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Prove a limit exists using formal definition

  1. Oct 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate the value of the limit and justify your answer with the ε-δ definition of the limit.
    lim (x->1) x2

    2. Relevant equations
    My professor gave us the hint that we have to take δ as 0<δ≤ k0 so that δ(ε)=min{k0,ε/ (k0+2)}

    I'm guessing that k0 is meant to be any number though it's usually 1?

    3. The attempt at a solution
    I'm trying to relate |f(x)-1|<ε to |x-a|<δ
    |x2-1| < ε
    -ε < x2-1 <ε
    -ε+1 < x2 <ε+1
    (-ε+1)1/2-1 < x-1 < (ε+1)1/2-1

    I'm not sure what to do from here. I'm guessing that
    δ=(-ε+1)1/2-1 or (ε+1)1/2-1
    but I have no clue where to go from there.

    I've tried doing this another way as well to make use of a k0 from my professor's hint

    |x2-1| < ε
    -ε < x2-1 <ε
    -ε+1 < x2 <ε+1
    -ε+1 < x x < ε+1 where if we assume that |x|<k0=1, we can then assume that -ε+1 < k0x < ε+1
    (-ε+1)/k0 - 1 < x-1< (ε+1)/k0 - 1
    This still doesn't give me what my professor hinted at though. (I don't know what to do after that.)
    Also, what does the min{_____,_______} mean?
     
  2. jcsd
  3. Oct 1, 2014 #2

    Mark44

    Staff: Mentor

    k0 might be 1. It's just some unspecified number.

    min{..., ...} means the minimum of the numbers in the list inside the braces.
     
  4. Oct 2, 2014 #3
    I think I managed to connect the dots together. I shouldn't have gotten rid of the -1 in |x2-1|. Instead, I should have just turned it into (x-1)(x+1) and move on from there.

    Assuming that x+1<1 and δ<1,
    -1<x-1<1 so if we add +2 to both sides, we get 1<x+1<3
    |x+1||x-1|<3|x-1|<ε
    |x-1|<ε/3

    But if I assume that δ<1, then δ=min{1,ε/3} won't make sense?
     
  5. Oct 2, 2014 #4

    Mark44

    Staff: Mentor

    Just assume that δ < 1. That makes |x - 1| < 1, which gives you bounds on |x + 1|, as you show below.
    Now, choose δ = min{1, ε/3}. Typically, someone (else) would pick a small value for ε, so δ will typically be much smaller than 1.
     
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