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Delta-epsilon proofs

  1. Oct 19, 2005 #1
    I am currently having a lot of troublw with delta-epsilon proofs--can someone please help explain how they work and how to do them

  2. jcsd
  3. Oct 19, 2005 #2
    This is a VERY basic idea to start helping you:

    For every X, there should be an E that delimitates X in the X-E and X+E range, and for that range (which is characterized by E) there should be too a d that would delimitate X's function/relation between f(x)-d and f(x)+d.

    Now, you have to find E and d that would satisfy a small enough approach, which might be the tricky part. The basic idea is to find a relation between the ranges marked by the E and d in x and f(x), respectively.

    Surely someone can expand this ideas. I'll try to go deeper into this later.

    Good luck!
  4. Oct 20, 2005 #3


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    I assume that by "[itex]\delta, \epsilon[/itex] proofs", you mean proving that a limit is correct by using the basic definition:
    " [itex]lim_{x->a}= L[/itex] if and only if, for any [itex]\epsilon[/itex]> 0, there exist [itex]\delta[/itex]> 0 such that is |x-a|< [itex]\delta[/itex] then |f(x)-L|< [itex]\epsilon[/itex]".
    A fairly standard way to do that is to start from |f(x)-L|< [itex]\epsilon[/itex] and do whatever algebra is necessary to reduce to |x-a|< something. Then you can take whatever is on the right side as [itex]\delta[/itex].
    For example: show that the limit, as x- 2, of f(x)= 3x- 4 is 2:
    We must have |f(x)-L|= |3x-4-2|<[itex]\epsilon[/itex]. |3x- 4- 2|= |3x-6|= 3|x-2|. So that is the same as 3|x-2|< [itex]\epsilon[/itex] or |x-2|<[itex]\frac{\epsilon}{3}[/itex]. Clearly there exist such a [itex]\delta[/itex], just take it equal to [itex]\frac{\epsilon}{3}[/itex]. Strictly speaking, an actual proof would be to do everything backwards: Take [itex]\delta= \frac{\epsilon}{3}[/itex], the work back to |f(x)-L|< [itex]\frac{\epsilon}{3}[/itex]. Since every step we took was clearly reversible, normally, you don't have to write that.
    Last edited by a moderator: Oct 27, 2005
  5. Oct 20, 2005 #4

    matt grime

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    I would suggest you start by pretending that


    and seeing what that states about

    |f(x)-f(y)| (where f(y)=L)

    so if we were looking at the limit as x tends to 1 of x^2 (ie y=f(y)=1)

    then |x-1|<d means that |x^2-1|=|x-1||x+1| < d|x+1|

    Since we are free to pick d we may as well assume d<1 (after all if something is true whenever |x-1|<d for some d then it is certainly true for any smaller d and so there is no harm in assuming our d is less than 1) for then |x-1|<1 means that 1+x is less than 2 (think about it |x-r|<s means x is no more than s away from r so x is in the range r-s to r+s).

    since |x+1|<2 for d<1 it follows that for d<1 |x-1|<d implies |x^2-1| < 2d

    so given e>0 pick d=2/e (or 1 id 1 is less than 2/e) and we are done.

    my webpage has somewhere some method of doing any analysis problem of this type. it is my rewriting for my students of something that tim gowers www.dpmms.cam.ac.uk/~wtg10 wrote in his discussion of mathematics a la polya
  6. Oct 20, 2005 #5


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  7. Oct 24, 2005 #6
    I like this example but I think you left out something. f(a) might not be equal to L if there was a discontinuity there. So we should say that 0 < |x - a| < [itex]\delta[/itex] so that x can't be equal to a.
  8. Oct 25, 2005 #7


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    What are the other ways ?
  9. Oct 26, 2005 #8


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    Yes, I am forever forgetting that "0< "!

  10. Oct 27, 2005 #9
    From a beginner:

    The definition of a limit basically says that if L is a limit then you can choose x values that will make f(x) get as close to the limit L as someone demands. So, if someone challenges you to get f(x) within .0001 of the limit, then you will be able to find such an x value.

