Explain the delta-epsilon defintion of a limit.?

[global]

I already know the definition.The limit of f(x) as x approaches p from above is L if, for every ε > 0, there exists a δ > 0 such that |f(x) − L| < ε whenever 0 < |x − p |< δ.

However if this definition of a limit explains the behavior of a function around a point why is the inequality
|f(x) − L| < ε not 0<|f(x)-L|<ε. I get why this is a triple inequality 0 < |x − p |< δ, because you are specifying the behavior around x but not at x. For this same reason shouldn't |f(x) − L| < ε be 0<|f(x)-L|<ε. Why forget the <0 restriction?

Also, I already know what the definition means. The limit statement means that no matter how small ε is made, δ can be made small enough.In other words F(x) get get as close as it wants to L as x can get as close as it wants to a point p. However would the limit exist if as ε gets smaller, δ increases. In other words would the limit exist if F(x) gets closer to L with at the same time x getting further away from point p.

 

[Stephen Tashi]

However if this definition of a limit explains the behavior of a function around a point why is the inequality
|f(x) − L| < ε not 0<|f(x)-L|<ε.

Suppose $f(x)$ is the constant function $f(x) = L$. It is equal to $L$ at values of $x$ "all around" $p$. Would you want the limit of a constant function to fail to exist?

 

[arildno]

"Why forget the <0 restriction?"

Don't you think the constant function, f(x)=C ought to be proven to be continuous when we choose our L to equal C?

 

[arildno]

You are too fast, Stephen..

 

[global]

But the definition itself says that for every E>0 their is a delta>0. So does the formal definition of a limit not apply for constant functions

 

[johnqwertyful]

But the definition itself says that for every E>0 their is a delta>0. So does the formal definition of a limit not apply for constant functions

It does apply. Why wouldn't it?

 

[global]

If the function is constant, how do you chose an E>0. Every point on the function will always be the same value. I get why 0 < |x − p |< δ because you are taking about the points around x but not x itself. I understand what you guys said about how there are other points around x that have the same value of the limit so that why |f(x)-L|<e is not written like this 0<|f(x)-L|<e. But if this was truly so, why does the definition itself say for every E>0. By writting that doesn't that mean that function can't be equal to the value of the limit, even if it is at another point, so that means it should be written as 0<|f(x)-L|<e and not |f(x)-L|<e. If |f(x)-L| is not truly restricted to zero shouldn't the definition say for every E greater and equal to zero.

 

[Mandelbroth]

If the function is constant, how do you chose an E>0. Every point on the function will always be the same value. I get why 0 < |x − p |< δ because you are taking about the points around x but not x itself. I understand what you guys said about how there are other points around x that have the same value of the limit so that why |f(x)-L|<e is not written like this 0<|f(x)-L|<e. But if this was truly so, why does the definition itself say for every E>0. By writting that doesn't that mean that function can't be equal to the value of the limit, even if it is at another point.

If $0=a<b$, does that still mean that $b>0$? I don't know what you are saying.

 

[WannabeNewton]

What? I'm having the same confusion regarding your statement that Mandelbroth has. Let $\epsilon > 0$ be arbitrary and take any $\delta > 0$ you like because regardless, for any $x$ in the domain of the constant map such that $|x - a| < \delta$, we will trivially have $|f(x) - f(a)| = 0 < \epsilon$.

 

[chiro]

It just means the closer some arbitrary point gets to the limit point, the closer the map of the arbitrary point gets to the map of the actual limit point. This is with respect to same region (in terms of distance) that surrounds the limit point and the map of these points.

In a multi-dimensional scenario, the |.| is a norm instead of an absolute value.

If there is no region where the above exists, then the limit doesn't exist.

An easy way to think about it is start with a 2D case (since you have actual geometry) and draw two graphs: one is an x,y domain graph and the other is a 1-D limit.

Pick a limit point for the 2D graph and draw a few circles of various radii around it. For one of these circles, there will be a region where for all points in that region, the map of all of these points will move closer and closer to the limit itself as you move closer and closer to limit point (not the map itself).

Note that a limit does not include the limit point which is why there is an inequality. If the map of the actual limit point actually is equal to the limit itself, then you have continuity at that point.

