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Limit approaching infinity

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data

    lim (SQRT (8x^3 + 5x + 10) ) / x^2

    x -> infinity

    2. Relevant equations



    3. The attempt at a solution

    I tried factoring out x^3, but that didn't help anything.
     
  2. jcsd
  3. Sep 21, 2008 #2
    [tex]\frac{A+B+C}{D} = \frac{A}{D} + \frac{B}{D} + \frac{C}{D}[/tex]
     
  4. Sep 21, 2008 #3
    The square root sign is over the (8x^3 + 5x + 10)
     
  5. Sep 21, 2008 #4
    = lim SQRT ((8x^3 + 5x + 10) / x^4)
    does this help?
     
  6. Sep 22, 2008 #5
    1 < x^5/8 - 5/8x^2 is true for large values of x (notice that the second term goes to 0 x approaches infinite). So we then have:


    8x^2 + 5 < x^7

    And because of that, we have (because the term 10/x goes to 0 as x approaches infinite):
    8x^2 + 10/x + 5 < x^7

    x^2 < x^7/8 - 10/8x - 5/8

    8x^2 < x^7 - 10/x - 5

    8x^2 + 5 < x^7 - 10/x

    x(8x^2 + 5) < x^8 - 10

    8x^3 + 5x + 10 < x^8

    SQRT ((8x^3 + 5x + 10)< x^4



    So, we see that the denominator goes to infinite faster than the numerator. What is the limit then?
     
  7. Sep 22, 2008 #6
    If you can get f(x)=(8x^3 + 5x + 10) / x^4 ->0, as x->+[tex]\infty[/tex] with no problem, then Sqrt(f(x))-> Sqrt(0), as x->+[tex]\infty[/tex]
    In this case you may want to show that if f(x)>=0 and f(x)->L, then Sqrt(f(x))->Sqrt(L), as x->p , where p is a limit point of it's domain.

    There should be many other ways to explain the limit, based on the continuity of sqrt(x) maybe.
     
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