# Explain the delta-epsilon defintion of a limit.?

1. Aug 12, 2013

### 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.

2. Aug 12, 2013

### Stephen Tashi

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?

3. Aug 12, 2013

### 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?

4. Aug 12, 2013

### arildno

You are too fast, Stephen..

5. Aug 12, 2013

### 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

6. Aug 12, 2013

### johnqwertyful

It does apply. Why wouldn't it?

7. Aug 12, 2013

### 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.

Last edited: Aug 12, 2013
8. Aug 12, 2013

### Mandelbroth

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

9. Aug 12, 2013

### 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 $|f(x) - f(a)| = 0 < \epsilon$.

10. Aug 12, 2013

### 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.

11. Aug 12, 2013

### johnqwertyful

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.

12. Aug 12, 2013

### johnqwertyful

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.

13. Aug 12, 2013

### global

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.

14. Aug 13, 2013

### 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.

15. Aug 13, 2013

### johnqwertyful

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.

16. Aug 13, 2013

### Stephen Tashi

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.

17. Aug 13, 2013

### johnqwertyful

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=...".

Last edited: Aug 13, 2013
18. Aug 13, 2013

### global

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.

19. Aug 13, 2013

### 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 $L$.

20. Aug 13, 2013

### global

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.