 

[johnqwertyful]

If the function is constant, how do you chose an E>0. Every point on the function will always be the same value. I get why 0 < |x − p |< δ because you are taking about the points around x but not x itself. I understand what you guys said about how there are other points around x that have the same value of the limit so that why |f(x)-L|<e is not written like this 0<|f(x)-L|<e. But if this was truly so, why does the definition itself say for every E>0. By writting that doesn't that mean that function can't be equal to the value of the limit, even if it is at another point, so that means it should be written as 0<|f(x)-L|<e and not |f(x)-L|<e. If |f(x)-L| is not truly restricted to zero shouldn't the definition say for every E greater and equal to zero.

YOU don't choose epsilon, you choose DELTA.

So for a constant function, you can choose delta=1 (or any number you want).

Then if |x-a|<δ=1, |f(x)-c|=|c-c|=0<ε This is true REGARDLESS of how small ε is.

You must understand what the definition is formally

for all ε>0 there exists a δ>0 such that 0<|x-a|<δ implies |f(x)-L|<ε.




Lets have a specific example. Lim x->1 f(x)=2 where f(x)=2

Give me an epsilon. I don't care how small. I give you δ=1. Then if x is in the interval (0,1) U (1,2) f(x)=2 (f(x)=2 EVERYWHERE). So |f(x)-2|=|2-2|=0<ε Thus showing the limit is 2.

 

[johnqwertyful]

It just means the closer some arbitrary point gets to the limit point, the closer the map of the arbitrary point gets to the map of the actual limit point. This is with respect to same region (in terms of distance) that surrounds the limit point and the map of these points.

In a multi-dimensional scenario, the |.| is a norm instead of an absolute value.

If there is no region where the above exists, then the limit doesn't exist.

An easy way to think about it is start with a 2D case (since you have actual geometry) and draw two graphs: one is an x,y domain graph and the other is a 1-D limit.

Pick a limit point for the 2D graph and draw a few circles of various radii around it. For one of these circles, there will be a region where for all points in that region, the map of all of these points will move closer and closer to the limit itself as you move closer and closer to limit point (not the map itself).

Note that a limit does not include the limit point which is why there is an inequality. If the map of the actual limit point actually is equal to the limit itself, then you have continuity at that point.

If he's having trouble understanding the epsilon-delta definition in single variable case, I don't think bringing in multivariable cases and norms is going to help him, might even confuse him even further.

 

[global]

YOU don't choose epsilon, you choose DELTA.

So for a constant function, you can choose delta=1 (or any number you want).

Then if |x-a|<δ=1, |f(x)-c|=|c-c|=0<ε This is true REGARDLESS of how small ε is.

You must understand what the definition is formally

for all ε>0 there exists a δ>0 such that 0<|x-a|<δ implies |f(x)-L|<ε.

Lets have a specific example. Lim x->1 f(x)=2 where f(x)=2

Give me an epsilon. I don't care how small. I give you δ=1. Then if x is in the interval (0,1) U (1,2) f(x)=2 (f(x)=2 EVERYWHERE). So |f(x)-2|=|2-2|=0<ε Thus showing the limit is 2.

Really, you choose a delta and not epsilon? Because many other websites say you choose a epsilon and from that epsilon get a delta. Many sites have said that delta is a function of E, where E is the independent variable. If that was true don't you choose a epsilon and then from that info generate a known delta. Also from creating this function, one can show for any E you can get a delta. Also, even the definition says that for any E>0 you can find a delta>O such that if 0<|x-P|<delta then |f(x)-L|<E. Doesn't the definition itself say that you have a given epsilon and then from that fact generate a delta. Could you guys also clear up these questions, do you have a give E and choose a delta, or do you have a give delta and choose a E. Is delta a function of E or is E a function of delta. Besides these questions, could you guys also answer the orginal questions in the post. Also does this relationship between delta and epsilon have to be a function or can it just be a simple relation. For example y= x^2 is a function but x^2+y^2=5 is a relation and not a function, however for each case x and y are clearly related to each other. For this reason does delta an epsilon have to be related by only a function or can it be just a relation.

 

[johnqwertyful]

Usually it's functions, but you can say in a proof "choose δ so that δ$^{2}$+5δ+2=7ε" or something like that. This is rare.

The idea is that you are GIVEN an epsilon. Then you must find a delta, which may be a function of epsilon (but doesn't HAVE to be, in the case of a constant function). Epsilon is the independent variable, you define a function (or relation) on it.

