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Prove the limit is true using the epsilon,delta definition of a limit.

  1. Aug 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove the following:
    lim x2 - x = 0
    x→1


    2. Relevant equations

    If 0<|x-a|<δ then |f(x)-L|<ε

    3. The attempt at a solution
    Part I - Set up
    0<|x-a|<δ |f(x)-L|<ε
    If 0<|x-1|<δ then |(x^2 - x) - 0|<ε
    x|x-1|<ε
    x|x-1|<ε
    |x-1|<ε/x
    δ=ε/x

    Part II - Proof
    Given ε < 0, choose δ=ε/x
    If 0<|x-1|<δ then
    |(x^2 - x) - 0| = |x^2 - x| = x|x - 1| < xδ = x(ε/x) = ε
     
  2. jcsd
  3. Aug 1, 2013 #2
    No! Delta may not be dependent on x. Also your definition is incorrect, it is for every ε>0 there is a ... etc.

    Try this:

    |x||x-1| <δ|x| = δ |x-1+1|<δ^2+δ < δ if δ < 1,..., etc.
     
  4. Aug 1, 2013 #3
    I'm afraid you can't set ##δ=ε/x##. It needs to be a particular number not an arbitrary value. Try setting up ##|x||x-1|## and then apply triangle inequality to find the upper bound ##f(x)## can take. It can be assured that |x-1| will at least be smaller than 1.
     
  5. Aug 1, 2013 #4
    I'm not quite sure why you added 1 to |x - 1|
    |x||x-1| < δ|x| = δ|x-1+1| < δ^2 + δ < δ
    |x^2 - x| < δ|x-1+1| < δ^2 + δ < δ <== I'm not getting this part
     
  6. Aug 1, 2013 #5
    How can it be assured that |x - 1| will be smaller than 1? In a formal proof, how would I show this?

    I found the Triangle Inequality to be |A + B| ≤ |A| + |B|

    Set up
    0<|x - a|<δ |f(x) - L|<ε
    If 0<|x - 1|<δ then |(x^2 - x) - 0|<ε
    x|x-1|<ε
     
    Last edited: Aug 1, 2013
  7. Aug 1, 2013 #6

    Zondrina

    User Avatar
    Homework Helper

    So you have this :

    ##|x^2 - x| = |x||x-1| < δ|x|##

    Now this is quite a common trick and you should use it often. It allows you to use the triangle inequality on your second factor :

    ##|x| = |x-1+1| ≤ |x-1| + 1 < δ + 1##

    Therefore :

    ##|x^2 - x| = |x||x-1| < δ|x| < δ(δ+1)##

    Now at this point, it's very useful to place a bound on ##δ##, lets say ##δ ≤ 1## to make life simple. So we get :

    ##δ(δ+1) ≤ 2δ##

    Now you can find ##δ## in terms of ##ε##.

    What is your final value(s) for ##δ##?
     
  8. Aug 1, 2013 #7
    Here is what I have now.
    ##|(x^2 - x) - 0| = |x - 1| |x|##
    ##= |x - 1| |(x - 1) + 1| ≤ |x - 1| (|x - 1| + |1|)##
    ##= |x - 1|^2 + |x - 1| < |x - 1| + |x - 1| = 2|x - 1|##
    assuming
    ##|x - 1| < 1##
    Given
    ##ε > 0, δ = min{1, ε/2}##
    Then
    ##0 < |x - 1| < δ##
    ##|(x^2 - x) - 0| < 2|x - 1| < 2(ε/2) = ε##
     
    Last edited: Aug 1, 2013
  9. Aug 2, 2013 #8
    It looks fine to me. ##δ=min(1,\frac{ε}{2})##
     
  10. Aug 2, 2013 #9

    Zondrina

    User Avatar
    Homework Helper

    I'm not so sure I agree with the third line :

    ##= |x - 1|^2 + |x - 1| < |x - 1| + |x - 1| = 2|x - 1|##

    How were you able to deduce ##|x-1|^2 < |x-1|##. This is certainly not true in general. Your value for ##δ## is indeed correct though.
     
  11. Aug 2, 2013 #10
    Yes it is true if delta<1.
     
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