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Homework Help: Help! Espilon-Delta limits

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data
    I wanna ask a question about Epsilon-Delta definition...
    I have already read a tons of definitions about Epsilon-Delta limit proof. But i am still stucking in some places...
    E.g.) Prove it by using Espilon-Delta method, lim (x->4) x^3=64



    2. Relevant equations

    |x-4|<delta delta>0
    |x^3|<espilon espilon>0

    3. The attempt at a solution

    I am kinda know what I should do, I changed |x^3-64| into |(x-4)|(x^2+4x+16)| and my objective afterward is to ensure| (x^2+4x+16)| is "small".
    But i just dunno what I should do afterwards... Should I just Sub x=4 and then find out |(x^2+4x+16)|<48?

    Plx help me :( ThX!! :)
     
  2. jcsd
  3. Sep 24, 2010 #2
    The point is not in showing that | (x^2+4x+16)| is "small, but that it is "bounded" when x is close to 4. So, restrict yourself to the region, for instance, 3<x<5. What can be said about | (x^2+4x+16)| in this region? Is it always smaller than some fixed number?
     
  4. Sep 24, 2010 #3
  5. Sep 24, 2010 #4
    arkajad,
    Alright I got it. So u mean is to find out the max. number when x is "close" to 4? Afterwards we can state that |x-4|<1 (Because x is "close" to 4) and we can find the max. number of x is 5. we subsitute 5 into the equation and we have (5)^2+(4)(5)+16=61. And it would be espilon/61 afterwards?

    If its espilon/61, we write this "delta=min{1,e/61}"? What does "delta=min{1,e/61}" actually means? ThankS

    madah12,
    The link is extremely useful for me to solve this problem. I found out the "solution"
    (Which i dunno its right or not...) right there. Many thanks.
     
  6. Sep 24, 2010 #5
    min{1,e/61} means eiher 1 or e/61, whichever happens to be smaller.
     
  7. Sep 24, 2010 #6
    So what do u mean is that if i pick e=62, the output will be 1 and if i pick e=60, the output will be 60/61? min{1,e/61}
     
  8. Sep 24, 2010 #7
    Yes, that's it. But are interested in "however small is epsilon" part, and for epsilon small enough (e<61) you will never use 1 as an output.
     
  9. Sep 24, 2010 #8
    Ohhhh ya I got it! Thx for the explanation! It helped me muCH!
     
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