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Math Help - Proof

  1. #1
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    Proof

    Working on practice problems.


    Is anybody able to complete this?


    Give an epsilon - delta proof of the limit fact




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  2. #2
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    Re: Proof

    Quote Originally Posted by MathRaven View Post
    Working on practice problems.


    Is anybody able to complete this?


    Give an epsilon - delta proof of the limit fact




    To prove \displaystyle \begin{align*} \lim_{x \to 1} (6x - 5) = 1 \end{align*} you need to show that \displaystyle \begin{align*} 0 < |x - 1| < \delta \implies |(6x - 5) - 1| < \epsilon \end{align*}.

    \displaystyle \begin{align*} |(6x - 5) - 1| &< \epsilon \\ |6x - 6| &< \epsilon \\ 6|x - 1| &< \epsilon \\ |x - 1| &< \frac{\epsilon}{6} \end{align*}

    So by letting \displaystyle \begin{align*} \delta = \frac{\epsilon}{6} \end{align*}, we have

    \displaystyle \begin{align*} |x - 1| &< \delta \\ |x - 1| &< \frac{\epsilon}{6} \\ 6|x - 1| &< \epsilon \\ |6x - 6| &< \epsilon \\ |(6x - 5) - 1| &< \epsilon \end{align*}

    as required. Q.E.D.
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