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Math Help - limit prooving question..

  1. #1
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    limit prooving question..

    the question and where i got stuck in this link:

    http://img187.imageshack.us/img187/8868/44429492ix7.gif
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  2. #2
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    We have to prove the following statement:

    \left| x-5 \right|<\delta \implies \left| x^{2}-25 \right|<\epsilon .

    -----

    Observe that \left| x^{2}-25 \right|=\left| x+5 \right|\left| x-5 \right|. Now pick \delta=1, and \left| x+5 \right|\left| x-5 \right|\le 11\left| x-5 \right|<\epsilon .

    Hence, conditions are fulfilled if you take \delta =\min \left\{ 1,\frac{\epsilon }{11} \right\}.\quad\blacksquare
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  3. #3
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    my question i about how to choose the value for M

    ??
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  4. #4
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    i understand that our upper bound is 11

    so inorder for this to work (like your previous examples in the article you published
    we need only

    <br /> <br />
\delta = \frac{\epsilon }{11} <br />

    why to use minimum??

    the previoss werenot solved like that
    <br /> <br />
\delta =\min \left\{ 1,\frac{\epsilon }{11} \right\}.\quad\blacksquare<br />
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  5. #5
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    we get x as close as possible to 5

    |x-5|<1 ==>> -1<x-5<1 so x+5<11
    so M>=11
    lambda=e/11

    why minimum with 1 ??
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  6. #6
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    Krizalid's Avatar
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    It's just any number you want.

    I took x\in(4,6) because \delta=1 is a nice guy. You could've taken other delta, a ratio, 2, whatever.
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  7. #7
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    why you didnt do this min function in the previous examples
    in the article?
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  8. #8
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    usually ou ended this question by saying lambda=e/11

    why lambda=min{1,e/11}

    why 1? why any number i want?
    why minimum?
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  9. #9
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    did i prove it correctly
    i solved it differently from the article
    is that ok?

    http://img224.imageshack.us/img224/9929/93418507so5.gif
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