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Epsilon-delta definition of limit

by spandan
Tags: calculus, epsilon-delta, limit, limits
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spandan
#1
Dec12-08, 10:55 PM
P: 9
I'm studying limits now (for the first time) and though have understood the intuitive concept of limit, I didn't get at all the epsilon-delta concept.

What is epsilon and delta? What is x-2? I didn't get anything at all.

So please explain me these in detail.


Thanking you in advanced for your kind help
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Fredrik
#2
Dec13-08, 12:08 AM
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Consider the sequence 1/2, 1/3, 1/4,... If you pick a small number epsilon, I can always (no matter what small number you picked) find an integer N such that 1/n is smaller than your epsilon for all n≥N. That's what we mean when se say that the limit of the sequence is 0.

Consider the function f defined by f(x)=x. (Let's keep it as simple as possible). If you pick a small number epsilon, I can always (no matter what number you picked) find a positive number delta such that |f(x)-0| is less than epsilon for all x such that |x-0| is less than delta. That's what we mean when we say that the function goes to 0 as x goes to 0.

The point is that by choosing x "close enough" to 0 (the value that x goes to), we can make f(x) be "close enough" to 0 (the limit of f(x)).

I don't know what you mean by "x-2".
snipez90
#3
Dec13-08, 01:53 AM
P: 1,105
f(x) approaches a limit L (f(x) -> L) as x approaches a (x -> a) if we can make f(x) as close to L as we want provided that x is sufficiently close, but unequal to, a. This is probably the intuitive notion you are familiar with. Note we are not concerned with whether f is even defined at a since we only care about behavior of f as x gets arbitrarily close to a.

This graph helps connect the intuitive definition with the formal definition: http://www.math.ucdavis.edu/~kouba/C...ry/precise.gif

Let's be very specific to begin with. It is not hard to convince yourself that the function f(x) = [tex]\sqrt{|x|}cos(x)[/tex] approaches 0 as x approaches 0. We want to make f(x) close to 0, so why not make f(x) within [tex]\frac{1}{10}[/tex] of 0?

We want
[tex] \frac{-1}{10} < \sqrt{|x|}cos(x) < \frac{1}{10} [/tex],
which is the same as
[tex]|\sqrt{|x|}cos(x)| < \frac{1}{10}[/tex].
But [tex]|cos(x)| \leq 1[/tex],
so [tex]|\sqrt{|x|}cos(x)| = \sqrt{|x|}\cdot|cos(x)|\leq \sqrt{|x|} [/tex].

Now we need to make x sufficiently close to a.
By inspection, if we make x within [tex]\frac{1}{100}[/tex] of 0, i.e.,
[tex]|x| < \frac{1}{100}[/tex],
then [tex]|\sqrt{|x|}cos(x)| \leq \sqrt{|x|} < \sqrt{\frac{1}{100}} = \frac{1}{10}[/tex].

This means that if x is within 1/100 of 0, but unequal to 0 (again, don't be concerned with f(0) for the moment), then |f(x) - 0| < 1/10. Now try to generalize this example. Take another look at the definition and the picture, both provided here: http://www.math.ucdavis.edu/~kouba/C...ciseLimit.html

What epsilon was used in the above example? Could we have chosen any number to be epsilon? Which delta was chosen and how does it relate to the given epsilon? If you can relate each component of the specific example shown with those of the precise definition, then you prove the more general statement (bolded above).

JG89
#4
Dec13-08, 11:54 AM
P: 726
Epsilon-delta definition of limit

I liked Fredrik's explanation. It might also help to remember that when speaking about functions, the entire delta-neighborhood must be mapped into the epsilon-neighborhood.
Tac-Tics
#5
Dec13-08, 04:28 PM
P: 810
The epsilon-delta definition is very weird! Don't be discouraged if you have trouble understanding it at first.

First, let's get the definition out there so we have something to work with:

[tex]\lim_{x \to a}f(x) = L \text{ means } \forall \epsilon: \exists \delta: \text{ if } |x - a| < \delta \text{ then } |f(x) - L| < \epsilon[/tex]

That's definitely not an intuitive or simple definition. In one sentence, you have some of the hardest parts of logic put together: mixed quantifiers and implication (and the implication is often written "backwards" using the phrasing "whenever"). But nevermind all the details. Just get a feel for it. You never need the definition ever again once you've proven the continuity of the important classes functions. But it's a good workout for the brain!

