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One Sided Limit Theorom

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the limit as x->c of f(x) = L if and only if both one sided limits also = L


    2. Relevant equations
    Has to be an epsilon delta proof

    3. The attempt at a solution
    Being an if and only if, I have to do two cases : If A, then B. and if NOT A, then NOT B, logically.

    Case 1:
    Let lim x->c from the left be L, and lim x->c from the right be L.

    then if [tex] c - \delta < x < c then |f(x) - L| < \epsilon [/tex]

    and if [tex] c < x < c + \delta then |f(x) - L| < \epsilon [/tex]

    Case 2:

    Let lim x->c from the left = M, and lim x->c from the right = N.

    This is all I have really rationalized I am kind of stumped how to do a rigorous proof of this, I.e. I know how to do specific proofs but not a rigorous general proof. \

    Can anyone offer any help / a starting point =/ ?
     
  2. jcsd
  3. Oct 6, 2009 #2
    Okay, we have two things to prove: 1) if f(x) goes to L as x goes to c, then both one sided limits also approach L. 2) If both one sided limits go to L for x approaching c, then f(x) goes to L as x goes to c.

    For 1) we know that for every positive epsilon, there exists a positive delta such that [tex] |f(x) - L| < \epsilon [/tex] if [tex] |x - c| < \delta [/tex]. We can rewrite [tex] |x-c| < \delta [/tex] as [tex] c - \delta < x < c + \delta [/tex]. Now, say we're examining the right handed limit of f(x) as x->c, then we only look at the interval [tex] x < c + \delta [/tex], right? And with the left handed limits we only look at the interval [tex] c - \delta < x [/tex]. You should be able to prove it now.

    2) Look at the delta-interval of both one-sided limits and "put them together".
     
  4. Oct 24, 2009 #3
    Is it possible to continue the solution? I can find the (1).
     
  5. Oct 26, 2009 #4
    For 2), you're going to assume that both one-sided limits exist. So for every positive epsilon, there is a positive delta such that [tex] c - x < \delta \Rightarrow |f(x) - L| < \delta [/tex] and [tex] x - c < \delta \Rightarrow |f(x) - L| [/tex].

    Put the two delta inequalities together to get [tex] c - \delta < x < c + \delta \Rightarrow |c - x| < \delta \left [/tex], which we know implies [tex] |f(x) - L| < \epsilon [/tex].
     
  6. Oct 26, 2009 #5
    thnxxx mate.. I found it after a while... have a nice day...
     
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