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Intermediate Value Property

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  1. Jun 27, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-6-27_9-0-49.png '
    Here is the given problem

    2. Relevant equations


    3. The attempt at a solution

    a. For part a, I felt it was not continuous because of the sin(1/x) as it gets closer to 0, the graph switches between 1 and -1. Then I felt it might be continuous, therefore I am not sure.

    b. For part b, I felt it has the Intermediate Value Property (IVP), because I can do something with the IVT. Those were my thoughts and ideas.
     
    Last edited by a moderator: Jun 27, 2016
  2. jcsd
  3. Jun 27, 2016 #2

    BiGyElLoWhAt

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    Gold Member

    Can you show that the limit as approached positively is different from that as approached negatively? Or that it's different than f(0)? So in other words,
    ##\lim_{x \to 0} f(x) \neq f(0)##?

    For b, I'm not really sure. I thought one of the requirements was that f(x) was continuous...
    **
    Perhaps this will be of some use.
    http://math.stackexchange.com/quest...te-value-property-and-discontinuous-functions
    Looks like you need to look at the derivatives near zero.
     
  4. Jun 28, 2016 #3
    For part a you would have to do what BiGyElLoWhAt suggested. For b I believe you would have to prove that the function in either monotone increasing or decreasing. IVP says that for any x value between two other x values, the y value will be in between the y values for the other two x values.
     
    Last edited: Jun 28, 2016
  5. Jun 28, 2016 #4
    Screenshot_2016-06-28-13-20-07.png
    IVP theorem
     
    Last edited: Jun 28, 2016
  6. Jun 28, 2016 #5
  7. Jul 2, 2016 #6

    pasmith

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    Homework Helper

    If there exist sequences [itex]x_n[/itex] and [itex]y_n[/itex] such that [itex]\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n = 0[/itex] but [itex]\lim_{n \to \infty} f(x_n) \neq \lim_{n \to \infty} f(y_n)[/itex] then [itex]f[/itex] is not continuous at zero.
     
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