1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook