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Math Help - Help with Epsilon Delta continuous proof

  1. #1
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    Help with Epsilon Delta continuous proof

    Question: Prove that 2x^2-\frac{1}{x}\ is\ continuous\ on\ (2,3)

    ok so |f(x)-f(c)|=|\left(2x^2-\frac{1}{x}\right)-\left(2c^2-\frac{1}{c}\right)|\leq|2x^2-2c^2|+|\frac{1}{x}-\frac{1}{c}\displaystyle{|}\leq

    \leq2|x-c||x+c|+|\frac{1}{x}-\frac{1}{c}\displaystyle{|}

    when |x-c|<\delta<1 then

    |\frac{1}{x}-\frac{1}{c}\displaystyle{|}\leq|\frac{1}{x}|+|\fra  c{1}{c}|\leq|x|+|c|<1+2|c|

    and also |x+c|\leq|x|+|c|<1+2|c|

    so  2|x-c||x+c|+|\frac{1}{x}-\frac{1}{c}|<2|x-c|(1+2|c|)+(1+2|c|)=(1+2|c|)(2|x-c|+1)

    (1+2|c|)(2|x-c|+1)<\epsilon\rightarrow|x-c|<\frac{\epsilon}{2(1+2|c|)}-\frac{1}{2}<br />
    so here's where i get stuck the algebra says that is i set \delta=min\{1,\left(\frac{\epsilon}{2(1+2|c|)}-\frac{1}{2}\right)\}

    then i run into the problem that Delta is not 100% going to be greater than 0. namely if \frac{\epsilon}{2(1+2|c|)} is less than one half. so if anyone has any tips suggestions that would be amazing
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  2. #2
    MHF Contributor

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    You need to use the absolute value.
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  3. #3
    Junior Member
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    so you're saying that if i set delta equal to
    |\frac{\epsilon}{2(1+2|c|)}-\frac{1}{2}| then it works?

    maybe im dumb but i don't see the connection.
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