In real life, you can never have perfect precision. In any measurement you make, there will always be some amount of error. If you are careful, you can make the bounds of that error very small.

In the definition of a limit, epsilon and delta are error bounds. That is why the absolute values are there. When you say [tex] |x - a| < \delta[/tex], what you mean is that x and a are equal within an error tolerance of delta. Similarly with epsilon, f(x), and L.

A continuous function has to do with accuracy of input and output of the function. There needs to be a relationship between accurate output and accurate input. More specifically, a continuous function must be one that allows you get an output as accurate as you desire.... simply by inputting accurate enough data.

When coming up with proofs of continuity, you are usually looking for a value for delta in terms of epsilon that satisfy the logical implication. These proofs also make heavy use of the triangle rule for absolute value.

The
Count Iblis
#6
Dec13-08, 07:50 PM
P: 2,158
If the sequence a_n tends to a limit a, then we usually formulate this by saying that for every epsilon>0 there exists a N such that for all
n > N we have that |a_n -a| < epsilon.

A math prof. of a course I was following once said in class that he didn't understand why we always let the small n be larger than the big N
spandan
#7
Dec13-08, 11:57 PM
P: 9
I still don't get it very nicely. I would appreciate if you could put it in simpler words.

The [tex]\epsilon - \delta[/tex] definition of limit is given as follows:
Let [tex]f(x)[/tex] be defined in the neighbourhood of [tex]a[/tex]. Then [tex]f(x)[/tex] is said to tend to limit [tex]L[/tex] as [tex]x[/tex] approaches to [tex]a[/tex], or symbolically,
[tex]lim_{x \rightarrow a} f(x) =L[/tex]
If for every positive number [tex]\epsilon[/tex] - however small it be - there exists a corresponding positive number [tex]\delta[/tex] such that
[tex]0 < |x - a| < \delta \mbox{then} |f(x) - L| < \epsilon[/tex]
I have some questions:
1. Can [tex]x[/tex] be equal to [tex]a[/tex]?
2. Is is that we can select [tex]\epsilon[/tex] and [tex]\delta[/tex] ourselves? How? Can any small number serve as [tex]\epsilon[/tex]?

I am also confused about the "Existence of a funstion at a point" part:
A function [tex]f(x)[/tex] will exist at a point [tex]x = a[/tex] if [tex]\mbox{left hand limit = right hand limit}[/tex]
[tex]lim_{x \rightarrow a-h} f(x) = lim_{x \rightarrow a+h} f(x) \mbox{ for } |h| < \delta [/tex]
[tex]lim_{h \rightarrow 0} f(a-h) = lim_{h \rightarrow 0} f(a+h) \mbox{ for } |h| < \delta[/tex]
Explain this one to me please.


What is h?

Quote Quote by Fredrik
Consider the sequence 1/2, 1/3, 1/4,... If you pick a small number epsilon, I can always (no matter what small number you picked) find an integer N such that 1/n is smaller than your epsilon for all n≥N. That's what we mean when se say that the limit of the sequence is 0.
Say if I pick epsilon to be 0.0009, find me the integer N.
lubuntu
#8
Dec14-08, 01:31 AM
P: 473
The existence of a function at point has nothing to do with its limit at that point. You mean to say the limit at that point only exists if the left and right limits exist and agree.
Fredrik
#9
Dec14-08, 02:17 AM
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Quote Quote by spandan View Post
Say if I pick epsilon to be 0.0009, find me the integer N.
Since 1/0.0009=1111.111..., the smallest integer we can use for this purpose is 1112. So I pick N=1112, and you say "ha-ha, that N is useless when I choose epsilon to be 0.000007 instead". But then I calculate 1/0.000007=142857.142... and choose N=142858.