    In mathematical terms, if someone challenges you to get f(x) within .0001 of a limit L, that is the same as saying

    |f(x) - L| < .0001

    The absolute value just indicates that it doesn't matter whether f(x) is within L - .0001 or L + .0001. f(x) just has to be within .0001 of the limit in either direction. Written more generally,

    |f(x) - L| < E

    where E is any difference someone chooses, e.g. within .0001 of the limit, or within .000002 of the limit, or within .000000009 of the limit. In other words, no matter how small a number someone chooses, if L is a limit you can find an x such that f(x) is within that distance of the limit.

    The other part of the definition of a limit is:

    0<|x - a|< d

    That says that given:

    1) L is a limit

    2) and someone has challenged you to get f(x) within .0001 of the limit.

    then you can find x's that are close to 'a' that will result in f(x) being closer than .0001 to the limit. In fact, you can state a number d, e.g. .007, and all x's within .007 of 'a' will result in f(x) being within .0001 of the limit.

    For example, suppose someone says prove that:

    [itex]lim_{x\rightarrow3} (3x + 1) = 10[/itex]

    You can interpret that to mean, "If f(x) is required to be closer than some number E to the limit 10, show that there are values of x near 'a' that make that true." The values you are given for the problem are:

    a = 3
    L = 10
    f(x) = 3x + 1

    So, plugging those values into:

    |f(x) - L| < E


    |3x+1 - (10)| < E .......(1)


    |3x - 9| < E


    3|x-3| < E

    |x-3| < E/3

    To recap, I started at line (1), and the math I did says that when |f(x) - L| is smaller than E, then |x-3| < E/3. Written on the same line, it looks like this:

    if |(3x+1) - (10)| < E then |x - 3| < E/3

    Compare that line to the original definition of a limit:

    if |f(x) - L| < E then 0<|x - a|< d

    The two lines are identical given that f(x) = 3x +1, L = 10, and a = 3, AND if you choose d=E/3.
    (Well, they're identical as long as you declare the restriction x != a for the top line). So, if someone challenges you to get f(x) within some number E of the limit, the result says you can take the number and divide it by 3, and that will give you x's that meet the challenge. For example, if someone says I want you to give me an x such that f(x) is within .001 of the limit 10. Then, you can reply that all x's within .001/3 = .0003333 of 'a' will satisfy those requirements. Note that it doesn't matter what number the person gives you, e.g. if they demand that f(x) be within .000007 of the limit, all you have to do is divide the number by 3, and any x closer to 'a' than that will satisfy their demands.

    Since E/3 is a real number for any given real number E, you have proved that there is always some range of x's such that f(x) is within E of the limit.

    The methodology for doing the proofs is to start with:

    |f(x) - L| < E

    and make the substitutions for f(x) and L. Then, you need to manipulate the inequality, so that you end up with:

    |x-a| < something

    where 'a' is one of the givens. That way you can choose d to equal whatever is on the right side.
    Last edited: Oct 27, 2005
  11. Oct 28, 2005 #10
    Look at the graph below. In these proofs, all you are doing is trying to find the corresponding x's for the given f(x)'s. It's not earth shattering stuff.

    Attached Files:

  12. Apr 28, 2009 #11
    Perhaps this is a dumb question, but why do we use the inequality |f(x)-L|< E rather than |f(x)-L|≤ E ? If we are trying to get within Epsilon of our limit (L) than why not fit the mark exactly?
    Last edited: Apr 28, 2009
  13. Apr 28, 2009 #12
    It's like a science experiment. How precise do we need to make our instruments in order to achieve an error of less than 1%? The goal is not to achieve an error of 1% (at least I hope not!), it's to ensure that whatever error we get is less than 1%.
  14. Apr 28, 2009 #13


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    I personally think the former inequality conveys the idea of a limit better on some semi-unconscious level. But actually, the two definition are equivalent. That is to say, given a map f, we have

    (for all E>0, there exists a D>0 such that |f(x)-L|<E as soon as |x-a|<D) <==> (for all E>0, there exists a D>0 such that |f(x)-L|≤E as soon as |x-a|<D).

    This is very easy to prove and I encourage you to do so.

    Actually, it often happens that the second definition is used.
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