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Homework Help: Cubed root denominator limit question

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate lim x->8 (x-8)/(cubed root of x) -2)


    2. Relevant equations



    3. The attempt at a solution
    I multiplied both numerator and denominator by (cubed root of x) +2. In the denominator, I wrote it as ((x^1/3)-2)*((x^1/3)+2) which simplifies to (x^2/3)-4. I have ((x-8)*((cubed root of x) +2))/((x^2/3)-4). The answer should be 12 but I cannot reach this answer as I get an answer of 0 in the denominator
     
  2. jcsd
  3. Feb 14, 2010 #2
    Use L'Hôpital's rule...
     
  4. Feb 14, 2010 #3
    which is?
     
  5. Feb 14, 2010 #4
  6. Feb 14, 2010 #5
    I don't understand how to use it
     
  7. Feb 14, 2010 #6
    I found this question on another website and they factored x-8 which cancelled the ((x^1/3)-8) in the denominator and then you plug 8 in to the rest of the numerator to get 12. Is it not possible to just multiply numerator and denominator by (x^1/3)+2?
     
  8. Feb 14, 2010 #7
    Okay, as both the as both the numerator and denominator in your function equal 0, when x = 8, you end up with 0/0 which is undefined, so we must use L'Hôpital's rule. Differentiating the numerator with respect to (wrt) x and the numerator wrt x we get.

    [tex]\frac{1}{\frac{1x^{\frac{-2}{3}}}{3}}[/tex] = [tex] \frac{3}{x^{\frac{-2}{3}}} [/tex]

    when x = 8 the above expression equals 12.

    I hope this example helps you understand L'Hôpital's rule, which is a very powerful mathematical tool for functions. Take sinc x as another example which is (sin x)/x. As x approaches 0 sinc x approaches 0/0, but using L'Hôpital's rule, we can say that when x = 0 sinc 0 = 1.
     
  9. Feb 14, 2010 #8
    If you had [itex]\sqrt{x} - 2[/itex] with a square root, you could multiply by its conjugate [itex]\sqrt{x} + 2[/itex]. However, you have [itex]\sqrt[3]{x} - 2[/itex] with a cube root which don't really have a conjugate.

    So all you can do without using l'Hôpital's rule from Calc II is use the fact that x3 - y3 = (x - y)(x2 + xy + y2) and factor x - 8, or multiply [itex]\sqrt[3]{x} - 2[/itex] by something that will turn it into x - 8.
     
  10. Feb 14, 2010 #9
    do as bohrok suggests note that x-8 = (x^(1/3))^3 - 2^3 and use the formula he wrote
     
  11. Feb 14, 2010 #10
    That is an acceptable way for this problem, but some expressions will not factor out as easily as this. That is why I have emphasised learning L'Hôpital's rule as it is an easy way to find the limit for all most all expressions.
     
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