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Math Help - How to show there's no limit at a jump

  1. #1
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    How to show there's no limit at a jump

    For the function f:R -->R which is

    f(x): -1 when x<0
    1 when x>0
    0 when x=0

    prove that there is no limit at x=0. I'm told I can use a proof by contradiction.

    What do you guys think? I've seen I can do something with derivatives or something, but I haven't been taught that yet. I just learned delta-epsilon proofs. (I looked at the sticky, but it wasn't helpful to me).

    Thanks!

    Regards,
    Esther
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  2. #2
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    Quote Originally Posted by syok View Post
    For the function f:R -->R which is
    f(x): -1 when x<0
    1 when x>0
    0 when x=0
    prove that there is no limit at x=0.
    It is sufficient to note that \lim _{x \to 0^ -  } f(x) =  - 1\;\& \,\lim _{x \to 0^ +  } f(x) = 1

    If a limit exists at x=a it in necessary that \lim _{x \to a^ -  } f(x) = \lim _{x \to 0^ +  } f(x)
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