\epsilon> 0$ then, taking $\epsilon_1= \epsilon/5$ where $\epsilon$ is any given number, we have $|f(x)- L|< 5\epsilon_1= 5(\epsilon/5)= \epsilon$

Yes, if $|f(x)- L|< 5\epsilon$, for any $\epsilon> 0$ then, taking $\epsilon_1= \epsilon/5$ where $\epsilon$ is any given number, we have $|f(x)- L|< 5\epsilon_1= 5(\epsilon/5)= \epsilon$
Thanks!

Well, two days later and I have another question.

I tougth in creating another post, but since the question is similar and fairly simple, I will jsut continue on this post.

When introducing Integrals, Spivak reaches the following during a proof:

inf{U(f, P)} - sup{L(f, P´)} < ε , for any ε > 0.

Since this is true for all ε > 0, it follows that sup{L(f, P´)}= inf{U(f, P)}∴

First Comment : Althoug Spivak doesn't say it, we shoud keep in mind that inf{U(f, P)}$\geq$ sup{L(f, P´)}, otherwise one would have to use the absolute value, right?

Second comment, during the study of limits, we used |f(x) - L | < ε, for any ε > 0. So according with this part, for a limit to exist that should be a δ > 0, such that | f(x) - L | = 0. But this doesn't seem right.

Could you shed some light in this?

regards,

HallsofIvy
Homework Helper
Well, two days later and I have another question.

I tougth in creating another post, but since the question is similar and fairly simple, I will jsut continue on this post.

When introducing Integrals, Spivak reaches the following during a proof:

inf{U(f, P)} - sup{L(f, P´)} < ε , for any ε > 0.

Since this is true for all ε > 0, it follows that sup{L(f, P´)}= inf{U(f, P)}∴

First Comment : Althoug Spivak doesn't say it, we shoud keep in mind that inf{U(f, P)}$\geq$ sup{L(f, P´)}, otherwise one would have to use the absolute value, right?
Yes, that's the important case but Spivak doesn't have to say it because if it is negative, it is trivially less than any positive number.

Second comment, during the study of limits, we used |f(x) - L | < ε, for any ε > 0. So according with this part, for a limit to exist that should be a δ > 0, such that | f(x) - L | = 0. But this doesn't seem right.
No, that doesn't follow. The set of all positive numbers does not contain 0- no number in it is equal to 0 but the inf (greatest lower bound) is 0.

Could you shed some light in this?

regards,

I am not sure If I follow.

The set of all positive numbers doesn't coutain zero, however
| f(x) - L| = 0, would respect | f(x) - L| < ε for any ε > 0. right?

So, altough I am alredy a bit confused, I still don't understand why he can say that if

inf{U(f, P)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}= inf{U(f, P)}∴

But we cannot say that if | f(x) - L| < ε for any ε > 0, then |f(x) - L| = 0.

thanks,

I am not sure If I follow.

The set of all positive numbers doesn't coutain zero, however
| f(x) - L| = 0, would respect | f(x) - L| < ε for any ε > 0. right?

So, altough I am alredy a bit confused, I still don't understand why he can say that if

inf{U(f, P)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}= inf{U(f, P)}∴

But we cannot say that if | f(x) - L| < ε for any ε > 0, then |f(x) - L| = 0.

thanks,
Let me rearrange my line of thought, so I can make my doubt clearer in order for you to help me out.

So, I actually understand that :we cannot say that if | f(x) - L| < ε for any ε > 0, then |f(x) - L| = 0.

I guess this may be explained to the fact that for any ε > 0, the is a natural number n with $\frac{1}{n}$ < ε. Is this a suitable explanation?

Anyway, to my main question:

So, from | f(x) - L| < ε for any ε > 0, it doesn't follow that |f(x) - L| = 0, why can Spivak assure that if inf{U(f, P)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}= inf{U(f, P)}∴

I hope you have understood,

Regards,

HallsofIvy
Homework Helper
If it were true that $|f(x)- L|< \epsilon$ for every $\epsilon> 0$ then, yes, we would have to have f(x)= L. But that is NOT true. We are only saying that, given some $\epsilon> 0$, there exist $\delta> 0$ so that if $|x- a|< \delta$, then $|f(x)- L|<\epsilon$ for that particular $\epsilon$, not for all $epsilon$.

While the general statement "inf{U(f, P)}-sup{L(f,P)}<$\epsilon$" is true for all $\epsilon> 0$", the particular P depends upon the particular $\epsilon$.

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arildno
Homework Helper
Gold Member
Dearly Missed
If it were true that $|f(x)- L|< \epsilon$ for every $\epsilon> 0$ then, yes, we would have to have f(x)= L. But that is NOT true. We are only saying that, given some $\epsilon> 0$, there exist $\delta> 0$ so that if $|x- a|< \delta$, then $|f(x)- L|<\epsilon$ for that particular $\epsilon$, not for all $epsilon$.

While the general statement "inf{U(f, P)}-sup{L(f,P)}<$\epsilon$" is true for all $\epsilon> 0$", the particular P depends upon the particular $\epsilon$.
In my student days, I made the revolutionary, and highly applicable definition of the hypercontinuous function:
A function f(x) is hypercontinuous, if, for EVERY epsilon>0 and EVERY delta>0, there exists an L so that |f(x)-L| is less than epsilon.

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Limits, sure are a dificult concept to grasp completly.

Althoug, you seem quite confident about explanation, I don't undersant it. I have tought about it most of my day.

