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Math Help - Prove lim(x->3) x^2 = 9 by definition

  1. #1
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    Arrow Prove lim(x->3) x^2 = 9 by definition

    Hello!

    I know how to solve this task but I have certain things that I do not understand.

    Solution.

    [|x^2-9|< \epsilon if 0 < |x-3| < \delta

    Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.

    |x+3||x-3|< \epsilon if 0<|x-3|<\delta

    Using the triangle inequality:

    |x+3|=|(x-3)+6| \leq |x-3| + 6

    By multiplying the statement above with |x-3| (which is positive) we got:

    |x+3||x-3| \leq (|x-3| +6)|x-3|

    Now clearly 6+|x-3|< 6 + \delta and |x-3|<\delta

    By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:

    (6+|x-3|)|x-3|< (6 + \delta)\delta

    We can conclude that:

    |x+3||x-3| < (6 + \delta)\delta if 0 < |x-3| < \delta

    Now isn't \epsilon = (6 + \delta)\delta ??

    Why in my book they choose (6 + \delta)\delta \leq \epsilon ?

    And why they choose \delta \geq 1 and with that restriction
    (6 + \delta)\delta \leq 7\delta and 7\delta \leq \epsilon.

    Finally I don't understand why \delta = min(e/7,1).

    Could somebody possibly explain?

    Thanks a lot.
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  2. #2
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    Quote Originally Posted by p0oint View Post
    Hello!

    I know how to solve this task but I have certain things that I do not understand.

    Solution.

    [|x^2-9|< \epsilon if 0 < |x-3| < \delta

    Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.

    |x+3||x-3|< \epsilon if 0<|x-3|<\delta

    Using the triangle inequality:

    |x+3|=|(x-3)+6| \leq |x-3| + 6

    By multiplying the statement above with |x-3| (which is positive) we got:

    |x+3||x-3| \leq (|x-3| +6)|x-3|

    Now clearly 6+|x-3|< 6 + \delta and |x-3|<\delta

    By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:

    (6+|x-3|)|x-3|< (6 + \delta)\delta

    We can conclude that:

    |x+3||x-3| < (6 + \delta)\delta if 0 < |x-3| < \delta

    Now isn't \epsilon = (6 + \delta)\delta ??

    Why in my book they choose (6 + \delta)\delta \leq \epsilon ?

    And why they choose \delta \geq 1 and with that restriction
    (6 + \delta)\delta \leq 7\delta and 7\delta \leq \epsilon.

    Finally I don't understand why \delta = min(e/7,1).

    Could somebody possibly explain?

    Thanks a lot.
    If you reorganized your work, you will find it as follows:

    |x^2-9|=|x+3||x-3|<\epsilon
    |x^2-9|=|x+3||x-3|<|x-3|^2+6<\epsilon
    |x^2-9|=|x+3||x-3|<\delta^2+6<\epsilon
    |x^2-9|=|x+3||x-3|<(\delta+6)\delta<\delta^2+6<7\delta<\epsilon
    The reason for choosing \delta \leq 1 is that we know with the choice, |x^2-9|<\epsilon is still true. That does not mean that there is no other value of of \delta besides 0<\delta\leq 1. If 0<\delta\leq 1 will work, why do we care about larger value of \delta? In other words, if the small number is closer to the limit, why bother about other large \delta's that are farther from limit.

    Since we know 7\delta<\epsilon is true, we know \delta<\frac{\epsilon}{7} is also true. So we know it is safe to choose \delta=\frac{\epsilon}{7}, since we know with the choice, |x^2-9|<\epsilon will still remain true.

    If we know it will work by limiting \delta to 0<\delta\leq 1 to maintain  |X^2-9|<\epsilon , and we also know between 0 and 1, the is a number \frac{\epsilon}{7} that also can maintain |x^2-9|<\epsilon being true. Given the choices, of course we would choose \frac{\epsilon}{7} over 1, since the smallest \delta we could get is the one that can give us the value closest to the limit.
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  3. #3
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    Thank you. But how you are sure that

    <br />
(6 + \delta)\delta \leq \epsilon<br />

    is true?
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  4. #4
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    Quote Originally Posted by p0oint View Post
    Thank you. But how you are sure that

    <br />
(6 + \delta)\delta \leq \epsilon<br />

    is true?
    Yes, it's true.
    <br />
(6 + \delta)\delta can be as large as \epsilon, and it can never reach beyond the value of \epsilon. In other words, (6+\delta)\delta can only be made smaller than \epsilon if you choose to make \delta smaller. Because 7\delta is the nearest integer multiple of \delta, it's convinient to use 7 for dividing \epsilon. Suppose that \frac{\epsilon}{(6+\delta)\delta}=1, we can be 100% certain that when \delta =\frac{\epsilon}{7}, (6+\delta)\delta <\epsilon.

