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Epsilon-delta definition

  1. Oct 15, 2013 #1
    please i am new to math. I donot know exact meanings of epsilon-delta definition. i dont comprehend it. Would Anybody help me. thanks in advance
     
  2. jcsd
  3. Oct 15, 2013 #2

    vanhees71

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    What particularly are you not understanding? Do you mean the [itex]\epsilon[/itex]-[itex]\delta[/itex] definition of continuity of a function at a point?

    In this case it reads:

    A function [itex]f:D \rightarrow \mathbb{R}[/itex] is continuous at [itex]x_0 \in D[/itex] iff for any [itex]\epsilon>0[/itex] there exists a [itex]\delta>0[/itex] such that for all [itex]x \in D[/itex] with [itex]|x-x_0|<\delta[/itex]
    [tex]|f(x)-f(x_0)|<\epsilon.[/tex]
    This just says in a formal way that the graph of the function doesn't jump at [itex](x_0,f(x_0))[/itex].
     
  4. Oct 15, 2013 #3

    pwsnafu

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    ##0<|x-x_0|<\delta## actually.
     
  5. Oct 15, 2013 #4

    WannabeNewton

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  6. Oct 15, 2013 #5

    jbunniii

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    There's no reason to exclude the point ##x = x_0##. We trivially have ##|f(x_0) - f(x_0)| < \epsilon## for any ##\epsilon##.
     
  7. Oct 15, 2013 #6

    arildno

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    It sure is a reason.
    Otherwise, discontinuous functions would be deprived of limit values at the point of discontinuity. Thus, the limit concept would be conflated with the continuity concept.
    Think about it!
    :smile:
     
  8. Oct 15, 2013 #7

    jbunniii

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    But vanhees71 was giving the definition of continuity at ##x_0##, not the definition of the existence of a limit at ##x_0##. If the function is to be continuous at ##x_0##, then it must be defined at ##x_0## and have the correct value!
     
  9. Oct 15, 2013 #8

    arildno

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    Hmm..no read it again.
    What he posted was the definition in terms of the LIMIT concept in which L=f(x_0).
    In particular, he writes d>0
     
  10. Oct 15, 2013 #9

    jbunniii

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    One more remark and then I'll shut up. :tongue: Here's a discussion from Spivak's Calculus (chapter 6, before Theorem 2) which may clarify:

     
  11. Oct 15, 2013 #10

    arildno

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    And I do not see why we should make a special case for the continuity criterion relative to the general criterion for a limit?

    Why change a perfectly good criterion for continuity?
     
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