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Limits and Sequences

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    The function f is defined on a neighborhood N of [itex]\bar{x}[/itex]. Show that
    [itex] \lim_{x \rightarrow \bar{x}} f(x) = L [/itex] if and only if [itex] \lim_{n \rightarrow \infty} f(x_n) = L [/itex] when [itex] \{x-n\}[/itex] is a sequence of points in N with [itex]\lim_{n \rightarrow \infty} x_n = \bar{x} [/itex].


    2. Relevant equations



    3. The attempt at a solution
    I think I have the necessity, but I am having trouble tightening up the other direction. What I have is that if [itex]\lim_n \rightarrow \infty x_n = \bar{x} [/itex] then [itex]\forall \delta > 0, \exists N_1 : |x_n - \bar{x}| < \delta \text{ when } n > N_1[/itex] and that [itex]\forall \epsilon > 0, \exists N_2 : |f(x_n) - L| < \epsilon \text{ when } n > N_2 [/itex].
    Let [itex] N = \max\{N_1,N_2\} [/itex] there is an N so that if [itex]|x_n - \bar{x}| < \delta [/itex] then [itex]|f(x_n) - L| < \epsilon[/itex] for any [itex]\epsilon < 0[/itex]. Which is close to the definition of the limit of f, but how can I generalize it to [itex]|f(x) - L| < \epsilon[/itex] rather than [itex]|f(x_n) - L| < \epsilon[/itex] or is that not an issue?
     
  2. jcsd
  3. Sep 21, 2013 #2

    Office_Shredder

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    Try a proof by contradiction. Suppose that we have for every sequence
    [tex] \lim_{n\to \infty} f(x_n) = L [/tex],
    but that
    [tex] \lim_{x\to \overline{x}} f(x) \neq L [/tex].

    Can you use these two pieces of information to form a contradiction?
     
  4. Sep 21, 2013 #3

    vela

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    Is this really true? What if the sequence were given by ##x_n = \bar{x}## and f is defined such that f(x)=L if ##x=\bar{x}## and ##f(x)=L-1## for ##x\ne\bar{x}##? Clearly, the limit of the sequence f(xn) is L, but the limit of f(x) is L-1. Did you leave out a condition on f, like f being continuous?
     
  5. Sep 22, 2013 #4
    Vela, I checked again and the problem doesn't state that f is continuous, only that it is defined on a deleted neighborhood of [itex]\bar{x}[/itex]. Would the limit of the sequence f(xn) really be L in this case though? Intuitively it seems like it would still be L-1 because the x's are arbitrarily close to [itex]\bar{x}[/itex], but not equal to it, but maybe I'm not seeing it correctly.
     
  6. Sep 22, 2013 #5

    Office_Shredder

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    vela, your function has a limit of L-1 both for the sequences and for the arbitrary [itex] x \to \overline{x}[/tex].

    In fact the point of the question really is that it gives you a way of determining that a function is discontinuous. If you know f(x) = L, all you need to do is find a single sequence converging to x such that f(xn) is not equal to L in order to show that f is discontinuous. This will be the most common use of this result that you come across.
     
  7. Sep 22, 2013 #6

    vela

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    How can the limit be L-1 if ##f(x_n) = f(\bar{x}) = L## for all n?

    Isn't that what I just did with the sequence I suggested? And how are you supposed to use the result to show that f is discontinuous if it supposedly holds for all functions? The OP wasn't asked to prove the statement holds for only continuous functions but for seemingly all functions.
     
  8. Sep 22, 2013 #7
    The OP likely has some definition of what the expression [itex]\lim_{x\rightarrow\overline x}f(x)=L[/itex] means. This question is merely asking them to show that whatever definition they have is equivalent to having [itex]f(x_n)[/itex] converge to [itex]L[/itex] for any sequence [itex]x_n\rightarrow\overline x[/itex].

    We (those of us posting) know that this is equivalent to [itex]f[/itex] being continuous at [itex]\overline x[/itex], but the OP might not have seen this yet.

    To answer the OP's original question:
    Take Office_Shredder's advice. There is a fairly straightforward proof by contradiction you could try here.
     
  9. Sep 22, 2013 #8

    pasmith

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    The crucial point is this:

    It would have been helpful if the OP had said this initially, but now we know. It follows that your counterexample is not a counterexample, because none of the relevant sequences attain the value [itex]\bar x[/itex], and [itex]f(\bar x)[/itex] is itself undefined.
     
  10. Sep 22, 2013 #9

    vela

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    Ah, that's what a deleted neighborhood is. I never heard that term before. I thought it was a typo on the OP's part.
     
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