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Confused about continuity and limits

  1. Jun 20, 2012 #1
    Hi guys,

    I just started reading an introductory book on analysis. I'm up to the part where they talk about functions now, and I'm getting lost.

    The theorem that I'm having trouble envisioning is: Let f: D-> R and let c be an accumulation point of D. Then limx->cf(x)=L iff for each neighborhood V of L there exists a deleted neighborhood U of c such that f(U[itex]\bigcap[/itex]D) is contained in V.

    Why is it N*(c) rather than just N(c)? There's a picture in the book of the deleted point corresponding to L and...I think it's just confusing me more. First theorem in the book that I couldn't wrap my head around. :(
     
  2. jcsd
  3. Jun 20, 2012 #2

    pwsnafu

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    [STRIKE]Because [itex]c \notin D[/itex] in general. A set is closed iff it contains all of its accumulation points.[/STRIKE]

    Edit: Scratch that. Even if [itex]c \notin D[/itex], then [itex]c \notin U \cap D[/itex] and so it wouldn't be problem.

    What I should have said is that f may not be continuous at c.
    For example [itex]D = (-1,1)[/itex] and f(0) = 1 but f(0) = 0 elsewhere. We need a definition so that the limit is still 0 at x = 0, even though f(0) is not equal to 0.
    If we were using N(c), then when V = (-1/2,1/2), then [itex]f(0) \notin V[/itex].
     
    Last edited: Jun 21, 2012
  4. Jun 20, 2012 #3
    It seems straight forward that if the limit is L, then there is a neighborhood by definition, and if you delete a point, then it is still a neighborhood.
    (Though we may be assuming we're in a certain type of space, I'm not sure which development is in your book, so you may want to check that you've told us everything we need to know.)

    So my advice, try to look for the unusual part in the other direction that shows the L is the limit, or supposes it doesn't. If you pull apart that proof, you may gain a foothold in developing the intuition.
     
  5. Jun 21, 2012 #4

    HallsofIvy

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    The basic idea of "limit" is that if [itex]\lim_{x\to a} f(x)= L[/itex] then if x is very, very close to a, the f(x) is very, very close to L. But we are NOT concerned with what happens at x= a. (One important reason for that is that we want to use the limit to find derivatives which limits of "rate of change" calculations, [itex]\Delta y/\Delta x[/itex] which does not exist at the target point.)

    Let f(x)= 2x if x is NOT equal to 1 and f(1)= 3. If x is very very close to 1 but not equal to 1, f(x) will be very very close to 2. That is the result we want, not the "3" which just happens to be the value of f(1). To avoid that, we "delete" x= 1 from the neighborhood.
     
  6. Jun 21, 2012 #5
    Thanks guys! I get it now. You guys are spot on. I was thinking about the point at c and continuity instead of just focusing on the limit. Ha. Now I'm embarrassed. :redface:
     
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