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Math Help - Delta Epsilon Proof

  1. #1
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    Delta Epsilon Proof

    Given an Epsilon, Delta proof for the following statement:

    limit as x approaches 2 for f(x) = |x-2| =0
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  2. #2
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    This shouldn't be too hard. Just take \epsilon = \delta.

    |x-2|<\epsilon \Rightarrow |f(x)-0| = |x-2| < \delta for any \epsilon,\delta > 0
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  3. #3
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    Quote Originally Posted by UC151CPR View Post
    Given an Epsilon, Delta proof for the following statement: limit as x approaches 2 for f(x) = |x-2| =0
    Here is the basic idea:
    \left| {f(x) - 0} \right| = \left| {x - 2} \right| < \delta  = \varepsilon
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  4. #4
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    Would this be how to setup the proof:

    Let epsilon > 0.

    0 < |x-2|< delta

    ||x-2| - 0| < epsilon

    ||x-2| - 0| = ||x-2|| = |x-2|

    Now, |x-2| < delta ---> |x-2| < epsilon

    In order to achieve ||x-2|-0| < epsilon, delta < epsilon.

    Thus chosing 0 < delta < epsilon, we have |x-2|< delta and ||x-2| - 0| < epsilon.

    Hence the limit is proved.

    Does that work as a delta epsilon proof. If not, can you please show me how you would set this up mathematically.
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  5. #5
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    Quote Originally Posted by UC151CPR View Post
    Would this be how to setup the proof:

    Let epsilon > 0.

    0 < |x-2|< delta

    ||x-2| - 0| < epsilon

    ||x-2| - 0| = ||x-2|| = |x-2|

    Now, |x-2| < delta ---> |x-2| < epsilon

    In order to achieve ||x-2|-0| < epsilon, delta < epsilon.

    Thus chosing 0 < delta < epsilon, we have |x-2|< delta and ||x-2| - 0| < epsilon.

    Hence the limit is proved.

    Does that work as a delta epsilon proof. If not, can you please show me how you would set this up mathematically.
    You start with  \epsilon > 0. Then you say,  |x-2| < \epsilon. This is because we want the limit as x approaches 2.

    |x-2| represents the distance between x and 2. We would like to say, for ANY value of epsilon, that is, epsilon can be as small as we like, we can make |f(x)-L| as small as we like. |f(x) - L| is the distance between the value of the function and the limit, in this case, zero.

    The basic idea is that |f(x)-L| will be less than some quantity which depends only on the small quantity we chose earlier. So, the closer x becomes to 2, the closer f(x) will become to 0.

    Anyway, back to the problem. we say |x - 2| < epsilon.

    Now look at |f(x) - 0| = |f(x)|. But this is just |x-2|. What can we say about this quantity?

    Well, it's smaller than epsilon. So, take delta equal to epsilon, and the proof is complete.
    Last edited by harbottle; October 19th 2009 at 04:25 AM.
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  6. #6
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    So...

    Let epsilon > 0.

    0 < |x-2|< delta

    ||x-2| - 0| < epsilon

    ||x-2| - 0| = ||x-2|| = |x-2|

    Now, |x-2| < delta ---> |x-2| < epsilon

    delta = epsilon hence proving the limit?
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  7. #7
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    Quote Originally Posted by UC151CPR View Post
    So...

    Let epsilon > 0.

    0 < |x-2|< delta

    ||x-2| - 0| < epsilon

    ||x-2| - 0| = ||x-2|| = |x-2|

    Now, |x-2| < delta ---> |x-2| < epsilon

    delta = epsilon hence proving the limit?
    No. Start by assuming |x - 2| < epsilon. Now, look at |f(x) -0|. Using the information |x-2| < epsilon, find a delta , which depends on epsilon and only on epsilon, such that |f(x)-0| < delta.

    In this problem, it just so happens that delta = epsilon is that value.
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  8. #8
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    Can you show me how I would write this proof mathematically that way I can see the necessary steps/statements that I need to make for my other proofs?
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  9. #9
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    I did, in my first post.

    Let \epsilon > 0, and let \delta(\epsilon) = \epsilon.

    Then <br /> <br />
|x-2|<\epsilon \Rightarrow |f(x)-0| = |x-2| < \epsilon = \delta.<br />
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