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Limit question

  1. Jan 14, 2012 #1
    Determine the limit L for a given a, and prove that it is the limit.


    f(x) = x^4,
    a = a
    L = a^4.

    I get all the way to the following point...

    |x-a|x+a||x^2+a^2| < ε,

    I do not know how to find a "bound" for the product |x+a||x^2+a^2|. Can someone lead me in the right direction?

    thank you.
     
  2. jcsd
  3. Jan 14, 2012 #2
    If you were proving uniform continuity on all of ℝ you'd have a problem (it isn't), but here you're proving point-wise continuity, which means you can define your delta in terms of more than just epsilon. Think about why this is for a bit, it was a difficult concept for me to swallow at first.
     
  4. Jan 14, 2012 #3
    Are you saying that my delta value is not going to be in terms of epsilon?
     
  5. Jan 14, 2012 #4
    deleted comment.
     
    Last edited: Jan 14, 2012
  6. Jan 15, 2012 #5

    SammyS

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    Notice that (x+a)(x^2+a^2) = x3 + ax2 + ax +a3 .

    So, if δ < |a|, then you can find a bound on |x+a||x^2+a^2| .
     
  7. Jan 15, 2012 #6

    SammyS

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    I had a typo in that expression. It should have been:
    x3 + ax2 + a2x +a3

    I also suggested letting δ < |a| .

    In that case:
    If |x-a| < δ

    then -|a| < -δ < x-a < δ < |a|

    then (after a little work) you can show that |x|< 2|a|​

    This leads to |x2|< 4|a2|

    and |x3|< 8|a3| .

    Therefore |x3| + |ax2| +|a2x| + |a3| <    ?   
     
  8. Jan 16, 2012 #7
    Is the answer:
    ....< 8|a3| + 4|a3| + 2|a3| -|a3| or

    |x3| + |ax2| +|a2x| + |a3| - |a3| < 13|a3|...

    is this correct? If not am I close?
     
  9. Jan 16, 2012 #8

    SammyS

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    (Use the X2 button on the "Go Advanced" message box to display exponents. Otherwise use LaTeX.)

    13|a3| is not right.

    How did you get that?
     
  10. Jan 16, 2012 #9
    you gave me

    |x| < 2|a|
    |x2| < 4|a2| ,
    |x3| < 8|a3|, and
    |x3| + |ax2| + |a2x| + |a3| < .


    so
    |x3| < 8|a3|,
    |ax2| < 4|a*a2| ,
    |a2*x| < 2|a2*a|

    |a3| < ( dont know how to relate this inequality...)

    Then I added the above 4 lines together and got,

    |x3| + |ax2| +|a2*x| + |a3| < 8|a3| + 4|a*a2| + 2|a2*a| + ( dont know how to relate this inequality...)

    The right side is eqaul to 14|a3| + ( dont know how to relate this inequality...)
     
  11. Jan 16, 2012 #10

    SammyS

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    The way to deal with |a3| is simply that |a3| = |a3|


    So if
    |x| < 2|a|,
    |x2| < 4|a2| ,
    |x3| < 8|a3|,
    and
    |a3| is what it is:​
    then
    |x3| + |ax2| + |a2x| + |a3| < 8|a3| + 4|a2|∙|a| + 2|a|∙|a2| + |a3|​

    This will work fine as long as a ≠ 0.

    The case in which a = 0 can be handled pretty easily, after you have conquered the more general case. (It is also true that you can work out a method that handles all values of a, but the algebra will be tougher than that above and that looks as if it has given you enough of a problem.)
     
  12. Jan 16, 2012 #11
    Yeah to be honest when Im working with inequalities I am totally lost... Is there any place I can read up on inequalities so maybe I can understand them better and learn how to manipulate them better?

    and what is the more "general" way to do it?
     
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