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Math Help - Epsilon delta proof

  1. #1
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    Epsilon delta proof

    \[<br />
 \lim_{x \to 0} \frac{3x}{x^2-4x} = -\frac{3}{4}              \ \ \epsilon = 1.5<br />
\]<br />

    Now, the answer I already know is \delta = .5

    Ive tried at least 20 different things, but cannot figure out what I should be doing, or if I'm even allowed to do some of the things I have been doing. The book we use sucks, and barely covers this.


     \frac{3x}{x^2-4x}-(-\frac{3}{4}) \$< $ \ 1.5

    Which reduces to this after putting under a common denominator and reducing an x.

     \frac{3x}{4x-16} \$<$ 1.5

    At this point I can't figure out what to do, whether its 1.5 or just using epsilon.

    edit: and don't mind the $ signs, couldn't get the <, > signs up without it popping up.
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  2. #2
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    You need to show

    0 < \left|x - 0\right| < \delta \implies \left|\frac{3x}{x^2 - 4x} - \left(- \frac{3}{4}\right)\right| < \epsilon.


    Solve the inequality \left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon for |x - 0| = |x|


    \left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon

    \left|\frac{3x}{x^2 - 4x} + \frac{3}{4}\right| < \epsilon

    \left|\frac{12x}{4(x^2 - 4x)} + \frac{3(x^2 - 4x)}{4(x^2 - 4x)}\right| < \epsilon

    \left|\frac{12x + 3(x^2 - 4x)}{4(x^2 - 4x)}\right|<\epsilon

    \left|\frac{3x^2}{4x(x - 4)}\right| < \epsilon

    \left|\frac{3x}{4(x - 4)}\right| < \epsilon

    \frac{|x|}{4|x - 4|} < \frac{\epsilon}{3}.


    So we want to find a number M such that \frac{1}{4|x - 4|} \leq M.

    M|x| < \frac{\epsilon}{3}

    |x| < \frac{\epsilon}{3M}, i.e. \delta = \frac{\epsilon}{3M}

    If we let |x| < 5 then

    -5 < x < 5

    -9 < x - 4 < 1.

    So |x - 4| < 1

    4|x-4| < 4

    \frac{1}{4|x-4|} < \frac{1}{4}.


    So set M = \frac{1}{4} and define \delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}

     = \min\left\{\frac{\epsilon}{3\left(\frac{1}{4}\righ  t)}, 5 \right\}

     = \min\left\{\frac{4\epsilon}{3}, 5\right\}

    and we are set to write the proof.


    Proof:

    Let \epsilon > 0 and define \delta = \min\left\{\frac{4\epsilon}{3}, 5 \right\} . Then if  0 < |x| < \delta we have

    \left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| = \left|\frac{12x}{4(x^2 - 4x)} + \frac{3(x^2 - 4x)}{4(x^2 - 4x)}\right|

     = \left|\frac{12x + 3x^2 - 12x}{4x(x - 4)}\right|

     = \left|\frac{3x}{4(x - 4)}\right|

     = \frac{3|x|}{4|x - 4|}

     < \frac{3|x|}{4} since \frac{1}{4|x - 4|} < \frac{1}{4}

     < \frac{3\left(\frac{4\epislon}{3}\right)}{4} since |x| < \delta = \frac{4\epsilon}{3}

     < \epsilon.


    Therefore 0 < |x - 0| < \delta \implies \left|\frac{3x^2}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon, and thus

    \lim_{x \to 0}\frac{3x^2}{x^2 - 4x} = -\frac{3}{4}.
    Last edited by Prove It; September 6th 2010 at 06:38 PM.
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  3. #3
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    Great response thank you I've spent probably 5 hours on this over the weekend. However I'm still a little confused on a few parts.

    When we're at this point

    <br />
\frac{|x|}{4|x - 4|} < \frac{\epsilon}{3}<br />

    Is this where I can plug in numbers for  \epsilon because this is the only point where I see that I can plug in 1.5 for epsilon, and get .5 in return for delta. If this is the case do I only need to solve the numerator on this side to equal  |x| < \delta

    Outside that, why did you pick 5 for "If we let |x| < 5"

    And I don't even come close to understand the \delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}
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  4. #4
    Senior Member MacstersUndead's Avatar
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    Quote Originally Posted by Steel88 View Post
    Is this where I can plug in numbers for  \epsilon because this is the only point where I see that I can plug in 1.5 for epsilon, and get .5 in return for delta. If this is the case do I only need to solve the numerator on this side to equal  |x| < \delta
    You can't "plug in" epsilon.  \epsilon > 0 . The "game" (a poor one, admittedly), is that for any epsilon I choose, you can choose a delta such that |f(x) - f(a)| <  \epsilon

    Outside that, why did you pick 5 for "If we let |x| < 5"

    If we let |x| < 5 then

    -5 < x < 5

    -9 < x - 4 < 1.

    So |x - 4| < 1

    4|x-4| < 4

    \frac{1}{4|x-4|} < \frac{1}{4}.
    This inequality is later used in the proof. edit;// we can do this because we can make delta as small as we please.

    And I don't even come close to understand the \delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}
    Suppose I choose a epsilon. If you choose \delta = 5, and |f(x)-f(a)| < \epsilonyou are done.
    If not, you need to choose something lower (hence the min). In order to choose this delta without fail, (this is a game you want to win),
    then it must be calculated based on what epsilon I choose. (which in this case comes from the work established above)
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  5. #5
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    Okay, however I when I choose epsilon as 1.5, I end up with \frac{|x|}{4|x - 4|} < \frac{\ 1.5}{3} Which gives me .5 = delta, however I'm still left with \frac{|x|}{4|x - 4|} On the left, or if you substitute M its     M|x| < \frac{\epsilon}{3} or     \delta = \frac{\epsilon}{3M} Am I just supposed to ignore the denominator if I have |X| in the numerator, or just ignore the M otherwise?


    I'm just not seeing how when I choose the epsilon value where I get delta in return...In every other problem I've done with this the left side just collapses to equal the right and gives you the delta.
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  6. #6
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    I suggest you download and read the PDF attached in the following link, as they explain it much better than I could...

    http://www.mathhelpforum.com/math-he...ofs-47767.html
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