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Epsilon-delta definition of limits

  1. Jun 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Use the ε-δ definition of limits to prove that limx→2 x2 = 4.

    3. The attempt at a solution
    |x2 - 4| < ε

    0 < |x - 2| < δ

    |x - 2| |x + 2| < ε

    And that's where I get stuck, can I divide both sides by |x + 2| to yield

    |x - 2| < ε/|x + 2| = δ

    In which case, where do I go from there? Can I input 2 to get ε/4 = δ?
     
  2. jcsd
  3. Jun 11, 2015 #2

    SammyS

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    First you need to be able to state, using the ε-δ definition of this limit, what is meant by the statement, limx→2 x2 = 4 .

    Basically in ε-δ language we have:

    Given any ε > 0, there exists (indeed we can find what its value is) a δ > 0 (this δ generally depends on ε) such that for any x that satisfies 0 < |x - 2| < δ it is true that |x2 - 4| < ε .

    Often in figuring out what δ has to be, you will do some "scratch work" -- sort of working backwards. Then you turn things around and write up a nice proof, and in the process look like a genius.

    Your statement ε/|x + 2| = δ can be thought of as the result of some of that scratch work. However, there is the problem that x is in that expression.

    One way around this (for this particular limit) is put a bound on |x + 2| . To explain that, think about this ε-δ game that's being played here. It's really only a challenge when ε is small. For instance, if δ =1, then you would need an ε ≥ 5 . In other words, a δ of 1 will work just fine for any ε that's 5 or larger.

    What is the largest value |x + 2| has for those values of x satisfying 0 < |x - 2| < 1 ?
     
    Last edited: Jun 11, 2015
  4. Jun 11, 2015 #3
    Now I'm completely lost. Did you arbitrarily choose δ to equal 1? If δ = 1, why does it follow that ε ≥ 5? I get:

    -δ < x - 5 < δ

    5 - δ < x < δ + 5

    δ = 1 ⇒ 4 < x < 6

    ε = δ|x + 2|

    = |x +2| (where x is between 4 and 6).

    Edit: I don't know where I got 5 from, it should read:

    -δ < x - 2 < δ

    2 - δ < x < δ + 2

    δ = 1 ⇒ 1 < x < 3

    ε = δ|x + 2|

    = |x +2| (where x is between 1 and 3).
     
  5. Jun 11, 2015 #4

    RUber

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    Given some ##\epsilon >0 ## you need to show that there exists a ##\delta >0## such that if ## 2-\delta \leq x \leq 2+\delta ## then ## 4-\epsilon \leq x^2 \leq 4+ \epsilon ##.
    Normally you are looking to define ##\delta ## as a function of ##\epsilon.##
     
  6. Jun 11, 2015 #5
    How? I get no further than what I've written above.
     
  7. Jun 11, 2015 #6

    RUber

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    Try squaring the limits on x, then setting the stuff that is added or subtracted from 4 equal to epsilon and then solving for delta in terms of epsilon.
     
  8. Jun 11, 2015 #7
    If I understood you correctly:

    4 - ε < x2 < ε + 4

    2 - δ < x < 2 + δ ⇒ (2 - δ)2 < x2 < (2 + δ)2

    (2 - δ)2 = 4 - ε ⇒ ε = 4δ - δ2

    (2 + δ)2 = 4 + ε ⇒ ε = 4δ + δ2

    4δ - δ2 = 4δ + δ2 ⇒ 2δ2 = 0 ⇒ δ = 0

    But δ > 0 so this can't be right.
     
  9. Jun 11, 2015 #8

    SammyS

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    One thing you could do, since you edited that, use the strike-though or delete the unwanted part & leave a little note.


    Now use some of that scratch work from the Original Post, together with these results.

    To clean that up: You will never use δ > 1, Right? -- Note: δ will usually be much smaller than this.

    So that x will be between 1 and 3 as you said. That means that | x+2 | is between 3 and 5, but the main thing is that it's less than 5.

    So how does this help ?

