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Homework Help: Formal definition of the limit

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    prove the statement using epsilon delta definition of the limit

    lim x->2 (x^3)=8
    2. Relevant equations
    lim x->a f(x)=L is true when
    |f(x)-L|<e
    whenever
    0< |x-a|<d


    3. The attempt at a solution

    |x^3 - 8| < e
    |(x-2)| |(x^2+2x+4)|<e
    x^2+2x+4 has no real roots there for only has one sign f(0)=4 so it is always positive
    so |x^2+2x+4|=x^2+2x+4
    |(x-2)||(x^2+2x+4)| <C(x-2)<e
    where
    |(x^2+2x+4)|<C
    (x-2) < (e/C) =d

    I usually solve most problems like this but I can't find the C
    I know that we usually say that d is a small distance so |x-a| <1
    |x-2|<1
    1<x<3
    but then what exactly?
     
  2. jcsd
  3. Sep 21, 2010 #2

    lanedance

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    don't you want to find C such that [itex] |(x^2+2x+4)| \leq C [/itex] for all x?
     
  4. Sep 21, 2010 #3
    yes I am asking how to start that.
    also Does it have to be for all x? I mean aslong as I can find a C that works for x that are near two it is enough
     
  5. Sep 21, 2010 #4
    how can I relate or change |x-2|<1 to something like x^2+2x+4<C?
    if I square it
    x^2-4x+4<1
    x^2+2x+4<1+6x
    I will get a variable not a constant
    if I cube it
    (x-2)(x-2)^2<1
    x^2-4x+4<1/(x-2)
    x^2+2x+4<6x+ 1/(x-2)
    x^2+2x+4< (6x^2-12x+1)/(x-2)
    x^3-8<6x^2-12x+1
    this goes in rounds too
    I don't know If I got this right but if d is less than 1 then the maximum value of x that can be inputted into x^2+2x+4 is 3
    and cant I get C to be (6x^2-12x+1)/(x-2) of 3 = 19
    so can't I just say that as long as delta <1 and e/19 the limit is true?
    Also what I don't understand about these questions are what delta is necessary for it to be true?I mean I just speculated the solution but what does it mean that delta has to be less than e/19?
     
    Last edited: Sep 21, 2010
  6. Sep 22, 2010 #5
    Is it because I didn't show enough work that I don't get any replies?
     
  7. Sep 22, 2010 #6

    lanedance

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    so we need to show that for any given [itex] \epsilon> 0 [/itex] we can choose a [itex] \delta > 0 [/itex] such that for any x satisfying [itex] |x-2| < \delta [/itex] then

    [tex]|x^3 - 8 | < \epsilon [/tex]

    so expanding as you have
    [tex]|x - 2 |.|x^2+2x+4| < \epsilon [/tex]

    now it should be clear (apologies in the other post I put a less than)
    [tex]x^2+2x+4 \geq 3 [/tex]

    thus we have
    [tex] 3|x - 2 | \leq |x - 2 |.|x^2+2x+4| < \epsilon [/tex]

    hopefully you can take it from here
     
  8. Sep 22, 2010 #7

    lanedance

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    another way to do it would be to work in terms of a variable say d, such that [itex] -\delta < d < \delta [/itex], then you can write the limit as

    [tex] |(2+d)^3 - 8| < \epsilon [/tex]

    expanding that should allow you to solve for d, making the requirerd relation between epsilon & delta clearer
     
  9. Sep 22, 2010 #8
    3|x-2| <e
    |x-2|< e/3
    so d = e/3 might work?
    or did I get it wrong?
    I also tried your other method and got |d^3+6d^2+12d|<e but i dont see anything from it.
     
    Last edited: Sep 22, 2010
  10. Sep 22, 2010 #9

    lanedance

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    that'd do it

    the other way was
    [tex] |(2+d)^3 - 8| < \epsilon [/tex]
    [tex] |(2+d)(4+4d +d^2) - 8| < \epsilon [/tex]
    [tex] |(8+8d+2d^2+4d+4d^2+d^3) - 8| < \epsilon [/tex]
    [tex] |(12d+6d^2+d^3)| < \epsilon [/tex]
    [tex] |d| |(12+6d+d^2)| < \epsilon [/tex]
    [tex] |d| |(12+6d+d^2)| < \epsilon [/tex]

    similarly
    [tex] |(12+6d+d^2)| \geq 3 [/tex]
     
  11. Sep 22, 2010 #10

    ehild

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    We prove that the limit is 2 if - for any epsilon - a value for d is found so that |x^3-8|<epsilon if |x-2|<d. Unfortunately, we have a complicated expression for d. But remember, both epsilon and d are small numbers, much smaller than 1. So 12+6d+d^2 is certainly overestimated by replacing d with 1. So it is quite sure that d= epsilon/19 will do.


    ehild
     
  12. Sep 22, 2010 #11

    lanedance

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    not that it changes the idea, but why go to e/19, when you can show it with the looser constraint e/3?
     
  13. Sep 23, 2010 #12

    ehild

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    To get d you need the stricter constraint.

    Just try. Let be e=0.1. What should be d so as |(2+d)^3-8|<e?

    If you choose d=0.1/3=0.03333 and calculate 2.03333^3-8 it is
    0.4 which is much higher than e=0.1. d=e/12 is much better, but the result is still higher than e.


    ehild
     
  14. Sep 23, 2010 #13

    lanedance

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    yeah apologies, miscalulation on my part
     
  15. Sep 23, 2010 #14
    I did figure what epsilon is but I have problems writing it as a formal proof any hints about that?
     
  16. Sep 23, 2010 #15

    ehild

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    You have to find out d can be if you know epsilon. As the definition of the limit is:

    "lim x->a f(x)=L is true when
    |f(x)-L|<e
    whenever
    0< |x-a|<d "

    Suppose that |x^3-8| <epsilon, a small number (less that 1)
    Show you can find some value for d so that in case |x±2|<d |x^3-8| <epsilon is true.

    You choose x the farthest from 2 possible: x=2-d and 2+d. Find out the relation which has to be true between d and epsilon.

    ehild
     
  17. Sep 23, 2010 #16
    Given e>0 we choose d =min{1,e/19} such that to show that 0<|x-2|<d whenever |x^3-8|<e
    Thus
    |x-2||x^2+2x+4|<e
    '.' d<1 , |x-2|<d
    ,', |x-2|<1
    .'. -1<x-2<1
    .'. 1<x<3
    defining f(x) = x^2+2x+4 = (x+1)^2+3
    '.' the vertex of the parabola formed by is at x=-1, the function is increasing for all x>-1,.'. f(1)<x<f(3) 7<f(x)<19
    7<x^2+2x+4<19(1)
    assuming a constant C=19
    |x-2|<d
    x^2+2x+4 |x-2|<19|x-2|<e
    |x-2|<e/19
    choose d min ={1,e/19}
    e=19d
    then for every
    0<|x-2|<d
    |x-2|x^2+2x+4<19d=19*e/19=e
    thus by definition limit x->2 x^3=8
    ---------------------------------------------
    What are the unnecessary details? what should be added or deleted?
    or should I do a completely different format?
     
  18. Sep 23, 2010 #17

    ehild

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    It is the opposite: Given arbitrary small e>0 we choose d such that whenever 0<|x-2|<d, |x^3-8|<e. It is all right otherwise.

    ehild
     
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