The meaning of the statement "the limit of this sequence is 0" is that I will always win this game if I get to choose my number (N) after you choose yours (epsilon).
Fredrik
#10
Dec14-08, 02:46 AM
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Quote Quote by spandan View Post
I have some questions:
1. Can [tex]x[/tex] be equal to [tex]a[/tex]?
2. Is is that we can select [tex]\epsilon[/tex] and [tex]\delta[/tex] ourselves? How? Can any small number serve as [tex]\epsilon[/tex]?
1. The function f doesn't have to be defined at a, so if we allow x=a, the inequality [itex]|f(x)-L|<\epsilon[/itex] doesn't always make sense. So [itex]0<|x-a|<\delta[/itex] is right.

2. All positive numbers serve as [itex]\epsilon[/itex].

Quote Quote by spandan View Post
I am also confused about the "Existence of a funstion at a point" part:
That should be "the limit of the function f as x goes to a exists if..."

Quote Quote by spandan View Post
What is h?
Any number satisfying [itex]0<h<\delta[/itex].

f(x) is said to go to y as x goes to a from the right (x→a+) if

[tex]\forall\epsilon>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } \exists \delta>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } 0<x-a<\delta\implies |f(x)-y|<\epsilon[/tex]

f(x) is said to go to y as x goes to a from the left (x→a-) if

[tex]\forall\epsilon>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } \exists \delta>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } 0<a-x<\delta\implies |f(x)-y|<\epsilon[/tex]
lion0001
#11
Dec15-08, 02:03 AM
P: 21
Quote Quote by spandan View Post
I'm studying limits now (for the first time) and though have understood the intuitive concept of limit, I didn't get at all the epsilon-delta concept.

What is epsilon and delta? What is x-2? I didn't get anything at all.

So please explain me these in detail.


Thanking you in advanced for your kind help
Consider this example , f(x)_as x approaches 1_= 2x + 3 = 5
We must prove that:
for each positive number ", there is a positive number delta such that

| (2x + 3 ) - 5| < epsilon for all x satisfying 0 < | x -1 | < delta

that is | 2(x - 1 ) | < epsilon , for all x satisfying 0 < | x -1 | < delta

now 2|x - 1 | < epsilon , now divide by 2 to get

| x - 1 | < (epsilon / 2) , here is the thing, look at this and look at

0 < | x - 1 | < delta

both look identical except for the " 0 < " in the delta part

since | x - 1 | < (epsilon/2)

and | x -1 | < delta ( note here i didnt include the " 0 < " part
which is fine because that just tell us that | x -1 | has to be greater than 0

so we can make | x - 1 | < delta = (epsilon/2)

we choose delta = (epsilon/2) , so the statement

[tex]\forall\epsilon>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } \exists \delta>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } 0<x-a<\delta\implies |f(x)-y|<\epsilon[/tex]

holds since for 0 < | x -1 | < (epsilon/2) we have

delta = (epsilon/2) ==> epsilon = 2*delta

|2(x - 1 ) | < 2(delta) here epsilon = 2 delta ,see above

|2(x -1 ) | < 2 ( epsilon /2 ) here we just simplify

| 2(x-1) | < epsilon as required


NOTE: look at this website, several examples in detail about proving with epsilon delta are given

http://tutorial.math.lamar.edu/Class...fnOfLimit.aspx
spandan
#12
Dec17-08, 09:29 AM
P: 9
Can I take epsilon = 600 billion as well?

And I remember my Calculus teacher mentioning that delta should be greater than or equal to epsilon. Why is this?
Fredrik
#13
Dec17-08, 11:59 AM
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Quote Quote by spandan View Post
Can I take epsilon = 600 billion as well?
Look at the definition again:

We say that f(x)→y when x→a if

[tex]\forall\epsilon>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } \exists \delta>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } 0<|x-a|<\delta\implies |f(x)-y|<\epsilon[/tex]

([itex]\forall[/itex] means "for all". [itex]\exists[/itex] means "there exists"). The statement "there exists a number [itex]\delta>0[/itex] such that [itex]0<|x-a|<\delta\implies |f(x)-y|<\epsilon[/itex]" is going to be true when [tex]\epsilon=6\cdot 10^{11}[/itex] for most functions f that you will encounter. But this tells you nothing, since we only say that f(x)→y when x→a if that statement holds for all [itex]\epsilon>0[/itex], including every member of (for example) the sequence 1/2n.

