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A function satisfying lim h->0[f(x+h-f(x-h)] = 0, and limit at 0 dne

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Give an example or show why it is impossible
    A function f: R→R satisfying lim h->0[f(x+h-f(x-h)] = 0, and limit at 0 dne
    2. Relevant equations

    n/a

    3. The attempt at a solution

    Been trying different piecewise defined functions, with none of them working. I do not really know where to go. I can sit here for hours trying different functions, but i decided to see if anyone on here could help me out.
     
  2. jcsd
  3. Mar 15, 2013 #2

    jbunniii

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    [edit] oops, misread the question. will update in a minute.
     
  4. Mar 15, 2013 #3

    Ray Vickson

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    So, in readable English are you saying that ##\lim_{h \to 0} f(x+h)## does not exist?
     
  5. Mar 15, 2013 #4

    jbunniii

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    Well, you know you need a function whose limit at zero does not exist. So start with a function which is only defined on ##(0, \infty)## and for which the limit does not exist as ##h \rightarrow 0##. Then consider how you can extend this function to ##(-\infty, 0)## in such a way that the bad behavior will cancel out when you form the difference ##f(x+h) - f(x-h)## at ##x = 0##.
     
  6. Mar 15, 2013 #5
    Would this work,

    f(x) = cos(1/x) x ≠ 0
    f(x) = 0 , x = 0
     
  7. Mar 15, 2013 #6

    jbunniii

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    Yes, that works. ##f(0+h) - f(0-h) = 0## for all positive ##h## but ##\lim_{x \rightarrow 0} f(x)## does not exist. You should also explain why ##\lim_{h \rightarrow 0} [f(x+h) - f(x-h)] = 0## for ##x \neq 0##.
     
  8. Mar 15, 2013 #7
    lim h→0 [f(x+h) - f(x-h)] = 0 because as h→0 we get lim f(x+0) - f(x-0) = lim f(x) - f(x) = 0

    right?
     
  9. Mar 15, 2013 #8

    jbunniii

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    That's true as long as ##f## is continuous at ##x##. So the key is that your function is continuous at all ##x \neq 0##.
     
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