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Infinite indice problem

  1. Apr 3, 2007 #1


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    if lets say : F(x) = 0.5^x , 0 < 0.5 < 1

    is lim(+infinte) f = 0 ?


    this is for one of my math questions :

    lim (+infinite) x^4 * 0.99^x = ?
  2. jcsd
  3. Apr 3, 2007 #2


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    Impossible to say, since you haven't said what f is, only F.

    And yes, we DO know that 0.5 lies between 0 and 1.

    What is meant by infinite indice is beyond me.
  4. Apr 3, 2007 #3
    WE have the sequence 1/2, 1/4, 1/8...,obviously the nth term gets as close to zero as we like, and so 0 is the least upper bound and since it is never less than zero, the limit is 0.
  5. Apr 4, 2007 #4


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    ok then

    lim (+infinite) x^4 * 0.99^x = 0

    i just had a little dougth

    BTW : arildno ;
    - for the F and f problem
    > i just made a caps mistake
    - for the 0 < 0.5 < 1
    > it was because the F(x) = 0.5^x was an example for any function F(x)= a^x, where a is a real number that respects (0 < a < 1)
    - for "infinite indice"
    > it was just to make short
  6. Apr 4, 2007 #5


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    No, 0 is the greatest lower bound. 1/2 is the least upper bound.
  7. Apr 4, 2007 #6


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    If you mean [itex]\lim_{x\right arrow +\infnty} x^4 * 0.99^x[/itex] then it is true that 0.99x goes to 0 but it does NOT follow from that alone that the whole thing goes to 0 because x^4 goes to infinity.

    It happens that the limit of x40.99x is 0 because 0.99x goes to 0 faster than x4 goes to infinity0- but that has to be shown.

    If, as your use of "x" rather than "n" indicates, you intended this to be a continuous limit, then x is not an "index" at all. (There is no such word as "indice" in English. "Indices" is the plural of "index".
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