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Limit of a Greatest Integer Function using Squeeze help

  1. Oct 23, 2009 #1
    my midterm is in 4 hours and this actually the only thing i need help with.

    1. The problem statement, all variables and given/known data
    prove using squeeze theorem that lim(x-> +inf) (x^2 - [[x^2]])/x = 0


    2. Relevant equations
    g(x)<=f(x)<=h(x) [squeeze theorem]


    3. The attempt at a solution
    on the assignment i didn't know we had to use squeeze, so i just plugged in +inf and got 0, but we had to use squeeze. i wasn't there for his explanation of it.

    all i need help with is setting up the inequality for the squeeze theorem and i'm fine, i'm just drawing blanks for all of this. help would be sooooo amazing. again, this is the only thing i'm confused on for my midterm.
     
  2. jcsd
  3. Oct 23, 2009 #2

    lanedance

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    what do the double brackets mean?
     
  4. Oct 23, 2009 #3
    the greatest integer function/the floor function.

    so if its 5.5, the [[]] make it 5, if its -6.3 the [[]] make it -7
     
  5. Oct 23, 2009 #4
    i've been trying, and the only result i can get, and i don't know if its true is:

    x-1 < [[x]] <= x

    (x^2 - (x^2 - 1))/x < (x^2 - [[x^2]])/x <= (x^2 - (x^2))/x

    -1/x < f(x) <= 0/x

    -1/x < f(x) <= 0

    (lim x-->+inf) -1/x = 0
    (lim x-->+inf) 0 = 0

    .:. through squeeze theorem, (lim x-->+inf) f(x) = 0

    but does that squeeze theorem work if the inequality on the left is just < and not <=?
     
  6. Oct 23, 2009 #5

    lanedance

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    close, few mistakes though
    first f is always positive, consider the squeeze limts you get, they don't really make sense

    few changes here, first start with the assumption x>1 to reduce confusion later
    the first line is correct
    x-1< [[x]] <= x

    now square everything
    (x-1)2< [[x]]2 <= x2

    multiply by -1, reversing the order
    -x2<= -[[x]]2 < -(x-1)2

    add x2 and divide by x gives

    0=(x2 - x2)/x<= (x2-[[x]]2)/x < (x2-(x-1)2)/x

    no its fine (if you use the correct squeeze as above), its still squeezed between 2 functions that go to the same limit
     
    Last edited: Oct 23, 2009
  7. Oct 23, 2009 #6
    if you do it your way though:

    x-1 < [[x]] <= x

    (x-1)2 < [[x]]2 <= x2

    first of all, does [[x2]] = [[x]]2?

    then we goto:

    -x2 <= -[[x]]2 < -(x-1)2

    0/x <= f(x) < (x2 - x2 + 2x + 1)/x

    don't we?

    then you'd have 0 <= f(x) < 2
     
  8. Oct 23, 2009 #7
    should i start off with:
    for x>1
    x^2 - 1 < [[x^2]] < x^2

    and move from there?
     
  9. Oct 23, 2009 #8

    lanedance

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    yeah that looks good, i misplaced the bracket,

    so for x>1
    x - 1 < [[x]] <= x
    giving
    x2 - 1 < [[x2]] <= x2
    minus
    - x2 < -[[x2]] <= -x2+ 1
    then +x2
    0 = x2- x2 <x2 -[[x2]] <= x2-x2+ 1 = 1
    then /x

    0/x< (x2 -[[x2]])x <= 1/x as required
     
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