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Limit proof by contradiction

  1. Jan 3, 2012 #1
    Prove: If f approaches l near a and f approaches m near a, then l = m.

    ....Im skipping to the end of the proof...

    " to comlete the proof a particular ε>0 has to be choses for which the two conditions
    |f(x) - l|< ε and |f(x) - m|< ε cannot both hold if l=/=m."

    if l=/=m so that |l - m|> 0 , we can chose ε to be |l - m|/2... How did they decide this |l - m|/2??

    It follows that there is a δ>0 such that for all x,

    If 0<|x-a|<δ, then |f(x) - l|< |l - m|/2
    and |f(x) - m|< |l - m|/2.


    |l - m| = | l - f(x) + f(x) - m|≤ |l - f(x)| + | f(x) - m| < |l - m|/2 + |l - m|/2 = |l - m|, a contradiction.

    can some one explain to me the contradiction?
     
  2. jcsd
  3. Jan 3, 2012 #2
    You're using the definition of a continuous function, which requires that a sufficiently small change (δ) in x will produce less than a small change (ε) in f(x), and this can be achieved no matter how small ε is.

    The contradiction exhibited is |l - m| < |l - m| which is achieved by making ε small enough that even two function evaluations cannot bridge the gap between two limit values for the function.
     
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