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Homework Help: Lim x->infinity

  1. Feb 9, 2010 #1
    find the values for A,B such that

    Limx->infinity x*((x3+x2+ax)(1/3)-(x3-bx)(1/3)) = 3

    what i thought was

    t=1/x

    Limt->0 1/t*((1/t3+1/t2+a/t)(1/3)-(1/t3-b/t)(1/3)) = 3

    Limt->0 1/t*((1/t3+1/t2+a/t)(1/3)-(1/t3-b/t)(1/3)) = 3

    since here we have 0/0 i can use l'hopital's law, but it looks like its going to get really ugly whith too manu terms,

    also how can i solve for both A and B when i have only one equation
     
  2. jcsd
  3. Feb 9, 2010 #2

    LCKurtz

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    Before you let x --> oo, take the natural log of both sides. I think a little algebra will take you farther than L'Hospital's rule. At first glance, I would say you will have a hard time finding any value of a and b that work. But perhaps "none" is the answer.
     
  4. Feb 9, 2010 #3
    what will logs help, i have subtraction between the 2 roots
     
  5. Feb 9, 2010 #4

    LCKurtz

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    [tex]\ln{(A-B)} = \ln{\left (\frac A B\right)}[/tex]
     
  6. Feb 9, 2010 #5
    No no,
    log(a) - log(b) = log(a/b)
     
  7. Feb 9, 2010 #6

    LCKurtz

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    Aaargh!! Forgot to engage brain. In a danger zone. Still reeling from Super Bowl sunday. What can I say?
     
  8. Feb 10, 2010 #7
    I approached the question by first factoring out x1/3 from the bracketted expression. with the remaining surdic expression, i rationalised the numerator in terms of cube roots so the numerator i would have
    [(x2 + x + a) - (x2 - b)] = x + a + b in it.
    I rearranged the x4/3 in the numerator to 1/(1/x4/3) so i would have as the denominator f(x)/x4/3
    where f(x) was the expression i used to rationalise the numerator.

    using lim (f(x)/g(x)) = lim(f(x))/lim(g(x)), i evaluated the limit for the denominator and got 3 as the result.

    so that means lim (x->infinity) (x + a + b) = 9

    all linear functions i know of tend towards infinity as x tends towards infinity. thus, the limit is impossible no matter what value of a or b. at least that's what i got.
     
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