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Real Analysis Limits

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that if f: R->R is an even function, then lim x->0 f(x)=L if and only if lim x->0+ f(x)=L.


    2. Relevant equations



    3. The attempt at a solution

    So far I have:

    If f is an even function f(x)=f(-x) for x in domain of f.

    Then I am trying to apply the limit definitions, but am unsure of how to write the proof from here.
     
  2. jcsd
  3. Oct 24, 2012 #2

    Dick

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    Write down the definition of lim x->0+ f(x)=L. Now change x to -x. Doesn't it look like the definition of lim x->0- f(x)=L once you use that f is even?
     
  4. Oct 24, 2012 #3
    So lim x->0+ f(x)=L implies there exists a real number L s.t. epsilon>0 there exists delta>0 s.t. lf(x)-Ll<epsilon provided 0<x-a<delta.

    Then lim x->0+ f(-x)=L implies that there exists a real number L s.t. epsilon>0 there exists delta>0 s.t. lf(-x)-Ll<epsilon provided 0<x-a<delta.

    I have the definitions, but I don't understand the last part of your comment, can you clarify?
     
  5. Oct 24, 2012 #4

    Dick

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    'a' in your problem is 0. 0<x<delta, is the same as -delta<-x<0. What does the definition of lim x->0- f(x)=L look like?
     
  6. Oct 24, 2012 #5
    lim x->0- f(x)=L implies that there exists a real number L s.t. epsilon>0 there exists delta>0 s.t. lf(x)-Ll<epsilon provided 0<a-x<delta.

    I'm am getting confused with all these definitions though, can you help me organize the argument using the definitions?
     
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