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Epsilon Delta Proof

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data

    lim 3 as x->6
    lim -1 as x->2

    2. Relevant equations

    In the first weeks of a calculus class and doing these epsilon delta proofs.

    As i am looking at two of the problems i have been assigned:

    Lim 3 as x->6
    Lim -1 as x->2



    3. The attempt at a solution
    considering both are horzontal lines, if i give some ε then there is no intersection with the function and thus no δ. ε must be > 0 so not sure how our proof statement will work:

    Given ε>0, choose δ=ε. If 0<0<δ, then |0|<δ, then |3-3| = |0| < δ=ε thus |3-3|<ε whenever 0<|0|<δ. Therefore, by the definition of a limit, lim 3 = 3 as x->6
     
  2. jcsd
  3. Sep 5, 2012 #2

    SammyS

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    Hello freezer. Welcome to PF !

    One comment first: For a constant function, you can pick anything for δ, often just choose δ = 1.

    Now for your proof. What is the general ε - δ formulation for limx → a f(x) = L ?

    For any ε > 0, there exists a δ such that for any x for which 0 < |x - a| < δ, then |f(x) - L| < ε .

    You don't have any x the following statement of yours.
    ... If 0<0<δ, then |0|<δ, then |3-3| = |0| < δ ...​
     
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