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Ε-δ definition of a limit

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    use the elipson-delta of a limit to justify the answer of: show proof
    lim x->0 1/(x+1)

    2. Relevant equations

    lim x->0 1/(x+1)=1

    3. The attempt at a solution


    lim x->0 1/(x+1)

    I did some work but after that I don't know how to keep going.

    if 0<|x-c|<δ => |f(x)-<ε|, then lim x->c f(x)=L
    0<|x-0|<δ=> |1/(x+1)-1|
    scratchwork: |1/(x+1)-1|<ε
    |x/x+1|<ε
    |x|/|x+1|<ε
    |x|<|x+1|ε
    restrict |x|<1/2
    -1/2<x<1/2=>1/2<x+1<3/2=>1/2 <|x+1|
    |x|<|x+1|ε
    |x|<1/2 ε
    δ=min{1/2,1/2ε}
    how do I find the proof ??
     
  2. jcsd
  3. Oct 1, 2011 #2

    SammyS

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    For the most part, reverse the steps in the scratch-work part.

    Of course, start with
    Let ε > 0​
    Then define: δ=min{1/2,1/2ε}

    δ ≤ 1/2 should lead back to 1/2 <|x+1| if 0 < |x-0| < δ . What does that say about 1/|x+1|

    δ ≤ ε/2 and 0 < |x-0| < δ give |x| < ε/2
    This along with your conclusion about 1/|x+1|, should pretty much do the trick. ​




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