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Math Help - question regarding continuity of sign function

  1. #1
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    question regarding continuity of sign function

    Hi! I need to show that the sign function:
    f(x)=\left\{\begin{array}{lr}1:&x>0\\0:&x=0\\-1:&x<0\end{array}\right\}
    is discontinuous.
    Is it true that \lim_{x\to 0^-}f(x)=1 and \lim_{x\to 0^+}f(x)=-1 so therefore \lim_{x\to 0}f(x) does not exist? Im confused because doesnt the limit of f(x) at x=0 exist and is equal to 0?
    Does that somehow mean that f(x) is continuous at 0 but discontinous at 0?
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  2. #2
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    Quote Originally Posted by dannyboycurtis View Post
    Hi! I need to show that the sign function:
    f(x)=\left\{\begin{array}{lr}1:&x>0\\0:&x=0\\-1:&x<0\end{array}\right\}
    is discontinuous.
    Is it true that \lim_{x\to 0^-}f(x)=1 and \lim_{x\to 0^+}f(x)=-1 so therefore \lim_{x\to 0}f(x) does not exist? Im confused because doesnt the limit of f(x) at x=0 exist and is equal to 0?
    Does that somehow mean that f(x) is continuous at 0 but discontinous at 0?

    Yes, no and no: the limit of a function at some point exists iff both one-sided (or lateral) limits exist at that point and are equal, so yes.
    The limit of f(x) at x = 0 doesn't exist since f(x)\xrightarrow [x \to 0^{-}] {} -1\,,\,\,and\,\,f(x)\xrightarrow [x \to 0^{+}] {} 1 , and so no.
    Finally, as \lim_{x\rightarrow 0}f(x)\, doesn't exist f(x) cannot be continuous at 0 and thus no.

    Tonio
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