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Thread: Epsilon-Delta Limit Definition

  1. #1
    Super Member Aryth's Avatar
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    Epsilon-Delta Limit Definition

    Use epsilon-delta proof to show that $\displaystyle \lim_{x\to 4} x^2 = 16$

    So far, I have:

    Let $\displaystyle \epsilon > 0$

    We know that:

    $\displaystyle |x^2 - 16| < \epsilon$

    and

    $\displaystyle |x - 4| < \delta$

    Choosing $\displaystyle \delta$:

    $\displaystyle |x^2 - 16| < \epsilon$
    $\displaystyle |(x + 4)(x - 4)| < \epsilon$
    $\displaystyle |x + 4||x - 4| < \epsilon$

    From here, I'm having trouble remembering what to do.
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  2. #2
    o_O
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    Arbitrarily assume $\displaystyle \delta \leq 1$ which implies $\displaystyle |x- 4 | < \delta \leq 1$.

    But this implies:
    $\displaystyle -1 < x - 4 < 1 \ \Leftrightarrow \ 7 < x + 4 < 9 \ \Rightarrow \ |x + 4| < 9$

    So we know: $\displaystyle |x+4||x-4| < 9 |x-4| < \epsilon$

    Now we can see what we should let $\displaystyle \delta$ be.
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  3. #3
    Super Member Aryth's Avatar
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    I forgot that it was reasonable to insist that x would be within 1 of 4, thanks for the help.
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  4. #4
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    Notice that you must say that $\displaystyle \delta$ is the smaller of $\displaystyle \epsilon/9$ and 1.
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