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Homework Help: Epsilon delta problem

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    evaluate lim2x^2 as x approaches 3 using formal definition (epsilon and delta) of limit


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 25, 2010 #2

    Mark44

    Staff: Mentor

    What have you attempted?
     
  4. Jan 25, 2010 #3
    I am having a hard time starting it. I'm not sure how to prove an exponential.
     
  5. Jan 25, 2010 #4

    Mark44

    Staff: Mentor

    You could start by showing what you intend to prove in this problem.
     
  6. Jan 25, 2010 #5
    epsilon> 0 and (|x-1|) <delta....delta equals ??
     
  7. Jan 25, 2010 #6

    Mark44

    Staff: Mentor

    delta = what so that what happens?
     
  8. Jan 25, 2010 #7

    Mark44

    Staff: Mentor

    Why |x - 1|? x is not approaching 1 in this limit problem.
     
  9. Jan 25, 2010 #8
    delta= a fraction of epsilon so that it is less than x^2 - 1?
     
  10. Jan 25, 2010 #9
    so it would be |x-3|...
     
  11. Jan 25, 2010 #10

    Mark44

    Staff: Mentor

    x^2 - 1 has nothing to do with this problem.

    What is the function you're dealing with, what value is x approaching, and what is the limit that you are trying to prove?
     
  12. Jan 25, 2010 #11
    evaluate lim2x^2 as x approaches 3 using formal definition (epsilon and delta) of limit
     
  13. Jan 25, 2010 #12

    Mark44

    Staff: Mentor

    And what is the limit that you are trying to prove? IOW
    [tex]\lim_{x \to 3} 2x^2 = ?[/tex]

    You need to know what the limit is before you can prove that this value is the limit using delta and epsilon.

    After you answer my question above, show me the limit definition in the context of your function and its limit. You will need to use delta and epsilon in this definition.
     
  14. Jan 25, 2010 #13
    2x^2 is a continuous function, so this limit is easily computable (to a trivial level)
    Once you know this limit, you may apply the concept of the limit: as f(x) gets closer to this limit, its values should lie somewhere in between (L - epsilon , L+ epsilon). You may find appropriate "deltas" by means of algebra (factor..et c), and you'll probably have to bound your interval (for example, say delta cannot be larger than 1)
     
  15. Jan 25, 2010 #14
    limit is 18.....I am just confused with this whole process.....
     
  16. Jan 25, 2010 #15
    You know that the limit is 18, now you have to find a sufficient interval for your "inputs" x so that any x you pick out of that interval (a - delta , a + delta ) or in this case ( 3 - delta, 3 + delta) will be "close enough" to your limit 18. We only know that the limit is 18 because we know that the function gets closer and closer to 18 as you get closer to 3. So appropriately, find an interval where you CAN see values that are extremely close to 18.

    How can you do this? well, just play around with algebra and inequalities.. You know that if 18 is the limit, then you can find that interval I talked about eariler, so for x in (a - delta, a + delta) , applying the function to such an x should yield |f(x) - 18 | < epsilon (remember that |a - b | just denotes distance and epsilon is just the length of an interval, this is the interval for which values are sufficiently close to 18 in a way you can call it the "limit" ).

    Now you know that |2x^2 - 18 | < epsilon must be true, from this true statement you may try to derive another true statement (algebraically). In layman terms, try to get the absolute value thing on the left look like |x - 3 | , so that you'll get | x - 3 | < something. Now you know that for x in some interval ( 3 - [something] , 3 + [something] ), |2x^2 - 18 | will be less than epsilon. Also, keep in mind that epsilon is a variable, it can be anything.
     
  17. Jan 26, 2010 #16
    I also found it very difficult to understand this epsilon-delta version of limits.

    Not because of me, but because of the explanations I've found always seem to beat around the bush. It's such as imple idea but the way people talk about it really obscures the intuition behind it.

    This is called the Weierstrass definition of a Limit, and it's such a smart way to look at the idea of a limit when you want to approach or inch up to a point that doesn't exist along a function.

    This video, http://www.youtube.com/watch?v=-ejyeII0i5c&feature=youtube_gdata
    and the next one in the playlist after it, part 2, should show you exactly what is going on.

    Also, here is a .pdf with a few examples of how to apply this technique in practice.

    http://docs.google.com/viewer?a=v&q=cache:_OYvmsulbDIJ:www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf+epsilon-delta+limit+example&hl=en&sig=AHIEtbQijZifL9dG46lTjmQMCpKpcrrY1g [Broken]

    If you study these two sources fully you should have no problem with ε - δ limits for a while, maybe real analysis is different, idk yet :tongue2:
     
    Last edited by a moderator: May 4, 2017
  18. Jan 26, 2010 #17
    Thanks!! youtube was a GREAT help!!!
     
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