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What is epsilon and delta?

  1. Dec 12, 2008 #1
    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
  2. jcsd
  3. Dec 13, 2008 #2


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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".
  4. Dec 13, 2008 #3
    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/CalcOneDIRECTORY/preciselimdirectory/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/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.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).
  5. Dec 13, 2008 #4
    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.
  6. Dec 13, 2008 #5
    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.

  7. Dec 13, 2008 #6
    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 :biggrin:
  8. Dec 13, 2008 #7
    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:
    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:
    Explain this one to me please.

    What is h?

    Say if I pick epsilon to be 0.0009, find me the integer N.
    Last edited: Dec 14, 2008
  9. Dec 14, 2008 #8
    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.
  10. Dec 14, 2008 #9


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    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).
    Last edited: Dec 14, 2008
  11. Dec 14, 2008 #10


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    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].

    That should be "the limit of the function f as x goes to a exists if..."

    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]
  12. Dec 15, 2008 #11
    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

    Last edited: Dec 15, 2008
  13. Dec 17, 2008 #12
    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?
  14. Dec 17, 2008 #13


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    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.

    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.
  15. Dec 17, 2008 #14
    The equation has to hold for all epsilon, including 600 billion. However, the challenge is in small numbers.

    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.
  16. Dec 24, 2008 #15
    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.
    [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.
    Last edited: Dec 24, 2008
  17. Dec 24, 2008 #16


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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.
  18. Dec 24, 2008 #17


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    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.
    Last edited: Dec 24, 2008
  19. Dec 24, 2008 #18
    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:
    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:

    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]

    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:
    Last edited: Dec 24, 2008
  20. Dec 24, 2008 #19


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    Anyone notice that when functions are not simple linear expressions, that the epsilon-delta limit proofs are usually not easy?
  21. Dec 24, 2008 #20


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    That's because linear functions are among the simplest of functions! That's why things like derivatives and the mean value theorem are so important -- they allow you to approximate complicated functions with linear functions. Another particular example is the limit of a quadratic expression; one controls a quadratic term like
    [tex]|x^2 - a|[/tex]​
    by factoring it into a product of linear terms
    [tex]|(x - \sqrt{a})| |(x + \sqrt{a}) |[/tex]​
    and then finding a way to simultaneously control both (linear) factors.
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