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Diffrentiating Piecewise Function

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex] L\in R [/tex] and define

    [tex]
    h(x) = \begin{cases} sin(1/x) & \text{ if } x \neq 0 \\ L & \text{ if } x = 0 \end{cases}
    [/tex]

    Prove that h is not continuous at 0

    2. Relevant equations

    My Professor gave us the hint "Prove by contradiction, taking [tex] \epsilon = 1/2 [/tex]"


    3. The attempt at a solution

    I didn't see how this hint was relevant, but it seems too easy. All I did was say since a function is continuous at a point if it is differentiable at that point, then take 0. then h(x) = L and by first principles

    lim h->0 f(x +h) - f(x) / h = L/0 which does not exist, and therefore the function is not continuous.

    Why would he give that hint, and the piecewise function, when it could be proven so simply? Thanks in advance.
     
  2. jcsd
  3. Oct 27, 2009 #2

    lurflurf

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    Homework Helper

    You are correct that differentiability implies continuity, but the converse is false. Lack of differentiability does not imply lack of continuity, consider |x| which is continuous when x=0 dispite failing to be differentiable. For continuity consider the family of punctured disks (-n pi,n pi)\0 where n is a positive integer.
     
    Last edited: Oct 27, 2009
  4. Oct 28, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Note that sin(1/x)= 1 for [itex]x= 1/(2n\pi)[/itex] for any integer n and that sin(1/x)= -1 for [itex]x= 1/((2n+1)\pi)[/itex] for any integer n.
     
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