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Math Help - Prove continuouty of 1/x (delta-epsilon)

  1. #1
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    Prove continuouty of 1/x (delta-epsilon)

    Hi guys, I've been trying to do this for a while but I'm not really getting anywhere. Hints would be much appreciated!



    The problem
    Prove that the function g(x)=1/x is continuous on \mathbb{R}\smallsetminus\{0\}, but cannot be defined at the origin 0 in such a way that the resulting function is continuous on \mathbb{R}.

    My attempt
    Just the first part for now - proving continuity on \mathbb{R}\smallsetminus\{0\}.
    They want an \epsilon-\delta proof. Here we go...


    Let a,b\in\mathbb{R}\smallsetminus\{0\}.
    Let \delta=\epsilon>0.
    Then \forall a\in\mathbb{R}\smallsetminus\{0\} and \forall\epsilon>0 we have:
    \mid a-b\mid<\delta\Rightarrow\mid g(a)-g(b)\mid=\mid \frac{1}{a}-\frac{1}{b}\mid=\mid\frac{b-a}{ab}\mid<\frac{\delta}{\mid ab\mid}
    Now I take the case where b is such that \mid a-b\mid>1 and we have...
    \frac{\delta}{\mid ab\mid}<\delta=\epsilon
    So it's proven for such b (I think?)

    Now I have no idea what to do about b such that \mid a-b\mid<1. Or was splitting it into two cases a bad idea? Am I going anywhere useful here?
    Please help!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by wglmb View Post
    Hi guys, I've been trying to do this for a while but I'm not really getting anywhere. Hints would be much appreciated!



    The problem
    Prove that the function g(x)=1/x is continuous on \mathbb{R}\smallsetminus\{0\}, but cannot be defined at the origin 0 in such a way that the resulting function is continuous on \mathbb{R}.

    My attempt
    Just the first part for now - proving continuity on \mathbb{R}\smallsetminus\{0\}.
    They want an \epsilon-\delta proof. Here we go...


    Let a,b\in\mathbb{R}\smallsetminus\{0\}.
    Let \delta=\epsilon>0.
    Then \forall a\in\mathbb{R}\smallsetminus\{0\} and \forall\epsilon>0 we have:
    \mid a-b\mid<\delta\Rightarrow\mid g(a)-g(b)\mid=\mid \frac{1}{a}-\frac{1}{b}\mid=\mid\frac{b-a}{ab}\mid<\frac{\delta}{\mid ab\mid}
    Now I take the case where b is such that \mid a-b\mid>1 and we have...
    \frac{\delta}{\mid ab\mid}<\delta=\epsilon
    So it's proven for such b (I think?)

    Now I have no idea what to do about b such that \mid a-b\mid<1. Or was splitting it into two cases a bad idea? Am I going anywhere useful here?
    Please help!
    When |x-x_0|<1,

    -1<x-x_0<1 \implies -x_0<x_0x-x_0^2<x_0 \implies x_0^2-x_0<x_0x<x_0^2+x_0 \implies \frac{1}{x_0x}<\frac{1}{x_0^2-x_0}

    Thus, let \delta = \min\{(x_0^2-x_0)\epsilon,1\}. Therefore,

    \left|\frac{1}{x_0}-\frac{1}{x}\right| = \left|\frac{x-x_0}{x_0x}\right| < \left|\frac{\delta}{x_0x}\right|<\left|\frac{\delt  a}{x_0^2-x_0}\right| = \left|\frac{(x_0^2-x_0)\epsilon}{x_0^2-x_0}\right| = \epsilon

    Thus, \frac{1}{x} is continuous.

    For the second part, all you have to prove is that \lim_{x\to0^-}\frac{1}{x}\neq\lim_{x\to0^+}\frac{1}{x}, which is obvious.
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  3. #3
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    Stunning, thank you so much!
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  4. #4
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    Quote Originally Posted by redsoxfan325 View Post
    When |x-x_0|<1,

    -1<x-x_0<1 \implies -x_0<x_0x-x_0^2<x_0 \implies x_0^2-x_0<x_0x<x_0^2+x_0 \implies \frac{1}{x_0x}<\frac{1}{x_0^2-x_0}
    Thus, let \delta = \min\{(x_0^2-x_0)\epsilon,1\}. Therefore,
    \left|\frac{1}{x_0}-\frac{1}{x}\right| = \left|\frac{x-x_0}{x_0x}\right| < \left|\frac{\delta}{x_0x}\right|<\left|\frac{\delt  a}{x_0^2-x_0}\right| = \left|\frac{(x_0^2-x_0)\epsilon}{x_0^2-x_0}\right| = \epsilon
    There are several mistakes in the above.
    It must be specified that x_0>0 because otherwise  -1<x-x_0<1 \implies -x_0<x_0x-x_0^2<x_0 is false.

    Here is another mistake: even if x_0>0 unless x_0>1 we do not know that x_0^2-x_0>0
    so \frac{1}{x_0x}<\frac{1}{x_0^2-x_0} may not be true.

    To correct these mistakes you need two cases: x_0\ge 1 and 0<x_0<1.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Plato View Post
    Here is another mistake: even if x_0>0 unless x_0>1 we do not know that x_0^2-x_0>0
    so \frac{1}{x_0x}<\frac{1}{x_0^2-x_0} may not be true.
    OK, the first mistake was something that I forgot to mention. The proof is analogous if x_0<0.

    And the second mistake I believe is taken care of by the absolute value signs in the actual delta-epsilon proof.
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  6. #6
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    Quote Originally Posted by redsoxfan325 View Post
    And the second mistake I believe is taken care of by the absolute value signs in the actual delta-epsilon proof.
    Actually no. That will not work.
    Cases are necessary in this well worn proof.
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