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Understanding Epsilon Delta Proofs

  1. May 23, 2009 #1
    I understand most of the logic behind the formal definition of a limit, but I don't understand the the logic behind an epsilon delta proof. The parts I'm having trouble with are these:

    1. How does proving that, the distance between the function and the limit is less than epsilon whenever the distance between x and a certain x value less than delta?

    2. How does reducing the epsilon side absolute value equation to mirror the delta side absolute value equation give us delta? (As inarticulate as that sounds, I can't word it any other way)

    Thank you for your help!
     
  2. jcsd
  3. May 23, 2009 #2

    CompuChip

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    I think that your question in 1 isn't really complete. I will just repeat the definition for you and hope that answers the first question.
    So we're interested in the limit of f(x) as x goes to a. The idea is, that you can pick any number, epsilon. Then I can give you an interval [a - delta, a + delta] around a, of size delta. As long as you choose an x inside the interval, f(x) will be closer than epsilon to the limit. This is formally capturing what a limit should be: we can get f(x) arbitrarily close to the limit, by getting x arbitrarily close to a.

    For the second question, the procedure you follow is merely a "draft". The formal proof should be the "draft" written down in reverse. So in the draft, you try to construct a sequence of inequalities on |f(x) - L| (with L the supposed limit), hoping that on the right hand side you get something with |x - a|. For example, suppose you can show that |f(x) - L| < 3 |x - a|^2. Now suppose that |x - a| < delta. Then |x - a|^2 < delta^2, so |f(x) - L| < 3 delta^2. You want it to be smaller than epsilon. So if you make sure that 3 delta^2 < epsilon, then that definitely happens because you have the sequence |f(x) - L| < 3 delta^2 < epsilon. You can get this, by choosing delta to be any (positive) value smaller than sqrt(epsilon / 3). So the reverse argument (the formal proof, which you write down and then throw away the draft) goes like this:
    Let epsilon > 0. Choose delta to be sqrt(epsilon / 3) (if you want to be sure, you can take it to be 1/2 * that, or something like that). Then,
    |f(x) - L| < .... < 3 |x - a|^2 < 3 delta^2 < 3 sqrt(epsilon / 3)^2 < epsilon.
    You have shown that for the given epsilon, there exists a delta (by explicitly giving one) such that the definition is satisfied.
     
  4. May 23, 2009 #3

    quasar987

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    For 5 years I've been answering people who ask the same question as you by referring them to this thread: https://www.physicsforums.com/showthread.php?t=54175&highlight=limit

    It contains a few reformulations of the definition of limit and a problem worked out in excruciating details.

    They usually say it helped.
     
  5. May 23, 2009 #4

    matt grime

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    I think you didn't quite write what you meant to there.

    Perhaps you'd be better using an example there?


    The idea of continuity is that if x is close to y then f(x) is close to f(y) - i.e. continuous functions do not tear things apart.


    f(x)=0 if x<0 and 1 otherwise is not continuous (at 0). If I choose x<0 and y>0 then no matter how close x and y are f(x) and f(y) are always distance 1 apart.

    Epsilon delta proofs are games. Given f, we play a game. I give you a target distance epsilon, and you have to try to find a delta so that x and y being at most delta apart means that f(x) and f(y) are at most epsilon apart.
     
  6. Aug 7, 2009 #5
    Thank you for the helpful responses. Quasar987, your explanation in the accompanying thread was very helpful. I did mistype in my original post and I'm still having trouble explaining what I don't understand, but I'm going to give this one more try:

    To find delta given epsilon, we try to simplify the absolute value equation on the epsilon side to resemble the absolute value equation on the delta side and somehow, this is supposed to show that the delta is a function of epsilon. I have difficulty understanding why, when the distance between f(x) and L on the epsilon equation is manipulated to resemble the distance between x - a, we have shown that epsilon is a function of delta.

    Thank you.
     
  7. Aug 7, 2009 #6

    jgens

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    As CompuChip noted already, this procedure is merely a draft - to formally complete the proof you need an equation where delta is a function of epsilon. However, despite this short-coming, doing epsilon-delta proofs in "draft" form is often desirable because it's often easier to simplify a complicated expression (opposed to making a simple expression more complicated). Hopefully some examples will illuminate this point: The first is a formal proof (try to understand the motivation for each step, even the definition of delta), while the second is a "draft."

    [tex]\lim_{x \to 0}\frac{1}{x+1} = 1[/tex]

    Proof: Let [tex]\epsilon > 0[/tex]. We define [tex]\delta[/tex] such that,

    [tex]0 < |x| < \delta = \mathrm{min}\left (\frac{1}{2}\, , \, \frac{\epsilon}{2}\right )[/tex]

    Then clearly we have that,

    [tex]\frac{1}{2} < x + 1 < \frac{3}{2}[/tex]

    Applying this inequality, we find that,

    [tex]0 < \left |\frac{1}{x + 1} - 1\right | = \left |\frac{x}{x + 1}\right | < 2|x| < \epsilon[/tex]

    Completing the proof.

    It's late now, so I'll give the second example tomorrow, but hopefully this illustrates why we use the "draft" form - it allows us to find a suitable delta (instead of just stabbing in the dark).
     
  8. Nov 16, 2009 #7
    Thank you for the help. I understand what an epsilon delta proof does.This is the difficulty I'm having:

    1. Why does finding delta as a function of epsilon prove that l x-a l is constrained on the interval L + E, L - E or rather,

    What does finding delta as a function of epsilon prove?

    2. How do you find delta as a function of epsilon when l f(x) - L l does not simplify to
    l x - a l?

    Thank you.
     
  9. Dec 2, 2009 #8

    CompuChip

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    Better late than never...
    It doesn't. You constrain x yourself! The claim is that IF x is constrained on an interval a-D, a+D, THEN automatically the function value f(x) is constrained to L-E,L+E.

    Well in particular that it exists for all epsilons. Which is what the definition says.

    Usually, make estimates. In the previous post by jgens, for example
    [tex]\left|\frac{x}{x+1}\right| < 2|x|[/tex]
     
  10. Dec 3, 2009 #9
    Thank you so much. Everything makes sense now. I feel like an idiot for not having figured this out sooner.
     
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