You are given an epsilon, then you must choose a delta so that the implication is true. However, to prove a limit you must prove it for arbitrary ε>0. The key of a limit is that you can make the function f(x) ARBITRARILY CLOSE to L so long as x is "close enough" to P.

 

[johnqwertyful]

Also, I already know what the definition means. The limit statement means that no matter how small ε is made, δ can be made small enough.

I'd like to comment on this. It's not that δ can be made small enough, it's that δ can be chosen so that the error can be made epsilon small.

The way it really clicked for me is that δ=d="distance" and ε=e="error". So by choosing x to be a certain "distance" from P, you can have a certain "error" in the function from the L. This is of course informal, but it made me understand it.

 

[Stephen Tashi]

YOU don't choose epsilon, you choose DELTA.

We shouldn't go off on a tangent about who "chooses". The definition of $lim_{x \rightarrow a} f(x) = L$ doesn't say anything about choosing or who does the choosing. It just says "For each $\epsilon > 0$ there exists a $\delta$...". In the liberal arts students are advised to understand a definition by "putting it in your own words", but this only leads to confusion in mathematics. Definitions mean what they say. Informal interpretations of definitions sometimes help students, but you can't debate the fine points of a definition by substituting the informal wording for the actual definiiton.

 

[johnqwertyful]

We shouldn't go off on a tangent about who "chooses". The definition of $lim_{x \rightarrow a} f(x) = L$ doesn't say anything about choosing or who does the choosing. It just says "For each $\epsilon > 0$ there exists a $\delta$...". In the liberal arts students are advised to understand a definition by "putting it in your own words", but this only leads to confusion in mathematics. Definitions mean what they say. Informal interpretations of definitions sometimes help students, but you can't debate the fine points of a definition by substituting the informal wording for the actual definiiton.

In a proof, you must find a δ(ε). I agree that you shouldn't put definitions in your own words, but in a proof you DO choose a δ. But I get your point and I do agree.

I've written many proofs as

"Let ε>0 be given. Choose δ so that δ=...".

 

[global]

The idea is that you are GIVEN an epsilon. Then you must find a delta, which may be a function of epsilon (but doesn't HAVE to be, in the case of a constant function). Epsilon is the independent variable, you define a function (or relation) on it.

I am confused, first you said you don't choose a epsilon but choose a delta, now you say choose a epsilon and then from that generate a delta. Do you choose a delta and then get a epsilon or do you choose a epsilon and then get a delta. Or does it really matter whether you choose a epsilon or a delta first.

 

[WannabeNewton]

$\epsilon$ is arbitrary. You have no control over what it can be. What you can control is what $\delta$ can be. The whole point of the limit is that no matter how arbitrarily small you make $\epsilon$, you can always find a $\delta$ that allows you to take an interval of radius $\delta$ around $a$ and, under $f$, fit it into an interval of radius $\epsilon$ around $L$.

 

[global]

Also, I already know what the definition means. The limit statement means that no matter how small ε is made, δ can be made small enough.In other words F(x) get as close as it wants to L as x can get as close as it wants to a point p. However would the limit exist if as ε gets smaller, δ increases. In other words would the limit exist if F(x) gets closer to L with at the same time x getting further away from point p.

Also nobody has commented on this part of the question yet? Even though the definition says of each E you can get a delta so that if 0<|x-P|<delta, then |f(x)-L|<e, what if when you choose a E the delta gets bigger, would the limit still exist? In other words would the limit exist if as f(x) gets closer to the value of the limit, x gets further away from the point P.

 

[global]

$\epsilon$ is arbitrary. You have no control over what it can be. What you can control is what $\delta$ can be. The whole point of the limit is that no matter how arbitrarily small you make $\epsilon$, you can always find a $\delta$ that allows you to take an interval of radius $\delta$ around $a$ and, under $f$, fit it into an interval of radius $\epsilon$ around $L$.

I am still confused on which variable you choose first. Some people say you choose epsilon first to get delta others says you choose delta first to get epsilon, which one is it?

 

[global]

Can somebody just start from the beginning, and answer all my questions in one reply in a clear and concise way. I am still just as confused as when i first started this thread.

 

[Stephen Tashi]

Can somebody just start from the beginning, and answer all my questions in one reply in a clear and concise way. I am still just as confused as when i first started this thread.

You haven't explained why you think (or thought) that the definition of a limit doesn't apply to constant functions. It does apply to them. What explanation do you need of that fact?

