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Strange limit question

  1. Jun 20, 2005 #1
    Let [itex]f: \mathbb{R} \rightarrow \mathbb{R}[/itex] be given. Let [itex]L[/itex] be a real number. State the condition for saying that as [itex]x[/itex] tends to [itex]a[/itex], the limit of [itex]f(x)[/itex] is not [itex]L[/itex]. The statement ought to begin with "Given there exists [itex]\epsilon > 0[/itex]".

    Best guess: [itex]\lim_{x \rightarrow a}f(x) \neq L[/itex] means, given there exists [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that [itex]|x-a|<\delta \Rightarrow |f(x) - L| > \epsilon[/itex]. I'm really not sure about this, however.
  2. jcsd
  3. Jun 20, 2005 #2
    Nonremovable singularity?
  4. Jun 20, 2005 #3
    Pardon me?
  5. Jun 20, 2005 #4


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    First, lets go over the rule for the limit:

    Given [itex]\epsilon>0[/itex], there exists a [itex]\delta>0[/itex] such that for all x satisfying:


    we have:


    Now, what happens if L is not the limit?

    Then there should exists an [itex]\epsilon>0[/itex] such that for some [itex]\delta>0[/itex] and all x satisfying:


    we have:


    Make sure you understand these two differences. If L is the limit, by choosing [itex]\delta [/itex] small enough I can make the difference between L and f as small as I want.

    If L is not the limit, then no matter how small I make [itex]\delta[/itex], I can always find an [itex]\epsilon[/itex] such that the difference between f and L will be larger.
  6. Jun 20, 2005 #5
    So I was correct? Great, thanks for your help saltydog.
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