The defition of lim:

"for every ε > 0, there is some δ > 0 , such that, for all x, if 0 < | x - a |< δ, then | f(x) - L | < ε".

So, there should be and δ$_{1}$ > 0, for wich | f(x) - L | < 10$^{-20}$, given that |x -a| < δ$_{1}$, and there should be another δ$_{2}$, for wich | f(x) - L | < 10$^{-1000}$, and in the end, since it is valid for any ε > 0, shouln't it exist an δ$_{3}$ for wich | f(x) - L | = 0, given that |x -a| < δ$_{3}$ ?
Because, this is the line of though we use to explain:

→ if inf{U(f, P)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}=inf{U(f, P)}∴?, or is it not this tipe of reasoning?

Thank you for your most valued explanations.

→ if inf{U(f, P)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}=inf{U(f, P)}∴?
I have actually tried to prove this by myself. Not even knowing if such thing of aplicable to the case.
Here it goes.

Let A = { ε : ε > 0}; Then 0 would be the greatest lower bound for the set A.
But, inf{U(f, P)} - sup{L(f, P´)} < ε, for any ε. Thus meaning inf{U(f, P)} - sup{L(f, P´)} is a lower bound for the set A. Consequenly, inf{U(f, P)} - sup{L(f, P´)}≤ 0, since 0 is the greatest lower bound. Hence sup{L(f, P´)} = inf{U(f, P)}, because on the other hand inf{U(f, P)} - sup{L(f, P´)} ≥ 0. ∴

By how much did I miss the target?

Regards,

I have actually tried to prove this by myself. Not even knowing if such thing of aplicable to the case.
Here it goes.

Let A = { ε : ε > 0}; Then 0 would be the greatest lower bound for the set A.
But, inf{U(f, P)} - sup{L(f, P´)} < ε, for any ε. Thus meaning inf{U(f, P)} - sup{L(f, P´)} is a lower bound for the set A. Consequenly, inf{U(f, P)} - sup{L(f, P´)}≤ 0, since 0 is the greatest lower bound. Hence sup{L(f, P´)} = inf{U(f, P)}, because on the other hand inf{U(f, P)} - sup{L(f, P´)} ≥ 0. ∴

By how much did I miss the target?

Regards,
That actually seems right to me.

Limits, sure are a dificult concept to grasp completly.

Althoug, you seem quite confident about explanation, I don't undersant it. I have tought about it most of my day.

The defition of lim:

"for every ε > 0, there is some δ > 0 , such that, for all x, if 0 < | x - a |< δ, then | f(x) - L | < ε".

So, there should be and δ$_{1}$ > 0, for wich | f(x) - L | < 10$^{-20}$, given that |x -a| < δ$_{1}$, and there should be another δ$_{2}$, for wich | f(x) - L | < 10$^{-1000}$, and in the end, since it is valid for any ε > 0, shouln't it exist an δ$_{3}$ for wich | f(x) - L | = 0, given that |x -a| < δ$_{3}$ ?
Because, this is the line of though we use to explain:

→ if inf{U(f, P)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}=inf{U(f, P)}∴?, or is it not this tipe of reasoning?

Thank you for your most valued explanations.
This is actually the line of reasoning used to explain that sup{L(f, P´)}=inf{U(f, P`)}

Since for any partition P, U(f,P) is greater than or equal to L(f,P') for any partition P', it follows that U(f,P) is an upper bound on {L(f,P)} From this, it follows that inf{U(f,P)} is greater than or equal to sup{L(f,P)} (Otherwise, we would be able to produce some P such that U(f,P) were not an upperbound for {L(f,P)}) Thus, inf{U(f,P)} - sup{L(f,P)} ≥ 0.

Now to the part I think you're confused about. inf{U(f,P)} - sup{L(f,P)} < ε for all ε > 0 because of the following.

Given any ε > 0, there exist two partitions P and P' such that U(f,P) - L(f,P') < ε. Since U(f,P) ≥ inf{U(f,P)}, and L(f,P) ≤ sup{L(f,P)}, we have that

ε > U(f,P) - L(f,P') ≥ inf{U(f,P)} - L(f,P') ≥ inf{U(f,P)} - sup{L(f,P)} ≥ 0

Since we can do this for all ε > 0, we have that

ε > inf{U(f,P)} - sup{L(f,P)} ≥ 0 for all ε > 0, so that inf{U(f,P)} = sup{L(f,P)} holds.

The difference between saying that Lim x-> a of f(x) is L and |a - b| < ε for all ε > 0 is that, in the first case, we are only guaranteed the existence of an interval centered around a so that |f(x) - L| < ε holds for all values in this interval, and it is crucial to note that the interval may or may not depend on ε. While in saying that |a - b| < ε for all ε > 0, we are saying that two constants are arbitrarily close to each other.

I apologize for my lack of LaTex, but I honestly have no clue how to use it. I hope I helped!

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..........The difference between saying that Lim x-> a of f(x) is L and |a - b| < ε for all ε > 0 is that, in the first case, we are only guaranteed the existence of an interval centered around a so that |f(x) - L| < ε holds for all values in this interval, and it is crucial to note that the interval may or may not depend on ε. While in saying that |a - b| < ε for all ε > 0, we are saying that two constants are arbitrarily close to each other.

I apologize for my lack of LaTex, but I honestly have no clue how to use it. I hope I helped!
Yes, you have helped me. It is much clearer know.

Thanks,