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Math Help - Epsilon Delta Proof

  1. #1
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    Epsilon Delta Proof

    Hi everyone.

    I've encountered an epsilon-delta proof problem. I think I'm kinda stuck here. Can someone please help?

    Prove:
    (x-->1)lim x^3 = 1

    I have already factor x^3 into abs(x-1)*abs(x^2+x+1), but I don't know how to continue.

    Thanks everyone.
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  2. #2
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    Re: Epsilon Delta Proof

    Quote Originally Posted by pohkid View Post
    Hi everyone.

    I've encountered an epsilon-delta proof problem. I think I'm kinda stuck here. Can someone please help?

    Prove:
    (x-->1)lim x^3 = 1

    I have already factor x^3 into abs(x-1)*abs(x^2+x+1), but I don't know how to continue.

    Thanks everyone.
    If \displaystyle 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon then \displaystyle \lim_{x \to c}f(x) = L.

    You want to show that \displaystyle \lim_{x \to 1}x^3 = 1, so you need to show that \displaystyle 0 < |x - 1| < \delta \implies |x^3 - 1| < \epsilon.

    Scratch work:

    \displaystyle \begin{align*} |x^3 - 1| &< \epsilon \\ |(x - 1)(x^2 + x + 1)| &< \epsilon \\ |x - 1||x^2 + x + 1| &< \epsilon \\ |x - 1| &< \frac{\epsilon}{|x^2 + x + 1|} \\ |x - 1| &< \frac{\epsilon}{\left|\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right|}  \end{align*}

    Now since we are making \displaystyle |x - 1| arbitrarily close to \displaystyle 0, we can restrict it to be no more than a certain distance from \displaystyle 0, say \displaystyle |x - 1| \leq 1.

    \displaystyle \begin{align*}|x - 1| &\leq 1 \\ -1 \leq x - 1 &\leq 1 \\ 0 \leq x &\leq 2 \\ \frac{1}{2} \leq x + \frac{1}{2} &\leq \frac{5}{2} \\ \frac{1}{4} \leq \left(x + \frac{1}{2}\right)^2 &\leq \frac{25}{4} \\ 1 \leq \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} &\leq 7 \\ \left|\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right| &\leq 7  \end{align*}

    So \displaystyle \frac{\epsilon}{\left|\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right|} will be at its minimum when \displaystyle \left|\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right| = 7.

    So we can let \displaystyle \delta = \min\left\{1, \frac{\epsilon}{7}\right\} and reverse the process, and you will have your proof.
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