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Finding Delta Given Epsilon with a Quadratic Function

  1. Jul 13, 2017 #1
    • Thread moved from the technical forums, so no Homework Help Template is shown
    Hey, everyone! I'm helping a friend through his calculus course and we've come across something that has stumped me (see: the title). When I learned calculus, our treatment of the epsilon-delta definition of the limit was, at best, brief. Anyway, here is the problem:

    Given ##\lim_{x \rightarrow 3} (x^2 -2) = 7## and ##\epsilon = 0.2##,
    find ##\delta## such that for all ##x, \quad## ##0<|x-3|<\delta \implies |x^2 - 9| < 0.2##.

    Here is my work so far:

    $$0 < |x-3| < \delta \quad, \quad |x^2 - 9| < 0.2$$
    $$0< |x-3| < \delta \quad, \quad |x+3||x-3| < 0.2$$
    $$0 < |x-3| <\delta \quad, \quad |x-3| < \frac{0.2}{|x+3|}$$

    ...And there is where I get lost. I've found a connection between the absolute value expressions, but I can't find any way to make progress from there. I'm sure this is the wrong approach for any problems of this form where ##f(x)## has a degree ##>1##, so any help is greatly appreciated!
     
  2. jcsd
  3. Jul 13, 2017 #2

    scottdave

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    I would not do this: |x+3||x-3|, you really only need the absolute value of the whole thing. But I would look at cases.
    So for the case when x² ≥ 9, then |x² - 9| < 0.2 is the same as just (x² - 9) < 0.2
    Then for the case when x² < 9, then |x² - 9| is the same as (x² - 9) > -0.2

    That should make it easier than having two factors with absolute values, and dividing by absolute values, etc.
     
    Last edited: Jul 14, 2017
  4. Jul 13, 2017 #3

    berkeman

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    Thread closed for Moderation.
     
  5. Jul 13, 2017 #4

    berkeman

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    Thread moved to the schoolwork forums and re-opened. @Cosmophile -- please check your PMs.
     
  6. Jul 14, 2017 #5

    scottdave

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    Thanks, I fixed it.
     
  7. Jul 14, 2017 #6

    SammyS

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    Good.

    By the way: Congrats on becoming an official Homework Helper !
     
    Last edited: Jul 14, 2017
  8. Jul 15, 2017 #7

    scottdave

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    Thank You.
    Thank You. I'm honored.
     
  9. Jul 15, 2017 #8

    SammyS

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    The method suggested by @scottdave can get you the maximum value that can be used for ##\ \delta \, ##, but that's not the usual method used in a proof for the case where ε has an arbitrary value.

    You have ##\displaystyle \ |x+3||x-3| < 0.2\,,\ ## giving ##\displaystyle \ |x-3| < \frac{0.2}{|x+3|}\ .\ ## So, if you can find some "upper bound" on ##\ | x+3|\,,\ ## then you can find a value for ##\ \delta\,.##

    Certainly, we must have ##\ \delta < 1\ .## That implies that x is between 2 and 4 and thus ##\ | x+3|\ ## must be between 5 and 7.

    Use this information to find a value for ##\ \delta\,.##
     
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