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Math Help - 1/x and continuity

  1. #1
    Senior Member I-Think's Avatar
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    1/x and continuity

    Prove f(x)=\frac{1}{x} is not uniformly continuous over (0,1)

    There are multiple ways to prove this, but I'm having trouble with a particular one

    It is known that a f:I\rightarrow{R} is uniformly continuous if and only if when x_n,p_n\in{I} and (x_n-p_n)\rightarrow{x}, then (f(x_n)-f(p_n))\rightarrow{0}

    I'm trying to use this result to prove the question
    So if (x_n-p_n)\rightarrow{0}

    f(x_n)-f(p_n)=\frac{1}{x_n}-\frac{1}{p_n}=\frac{p_n-x_n}{x_np_n}<br />
    But this would seem to converge to 0, by assumption.

    Can someone point out the flaw in my reasoning?
    Please and thank you
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by I-Think View Post
    Prove f(x)=\frac{1}{x} is not uniformly continuous over (0,1)

    There are multiple ways to prove this, but I'm having trouble with a particular one

    It is known that a f:I\rightarrow{R} is uniformly continuous if and only if when x_n,p_n\in{I} and (x_n-p_n)\rightarrow{x}, then (f(x_n)-f(p_n))\rightarrow{0}

    I'm trying to use this result to prove the question
    So if (x_n-p_n)\rightarrow{0}

    f(x_n)-f(p_n)=\frac{1}{x_n}-\frac{1}{p_n}=\frac{p_n-x_n}{x_np_n}<br />
    But this would seem to converge to 0, by assumption.

    Can someone point out the flaw in my reasoning?
    Please and thank you
    Consider that \displaystyle \frac{1}{n}-\frac{1}{n^2}\to 0 yet \displaystyle n-n^2\not \to 0.
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  3. #3
    Senior Member I-Think's Avatar
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    Just thought of this, given that the theorem works only if for any x_n, p_n in the interval I, and

    Would it fail in this case as if we have (x_n)\rightarrow{0} and (p_n)\rightarrow{0},we have (x_n-p_n)\rightarrow{0} but (f(x_n)-f(p_n)) does not converge to 0?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by I-Think View Post
    Just thought of this, given that the theorem works only if for any x_n, p_n in the interval I, and

    Would it fail in this case as if we have (x_n)\rightarrow{0} and (p_n)\rightarrow{0},we have (x_n-p_n)\rightarrow{0} but (f(x_n)-f(p_n)) does not converge to 0?
    Yes, I don't see why not. I don't see anything that doesn't serve to perfectly dissatisfy the hypotheses of uniform continuity.
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