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Find a δ δ such that | f ( x ) − L | < ϵ |f(x)−L|<ϵ for all x satisfying 0 < | x − a | < δ 0<|x−a

  1. Aug 18, 2016 #1
    I've pondered at this problem for a long time, and I don't know where I make a mistake, can anyone give me a hint?

    1. The problem statement, all variables and given/known data

    find a ##\delta## such that ##|f(x)-L|<\epsilon## for all x satisfying ##0<|x-a|<\delta##

    2. Relevant equations

    ##f(x)=x^4##; ##l=a^4##


    3. The attempt at a solution

    writing as the definition of limits

    if ##|x-a|<\delta ##, then ##|x^4-a^4|<\epsilon##

    factor the right side of equation I got

    ##|(x^2-a^2)(x^2+a^2)|<\epsilon##, then I can have ##|x^2-a^2|<\delta_1##

    chose ##\delta_1=min(1)##, and I got ##|x^2|-|a^2|\le|x^2-a^2|<1##, adding |a^2| to both side of the inequality I got ##|x^2|<1+|a^2|##, therefore ##|x^2|+|a^2|<1+2|a^2|## , and ##|x^2+a^2|\le|x^2|+|a^2|<2|a^2|+1##

    So I choose ##\delta_1=min (1, \frac{\epsilon}{2|a^2|+1})##

    Therefore I have, if ##|x-a|<\delta## , then ##|x^2-a^2|<min (1, \frac{\epsilon}{2|a^2|+1})##

    from my previous experience with ##\lim_{x \to a} x^2## , I know that I can pick ##\delta=min(1, \frac{\epsilon}{2|a|+1})##, under these conditions

    if ##|x^2-a^2|<\epsilon##, then ##|x-a|<\delta##

    so if I let ##min(1, \frac{\epsilon}{2|a^2|+1})## play the role of ##\epsilon##,

    and I got ##\delta=min(1, \frac{min(1,\frac{\epsilon}{2|a^2|+1})}{2|a|+1})##

    when I check the solution, it seems like I got the right idea but the solution assume ##\delta=min(1, \frac{min(1, \frac{\epsilon}{2(|a^2|+1)})}{2(|a|+1)})##

    for some reason, I suppose I miss out a factor of two somewhere, but I check it again and again and could not find where I make a mistake, can somebody give me a clue?
     
  2. jcsd
  3. Aug 18, 2016 #2

    Mark44

    Staff: Mentor

    Instead of factoring ##x^4 - a^4 = (x^2 - a^2)(x^2 + a^4)##, look at the right side of this equation as ##(x - a)(x + a)(x^2 + a^2) = (x - a)(x^3 + ax^2 + a^2x + a^3)##
     
  4. Aug 18, 2016 #3
    Thank you for your reply Mark

    Following your advice I got this,

    Second Attempt at the Solution,

    writing givens as the definition of limits

    if ##|x-a|<\delta## then ##|x^4-a^4|<\epsilon##

    factoring out the right side of the condition

    ##|(x-1)(x^3+a^2x+ax^2+a^3)|<\epsilon##

    choose ##\delta=min(1)## then I got, ##|x-a|<1##, using triangle inequality I got ##|x|-|a|\le|x-a|<1##, set ##|x|-|a|<1##, and I got ##|x|<1+|a|##

    rewrite the expression ##|(x^3+a^2x+ax^2+a^3)|\le|x^3|+|a^2|\times|x|+|a|\times|x^2|+|a^3|##

    Now, I sub ##|x|## in, and the final result is ##|(1+|a|)|^3+|a^2|\times|(1+|a|)|+|a|\times|(1+|a|)|^2+|a^3||##

    now, I just choose that ##\delta=min(1, \frac{\epsilon}{|(1+|a|)|^3+|a^2|\times|(1+|a|)|+|a|\times|(1+|a|)|^2+|a^3|})##

    When I check the result, it matched up with the solution the book has given out in this second method, I'm just curious on why I got a different answer in first method that I choose to find ##\delta##, it seems that I got the right methodology in the first one but just different answer and I don't know where I make my mistake, I don't know where the book got ##2(|a|+1)## and ##2(|a^2|+1)## instead of ##2|a|+1## and ##2|a^2|+1##, I check again and again to see where I miss the 1 but I just can't find it, Thank you for your time Mark
     
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