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Understanding a problem that uses episolon delta defintion

  1. Apr 10, 2014 #1
    let E = episolon and D = delta;

    the problem is as follows:

    let f(x) = (2x^2 - 3x + 3). prove that lim as x approaches 3 f(x) = 21,


    we write |f(x) - 21| = |x^2 + 2x - 15| = |x + 5||x - 3|


    to make this small, we need a bound on the size of |x + 5| when x is close to 3. For example,
    if we arbitarily require that |x - 3| < 1 then


    |x + 5| = |x - 3 + 8| <= |x - 3| + |8| < 1 + 8 = 9

    to make E f(x) within E of 21, we want to have |x + 5| < 9 and |x - 3| < E/9

    I don't understand how did he get E/9 |x - 3| < E/9 ????
     
  2. jcsd
  3. Apr 10, 2014 #2

    jbunniii

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    This is clearly false. Note that ##f(3) = 12##, not ##21##. And ##f## is continuous, so ##\lim_{x \rightarrow 3}f(x) = f(3)##. Is there a typo?
     
  4. Apr 10, 2014 #3

    jbunniii

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    Also, how did you get this:
    from ##f(x) = 2x^2 - 3x + 3##? Are you sure you are not mixing up two different problems?
     
  5. Apr 10, 2014 #4
    it should be 2x^2 + 2x + 6 your right ! but the same logic holds for the problem that I typed I dunno how did he get |x - 3| < E/9...
     
    Last edited: Apr 10, 2014
  6. Apr 10, 2014 #5
    srry I had a type your right it should be 2x^2 + 2x + 6 !
     
  7. Apr 10, 2014 #6

    jbunniii

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    That still doesn't have a limit of ##21## as ##x \rightarrow 3##. It's hard to help if you don't write down the correct problem!

    From this line:
    I am going to assume that you meant ##f(x) = x^2 + 2x + 6##, which does have the limit ##21## as ##x \rightarrow 3##. So, proceeding from that assumption:

    Clearly it's not going to be a problem to make ##|x-3|## as small as we like as ##x \rightarrow 3##. So as your narrative says, we just need to make sure that ##|x+5|## doesn't grow without bound as we shrink ##|x-3|## to zero.

    I assume you are OK with the logic that shows that if ##|x-3| < 1##, then ##|x+5| < 9##.

    So now our goal is to make ##|x+5||x-3| < \epsilon##. We already know we need ##|x-3|< 1## in order for the bound ##|x+5| < 9## to be valid. If we ALSO had ##|x-3| < \epsilon / 9##, then we could conclude that
    $$|x+5||x-3| < 9 \cdot \frac{\epsilon}{9} = \epsilon$$
    So how should we define ##\delta##?
     
  8. Apr 10, 2014 #7
    but why did u assume |x - 3| < E/9 out of nowhere?
     
  9. Apr 10, 2014 #8

    jbunniii

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    Because I recognized that was the factor I needed in order to get ##|x+5||x-3| < \epsilon##, given that ##|x+5| < 9##.

    So all that needs to be done is to show that the assumption can be achieved. In other words, we need a ##\delta## such that if ##|x-3| < \delta##, then both of the assumptions that we have made are satisfied, namely ##|x-3| < \epsilon/9## and ##|x-3| < 1##. How can I choose ##\delta## to guarantee this?
     
  10. Apr 10, 2014 #9
    oh I see oke so in order to achieve this we make E = min{1,E/9} right?
     
  11. Apr 10, 2014 #10

    jbunniii

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    I assume you mean D = min{1,E/9}. That is correct. :approve:
     
  12. Apr 10, 2014 #11
    can u give me a website that has alot of examples on epsilon delta proof of limits?
     
  13. Apr 11, 2014 #12

    jbunniii

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    Have you looked in the "Mathematics Learning Materials" section?

    https://www.physicsforums.com/forumdisplay.php?f=178 [Broken]
     
    Last edited by a moderator: May 6, 2017
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