Quote Quote by spandan View Post
And I remember my Calculus teacher mentioning that delta should be greater than or equal to epsilon. Why is this?
You probably just heard him wrong or misunderstood him. Consider the example [itex]f(x)=\sqrt x[/itex] and choose [itex]\epsilon=\frac 1 2[/itex]. We obviously want to define the limit in a way that means that this f(x)→0 as x→0, but do you think it's possible to chose a [itex]\delta\geq\epsilon=\frac 1 2[/itex] such that [itex]0<|x-0|<\delta \implies |f(x)-0|<\epsilon[/itex]? When x is slightly less than 1/2, f(x) is approximately 1.4 which is [itex]>\epsilon=0.5[/itex], so [itex]|f(x)-0|<\epsilon[/itex] doesn't hold.
Tac-Tics
#14
Dec17-08, 01:11 PM
P: 810
Quote Quote by spandan View Post
Can I take epsilon = 600 billion as well?
The equation has to hold for all epsilon, including 600 billion. However, the challenge is in small numbers.

And I remember my Calculus teacher mentioning that delta should be greater than or equal to epsilon. Why is this?
There is no need for delta to be greater than epsilon. They will be related to each other, but not necessarily in a simple way like this.
spandan
#15
Dec24-08, 03:50 AM
P: 9
So if I pick epsilon=0.00001, what would delta be? Say [tex]lim_{x \rightarrow 2} (2x^2 + 3x - 14) = 0 [/tex]. What would be delta? How do we find out? How are epsilon and delta related to each other? So delta is not smaller than or equal to epsilon?



This is how my book solves this question:
Find the limit [tex]lim_{x \rightarrow 3} (3x - 4)[/tex] and verify the result.
Solution:
[tex]lim_{x \rightarrow 3} (3x - 4)[/tex] = 3*3-4 = 5. To verify that result, we have to show that the corresponding to [tex]\epsilon > 0[/tex], there exists [tex]\delta[/tex] such that [tex]|(3x - 4) - 5| < \epsilon[/tex] whenever [tex]|x-3| < \delta[/tex].

Here we have to find [tex]\delta[/tex] in terms of [tex]\epsilon[/tex], considering [tex]\epsilon[/tex] is given. Now

[tex] |(3x-4)-5| < \epsilon \Rightarrow |3x-9| < \epsilon \Rightarrow |x-3| < (\epsilon / 3)[/tex]
Hence [tex]\delta \leq (\epsilon / 3)[/tex].





How is delta found out in terms of epsilon in this solution? Is |x - 3| equal to delta? How? I didn't get this solution. Please help me out.
Hurkyl
#16
Dec24-08, 04:22 AM
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Maybe an alternate description will help.

Fred and George are going to play a game. The game goes as follows:
(1) Fred writes a positive number. We will call that number [itex]\epsilon[/itex].
(2) George writes a positive number. We will call that number [itex]\delta[/itex].
(3) Fred writes down another number. We will call that number [itex]x[/itex].

Fred wins if both of the following are true:
(A) [itex]0 < |x - 2| < \delta[/itex]
(B) [itex]|(2x^2 + 3x - 14) - 0| \geq \epsilon[/itex]
otherwise, George wins.

The definition of limit says that if [itex]\lim_{x \rightarrow 2} (2x^2 + 3x - 14) = 0[/itex], then there is a strategy George can use that will let him win every time. If that limit doesn't exist or is not zero, then there is a strategy Fred can use that will let him win every time.
Fredrik
#17
Dec24-08, 08:39 AM
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Quote Quote by spandan View Post
This is how my book solves this question:
Find the limit [tex]lim_{x \rightarrow 3} (3x - 4)[/tex] and verify the result.
Solution:
[tex]lim_{x \rightarrow 3} (3x - 4)[/tex] = 3*3-4 = 5. To verify that result, we have to show that the corresponding to [tex]\epsilon > 0[/tex], there exists [tex]\delta[/tex] such that [tex]|(3x - 4) - 5| < \epsilon[/tex] whenever [tex]|x-3| < \delta[/tex].

