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Math Help - Delta-Epsilon proof

  1. #1
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    Delta-Epsilon proof

    Can you provide a formal proof for:

    lim X^2 = 9
    x->3
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  2. #2
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    Quote Originally Posted by CrazyAsian View Post
    Can you provide a formal proof for:

    lim X^2 = 9
    x->3
    For all \epsilon>0 consider the quadratic d^2 + 6 d - \epsilon.

    By Descartes rule of signs this has exactly one positive root, which I will call \delta, which depends only on \epsilon (if you are not happy with Descartes rule of signs here you can find the positive root explicitly if you prefer).

    Now suppose |x-3|< \delta, then:

    <br />
|x^2-9| = |x-3||x+3| < \delta |x+3|<br />
.

    But:

    |x+3| = |(x-3) + 6| < |x-3| + 6

    so:

    <br />
|x^2-9| = |x-3||x+3| < \delta [|x-3| + 6]< \delta^2 + 6 \delta<br />

    But by construction \delta^2 + 6 \delta = \epsilon.

    Hence we have proven that for all \epsilon>0 there exists a \delta>0 such that |x-3|< \delta implies that |x^2-9|< \epsilon, so we have proven that:

    <br />
\lim_{x \to 3} x^2 = 9<br />

    RonL
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  3. #3
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    Wow...

    I mean, yeah... Just what I was thinking.
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  4. #4
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    Quote Originally Posted by CrazyAsian View Post
    Can you provide a formal proof for:

    lim X^2 = 9
    x->3
    (First note that f(x)=x^2 is defined on an open interval).

    We need to show |x^2-3|<\epsilon for any \epsilon > 0 by choosing a \delta.

    Let use consider 0<\delta \leq 1. Then |x-3|< \delta by hypothesis. That means that |6|+|x-3| < \delta + |6| so |x+3|\leq |6|+|x-3| < \delta + 6. (Triangle inequality). Hence, |x^2-9| = |x+3|\cdot |x-3| < (\delta +6)\delta \leq (1+6)\delta = 7\delta.
    So choose,
    \delta = \min \left\{ 1,\frac{\epsilon}{7}\right\}.
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