    Example: For when (6+\delta)\delta = \epsilon, we can make table as follows:

    \delta-------> x=\delta+3--------> (6+\delta)\delta--------> \epsilon

    0.05 ---->3.05------------->0.3025---------->0.3025
    0.50 ---->3.70------------->3.2500---------->3.2500
    010 ----->4.00------------->7.0000---------->7.0000

    The following are the results when 7 \delta = \epsilon :

    \delta-------> x=\delta+3--------> (6+\delta)\delta--------> \epsilon
    0.05 ---->3.04------------->0.2590---------->0.3021
    0.50 ---->3.42------------->2.7551---------->3.2142
    010 ----->4.00------------->5.8775---------->6.8571

    Now, there should be no doubt the (6+\delta)\delta \leq \epsilon.
    Last edited by novice; November 5th 2009 at 08:00 AM.
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  5. #5
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    Ok I understand now. Just one thing:

    |x-3| < \delta

    and

    \delta \leq 1

    which makes

    |x-3| \leq 1

     2 \leq x \leq 4

     5 \leq x + 3 \leq 7

    And we can say |x+3| \leq 7 and it follows  |x+3||x-3| \leq 7|x-3|

    Because of |x+3|<7 then |x-3||x+3|<7|x-3| and because of |x-3| < delta then 7|x-3|<7delta so that |x-3||x+3|<7|x-3|<7delta

    Now we can set epsilon = 7 delta.

    But one thing why in my book 7 delta <= epsilon ?

    Is it because we can choose smaller values for delta then epsilon/7 ? for ex. delta= epsilon/14 so that epsilon/2 <= epsilon ?
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  6. #6
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    Quote Originally Posted by p0oint View Post
    Ok I understand now. Just one thing:

    |x-3| < \delta

    and

    \delta \leq 1

    which makes

    |x-3| \leq 1

     2 \leq x \leq 4

     5 \leq x + 3 \leq 7

    And we can say |x+3| \leq 7 and it follows  |x+3||x-3| \leq 7|x-3|

    Because of |x+3|<7 then |x-3||x+3|<7|x-3| and because of |x-3| < delta then 7|x-3|<7delta so that |x-3||x+3|<7|x-3|<7delta

    Now we can set epsilon = 7 delta.

    But one thing why in my book 7 delta <= epsilon ?

    Is it because we can choose smaller values for delta then epsilon/7 ? for ex. delta= epsilon/14 so that epsilon/2 <= epsilon ?
    For your last question: Yes.
    For the question above it, Yes and No.

    If you chose \delta arbitrarily, 7\delta \leq \epsilon is not necessarily true, but that's not the hypothesis.

    The hypothesis says, for all \epsilon > 0, there exist a \delta > 0 such that if the distance between x and 3 is less than \delta, then |x^2-9|<\epsilon. In other words, you must find \delta that is expressed in terms of \epsilon---not otherwise.

    Because of the hypothesis, yes, 7\delta < \epsilon is true. In other words, because of the hypothesis, we can choose \delta as small as we like depending on how precise we want, but we cannot choose \delta such that (6+\delta)\delta > \epsilon.
    Last edited by novice; November 5th 2009 at 12:57 PM.
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  7. #7
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    Quote Originally Posted by p0oint View Post
    Hello!

    I know how to solve this task but I have certain things that I do not understand.

    Solution.

    |x^2-9|< \epsilon if 0 < |x-3| < \delta

    Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.

    |x+3||x-3|< \epsilon if 0<|x-3|<\delta

    Using the triangle inequality:

    |x+3|=|(x-3)+6| \leq |x-3| + 6

    By multiplying the statement above with |x-3| (which is positive) we got:

    |x+3||x-3| \leq (|x-3| +6)|x-3|

    Now clearly 6+|x-3|< 6 + \delta and |x-3|<\delta

    By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:

    (6+|x-3|)|x-3|< (6 + \delta)\delta

    We can conclude that:

    |x+3||x-3| < (6 + \delta)\delta if 0 < |x-3| < \delta

    Now isn't \epsilon = (6 + \delta)\delta ??

    Why in my book they choose (6 + \delta)\delta \leq \epsilon ?

    And why they choose \delta \geq 1 and with that restriction
    (6 + \delta)\delta \leq 7\delta and 7\delta \leq \epsilon.

    Finally I don't understand why \delta = min(e/7,1).

    Could somebody possibly explain?

    Thanks a lot.
    The following method was tought by Professor George Thomas of Massachusetts Institute of Technology:

    Given: |x^2-9|< \epsilon if 0 < |x-3| < \delta

    Solution:

    |x^2-9|< \epsilon
    Expanding the LHS, |(x+3)(x-3)|< \epsilon

    Let 0<\delta < 1, then (x+3) will be in the neighborhood of 7, and

    |x-3|<1----------------------------------eq1

    Say (x+3)=6.999, then |6.999(x-3)|<\epsilon,
    so that

    |x-3|<\frac{\epsilon}{7}--------------------------eq2

    Between eq1 and eq2,

    \delta = min \{\frac{\epsilon}{7}, 1\}.
    Last edited by novice; November 7th 2009 at 12:33 PM. Reason: fix mistake
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  8. #8
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    Thanks.

    I invented one better method which does not involve arbitary numbers, but I wanted to learn other methods.

    I think you made a little mistake.

    It should be |x-3|<eps./7
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  9. #9
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    Quote Originally Posted by p0oint View Post
    Thanks.

    I invented one better method which does not involve arbitary numbers, but I wanted to learn other methods.

    I think you made a little mistake.

    It should be |x-3|<eps./7
    May I see your invention? Via Private Message if you prefer.
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