As to this business of who chooses delta or epsilon it is confusing because it is an attempt to deal informally with a formal principle of logic known as "universal generalization".

In applying that principle, to prove something true "for each W" of some sort , we begin by saying let "W be given" and also say it is a "given" thing of that sort. We prove the desired statement is true for W and conclude "Thus, for each W..." the desired statement is true. The requirement for using this method of proof is that the "given" W has no special properties other than it is the right sort of thing for the statement we are proving. (Universal Generalization is described in textbooks on symbolic logic.)

In the typical epsilon-delta proof, the principle of universal generalization is used. We say something equivalent to "Let the number $\epsilon > 0$ be given" or we just say "Let $\epsilon$ be a number greater than 0". Whether you want to call this "choosing epsilon" or not is not something I have strong feelings about! The person writing the proof obviously chooses the symbols to be used. Whether "you are given $\epsilon$" can also be debated. A more accurate description would be that "you give yourself" $\epsilon$.

 

[Mandelbroth]

I already know the definition.The limit of f(x) as x approaches p from above is L if, for every ε > 0, there exists a δ > 0 such that |f(x) − L| < ε whenever 0 < |x − p |< δ.

Let's start here. What you have written is wrong.

This definition has nothing to do with "from above." This is the definition of the limit. It is double sided. Also, since this is a definition, the $\forall\varepsilon>0~\exists\delta>0:\forall x(0<|x-p|<\delta\implies |f(x)-L|<\varepsilon)$ condition is necessary and sufficient. Thus, we must say "if and only if."

However if this definition of a limit explains the behavior of a function around a point why is the inequality
|f(x) − L| < ε not 0<|f(x)-L|<ε. I get why this is a triple inequality 0 < |x − p |< δ, because you are specifying the behavior around x but not at x. For this same reason shouldn't |f(x) − L| < ε be 0<|f(x)-L|<ε. Why forget the >0 restriction?

If we don't let the difference be 0, we can't make $f(x)$ arbitrarily close to $L$.

Also, I already know what the definition means. The limit statement means that no matter how small ε is made, δ can be made small enough. In other words F(x) get get as close as it wants to L as x can get as close as it wants to a point p. However would the limit exist if as ε gets smaller, δ increases. In other words would the limit exist if F(x) gets closer to L with at the same time x getting further away from point p.

The conditions you mention here have absolutely NOTHING to do with the existence of a limit of $F(x)$ at $p$.

 

[global]

1. I am sorry for wasting your guys time, but I am still confused. The definition says for every E>0 there is a delta>0. However the definition does not state that as E gets smaller delta has to get smaller, only that for each E>0 there is a delta >0 . Because the definition does not state this fact, that's mean that as E gets smaller, delta can also get bigger. If this situation did arise that as E gets smaller delta gets bigger, I am having a hard time visualizing whether or not the limit of the function at point P would exist.
2. The reason I don't understand why the formal definition of a function applys to constant functions is that the definiton itself says E has to be greater than zero. If e had to be greater than zero, doesn't this mean that the function can't equal the value of the limit at any other point. A constant function has the same value throghout, so how can E>O. For example say the limit(x approaches 4) of 5=5. Now say you want to prove this limit. Okay so say you have a given E=.01(that is range around the point 4,5). But how can you be .01 point (4,5) if every single point has the same value. All other values around (4,5), will always be the same value and therefore E will always be equal to zero. However the definition says for every E>0, so the definition does not apply to constant functions.

 

[Mandelbroth]

1. I am sorry for wasting your guys time, but I am still confused. The definition says for every E>0 there is a delta>0. However the definition does not state that as E gets smaller delta has to get smaller, only that for each E>0 there is a delta >0 . Because the definition does not state this fact, that's mean that as E gets smaller, delta can also get bigger. If this situation did arise that as E gets smaller delta gets bigger, I am having a hard time visualizing whether or not the limit of the function at point P would exist.

If you don't listen to us, we can't help you.

2. The reason I don't understand why the formal definition of a function applys to constant functions is that the definiton itself says E has to be greater than zero. If e had to be greater than zero, doesn't this mean that the function can't equal the value of the limit at any other point. A constant function has the same value throghout, so how can E>O. For example say the limit(x approaches 4) of 5=5. Now say you want to prove this limit. Okay so say you have a given E=.01(that is range around the point 4,5). But how can you be .01 point (4,5) if every single point has the same value. All other values around (4,5), will always be the same value and therefore E will always be equal to zero. However the definition says for every E>0, so the definition does not apply to constant functions.