    Let's see what that does for your relation between between ε and δ .

    You had ε = δ|x + 2| .

    But what do you know about |x + 2| ?

    (Well, we'll get to that scratch work in a minute.)
     
  10. Jun 11, 2015 #9

    RUber

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    Do one at a time. The result should end up being the same when you take the absolute value.
     
  11. Jun 11, 2015 #10
    Sorry about that, I'll do it next time:smile:

    3 < |x + 2| < 5

    ε = δ|x + 2]

    ε = 4δ ?

    I think I'm just getting more confused, hahah.
     
  12. Jun 11, 2015 #11
    I'm not following; do what one at a time? :)
     
  13. Jun 11, 2015 #12

    SammyS

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    All that's important here is the maximum possible value of |x + 2|.

    It will all work out shortly.

    So, if δ ≤ 1, what is the maximum value that |x + 2| can have?
     
  14. Jun 11, 2015 #13
    The maximum value is 4.999 or is it including 5? Making the maximum value 5 and making δ = ε/5?
     
  15. Jun 11, 2015 #14

    SammyS

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    That's good.

    Now, what's next ?

    Start with | x - 2 | < δ .

    But you chose δ = ε/5 , right?

    So | x - 2 | < ε/5 .

    Now how does | x - 2 | relate to | x2 - 4 | ? (You have something anout this in your scratch-work.)
     
  16. Jun 11, 2015 #15
    |x - 2||x + 2| < e/5 |x + 2|

    |x - 2||x + 2| < e/5 * 5

    |x^2 - 4| < e
     
  17. Jun 11, 2015 #16

    Fredrik

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    I don't want to interfere with a good discussion, but I have to go to bed, so if I don't post this now, it will have to wait until tomorrow. This is how I would describe what you should do:

    Let ##\delta## be an arbitrary positive real number (for now), and let x be an arbitrary real number such that ##|x-2|<\delta##. Now try to derive an inequality of the form ##|x^2-4|<f(\delta)## where ##f## is some function. If this succeeds, then we know that it will be sufficient to find a positive ##\delta## such that ##f(\delta)=\varepsilon##. If you find this equation, solve it for ##\delta##, and one of the solutions turns out to be positive, then you will have found a sufficiently small ##\delta##.
     
  18. Jun 11, 2015 #17

    RUber

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    OR

    If you solve either or both of these for delta, you will end up with very similar forms. Hint: Complete the square.
     
  19. Jun 11, 2015 #18

    SammyS

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    At this point, I would insert something like:
    We know that |x + 2| ≤ 5 so that (ε/5)⋅|x + 2| ≤ (ε/5)⋅5​
    Therefore, ... (what you have next)
    Now there are a few details to take care of. Then writing up a nice proof.

    What we have so far is the following. You have demonstrated that if δ ≤ 1, then choosing δ = ε/5 gives what is needed to prove that limx→2 x2 = 4 using the ε-δ definition of limit, namely:
    For any x satisfying 0 < |x - 2| < δ it follows that |x2 - 4| < ε ​

    How should we handle the fact that we restricted ourselves to δ ≤ 1 ? Instead of saying δ = ε/5, we simply say that δ = the minimum of { 1, ε/5 } .


    I take that this ε-δ proof stuff is rather new to you and that you're pretty unsure where some of the details here come from. We can go through some of that later, if you're up to it.

    For now, write up a nice proof.
     
  20. Jun 11, 2015 #19

    SammyS

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    This method is different from the method I was suggesting. With this approach, you find the maximum possible δ for any particular ε .

    In my opinion, neither method is better than the other. For some particular case one of them may be easier to implement than the other.
     
  21. Jun 12, 2015 #20

    Fredrik

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    The choice ##\delta=\frac{\varepsilon}{5}## doesn't work when ##\varepsilon## is large. But perhaps you can find a number ##r## such that the choice ##\delta=\min\big\{\frac\varepsilon 5, r\big\}## will work.
     
    Last edited: Jun 12, 2015
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