Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook