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Math Help - finding d (delta) for a given E(Epsilon)

  1. #1
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    finding d (delta) for a given E(Epsilon)

    Given lim as x-->-2 of 2x-2 is -8, what is the best choice of d such that |(2x-2)-(-8)| < 0.03 whenever |x+2|<d?

    --> I tried to solve |(2x-2)-(-8)| to get it in terms of |x+2| but I can't seem to get it.
    I had |2x+6| = 2|x+3| but that doesn't help much.

    Also:
    b) When the value of x is within 1/20 of 1, how close is the value of f(x) = 4x+5 to 9?

    --> I have:
    0 < |x-1| < 0.05
    |(4x+5)-9| < E

    But now I'm stuck....
    Last edited by janedoe; July 16th 2009 at 01:42 PM.
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  2. #2
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    Quote Originally Posted by janedoe View Post
    Given , what is the best choice of d such that whenever ?

    --> I tried to solve |(2x-2)-(-8)| to get it in terms of |x+2| but I can't seem to get it.
    I had |2x+6| = 2|x+3| but that doesn't help much.

    Also:
    b) When the value of x is within of 1, how close is the value of to 9?

    --> I have:
    0 < |x-1| < 0.05
    |(4x+5)-9| < E

    But now I'm stuck....




    Your pictures didn't take. Is this your limit

    \lim_{x \to -3} 2x-2 = -8
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  3. #3
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    Quote Originally Posted by Danny View Post
    Your pictures didn't take. Is this your limit

    \lim_{x \to -3} 2x-2 = -8

    No it's actually as x --> -2

    I'll repost so you can see the rest...
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  4. #4
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    Quote Originally Posted by janedoe View Post
    No it's actually as x --> -2

    I'll repost so you can see the rest...
    Well then, here's your problem

    <br />
\lim_{x \to -2} 2x-2 = -6

    not -8.
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