Here we have to find [tex]\delta[/tex] in terms of [tex]\epsilon[/tex], considering [tex]\epsilon[/tex] is given. Now

[tex] |(3x-4)-5| < \epsilon \Rightarrow |3x-9| < \epsilon \Rightarrow |x-3| < (\epsilon / 3)[/tex]
Hence [tex]\delta \leq (\epsilon / 3)[/tex].





How is delta found out in terms of epsilon in this solution? Is |x - 3| equal to delta? How? I didn't get this solution. Please help me out.
What part of it are you having problems with? The solution you posted clearly shows that if we choose δ=ε/3 (or smaller), we have |(3x-4)-5|<ε for all x that satisfies 0<|x-3|<δ, and by definition of the limit, that means that 3x-4→5 as x→3.

I don't see why you're asking if |x-3| is equal to delta. It seems that you have ignored the words "for all" in all of the replies you got. You should think about what they mean.
Edgardo
#18
Dec24-08, 10:26 AM
P: 685
If you have problems understanding the definition of
[tex]\lim_{x \rightarrow a} f(x) = L[/tex]
start with a really easy example.

Easy example:
Consider [tex]f(x) = x[/tex] and [tex]\lim_{x \rightarrow 5} x = ?[/tex]

Obviously the limit is 5, i.e. [tex]\lim_{x \rightarrow 5} x = 5[/tex] but how do you prove
that? (It is obvious if you draw it. Do you know the geometric meaning of the limit?)

To show that the limit is 5, you have to show the following:
[tex]|x-a|< \delta \Rightarrow |f(x)-L| < \epsilon[/tex]
(this is just taken from the definition. L is the limit)

In other words:
Proof:
Step 0. Choose an appropriate value for delta
Step 1. You start with [tex]|x-a|< \delta[/tex],
Step 2. then make some manipulations and
Step 3. arrive at [tex]|f(x)-L| < \epsilon[/tex]
End of proof
--

In our example [tex]\lim_{x \rightarrow 5} x[/tex],

we have a=5 and f(x)=x and we want
to show that L=5. For this, we have to show that:

[tex]|x-5|< \delta \Rightarrow |x-5|< \epsilon[/tex]
Remember the definition:
[tex]|x-a|< \delta \Rightarrow |f(x)-L|< \epsilon[/tex]


Step 1. You start with [tex]|x-5|< \delta[/tex],
Step 2. then make some manipulations
Step 3. and arrive at [tex]|x-5|< \epsilon[/tex].

How do you get from Step 1 to Step 3, i.e. from [tex]|x-5|< \delta[/tex] to [tex]|x-5|< \epsilon[/tex]?
See Step 0: Choose an appropriate value for delta.
An obvious choice is delta = epsilon.
Let's check that:

Proof:
Step 0: Choose delta = epsilon
Step 1: |x-5| < delta = epsilon
Step 2: No complicated manipulations necessary here.
Step 3: |x-5| < epsilon from Step 1, thus |f(x)-L| < epsilon.
End of proof

Note: If you WRITE DOWN THE PROOF always start with Step 0: Choose appropriate delta

Although working "backwards" is a good method to find delta in terms of epsilon
always start with Step 0 for the proof.


---

Another example:

f(x) = 7*x

What is [tex]\lim_{x \rightarrow 3} 7x[/tex] ?

[tex]\lim_{x \rightarrow 3} 7x = 21[/tex]

Proof:
Step 0: Let delta = 1/7 epsilon, then
Step 1: |x-3| < delta = 1/7 epsilon
Step 2: Manipulations
|x-3| < 1/7 epsilon
=> 7 |x-3| < epsilon
=> |7x-21|< epsilon

Step 3: We arrived at
|7x-21| < epsilon, thus
|f(x)-L| < epsilon
End of proof

----

Note that usually you have a delta that depends on epsilon.
In the first example we had
delta = epsilon
in the second we had
delta = 1/7*epsilon

In other examples you may have
delta = epsilon^1/2,
delta = epsilon ^3/2

or just delta = 8 (or some other constant).

------

Here's a nice video on the delta-epsilon definition of a limit:
http://www.youtube.com/watch?v=u06Yrvt2XLc&fmt=18


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