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Epsilon delta to prove continuity

  1. Aug 18, 2009 #1
    I have an example bit I can't quite follow it....?

    Use epsilon -delta definition of continuity to prove f(x) = 3x^2 - x is continuous at x=2

    Ep > 0 and delta > 0 in terms of Ep

    f(x) -f(2) = 3x^2 - x -(3*2^2 -2)

    f(x) - f(2) = 3x^2 -x - 10
    f(x) - f(2) = (3x + 5)(x - 2)

    So far so good - but now can someone explain what happens please......!!

    James
     
  2. jcsd
  3. Aug 18, 2009 #2

    arildno

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    Dearly Missed

    Well, we may, for example, rewrite 3x+5 as 3(x-2)+11.

    Now, set d=x-2

    We then have:
    |f(x)-f(2)|=|(3d+11)d|<=3|d|^2+11|d|<=14|d|, if |d| is tiny enough, specifically, when |d|<1 (i.e, when x is between 1 and 3)

    Now, can you make |d| so small that given any e, 14|d| will be smaller than e?
     
  4. Aug 18, 2009 #3

    tiny-tim

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    Hi James! :smile:

    (have a delta: δ and an epsilon: ε and try using the X2 tag just above the Reply box :wink:)
    You're probably wondering "wherever does δ come into it?? :confused:"

    You needed to start with "f(2 + δ) -f(2) = …" :wink:
     
  5. Aug 18, 2009 #4
    Thanks for the replies lads - I appreciate it! What I need is an explanantion of the whole epsilon delta thing really from start to finish.

    I don't understand it to be honest and I need to so I can apply it to other functions etc

    Many thanks

    James
     
  6. Aug 18, 2009 #5

    jgens

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    Last edited by a moderator: Apr 24, 2017
  7. Aug 18, 2009 #6

    arildno

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    Start with the first thing you don't understand.
     
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