You don't seem to understand what the inequality implies. $a>b=0$ does not imply that $a=0$. In fact, it means $a>0$.

 

[Stephen Tashi]

. If e had to be greater than zero, doesn't this mean that the function can't equal the value of the limit at any other point.

There is a difference between not commenting on what happens if $| f(x) - L| = 0$ and making the statement that this situation is not allowed to happen. The definition of limit doesn't address the situation $| f(x) - L | = 0$. It doesn't specify whether this can or cannot happen.

The definition of limit implies a definition for "not a limit".

The definition of $lim_{x \rightarrow p} f(x)$ is not $L$ is:
There exists an $\epsilon > 0$ such for for each $\delta > 0$ there exists an $x$ such that $0 < |x - p| < \delta$ and $| f(x) - L |$ is not $< \epsilon$.

(We consider the situation where $f(x)$ does not exist as a case where $|f(x) - L |$ is not less than $\epsilon$ because $| f(x) - L | < 0$ can only happen if $f(x)$ is a number.)

Take case of the constant function $f(x) = 2$ and $p = 1$ and see if you can find an example of an $\epsilon$ that proves $\lim_{x \rightarrow 1} f(x)$ is not $2$.

 

[global]

I do read each and every one the comments, however I still don't understand sorry.

 

[verty]

My translation into English is this:

Given ε > 0, there is a δ such that, whenever 0 < |x - a| < δ, |f(x) - L| < ε.

The free variables are f,a,L because this is a predicate that is true exactly when the limit of f at a is L. What you want to do is try different values of (f,a,l) to see which ones make the statement true.

Example: (y = 5, 2, 5).

Let δ = 1, then if 0 < |x - 2| < 1, |f(x) - 5| = |5 - 5| = 0 < ε.

So it is true for this constant function, abscissa and limit.

 

[global]

1. I am sorry for wasting your guys time, but I am still confused. The definition says for every E>0 there is a delta>0. However the definition does not state that as E gets smaller delta has to get smaller, only that for each E>0 there is a delta >0 . Because the definition does not state this fact, that's mean that as E gets smaller, delta can also get bigger. If this situation did arise that as E gets smaller delta gets bigger, I am having a hard time visualizing whether or not the limit of the function at point P would exist.

Thank you for the last comment, i understand that aspect, now can you explain this?

 

[Boorglar]

Thank you for the last comment, i understand that aspect, now can you explain this?

Think about it. The interval $0 < |x - a| < δ_1$ such that $|f(x) - L| < 0.01$ is obviously included in the interval $0 < |x - a | < δ_2$ such that $|f(x) - L| < 0.1$. So $δ_1$ cannot be larger than $δ_2$ unless you haven't chosen the largest possible values for $δ_1, δ_2$.

Here's an analogy:
In the Olympic games, the number of runners that arrived within 0.1 seconds after Usain Bolt cannot be larger than the number of runners who arrived within 1 second after him.

 

[global]

thanks for that comment, i think i understand it. One last question. Can anybody solve this problem
lim(x approaches 1) of x^2=1, find delta so that E=1. Can anybody solve this. Please show all of your work.

 

[Mandelbroth]

thanks for that comment, i think i understand it. One last question. Can anybody solve this problem
lim(x approaches 1) of x^2=1, find delta so that E=1. Can anybody solve this. Please show all of your work.

We have that $|x^2-1|<1$, so $x^2\in(0,2)$. Thus, we have $x\in(-\sqrt{2},0)$ or $x\in(0,\sqrt{2})$. Given our definition of a limit, it is optimal to choose $x\in(0,\sqrt{2})$ because we want $|x^2-1|<1$ for all $x$ satisfying $0<|x-1|<\delta$. Thus, we can choose $\delta\in(0,\sqrt{2}-1)$.

 

[johnqwertyful]

thanks for that comment, i think i understand it. One last question. Can anybody solve this problem
lim(x approaches 1) of x^2=1, find delta so that E=1. Can anybody solve this. Please show all of your work.

We won't do your homework.

 

[Mandelbroth]

We won't do your homework.

I think he's asking us to do an example so he can understand.