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Finding a Limit Using L'Hopital's Rule

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    lim x→∞ of ((x^3+2x-∏)^(1/3))-x


    2. Relevant equations
    L'Hopital's Rule: If it is an indeterminate form, take the derivative of the top and bottom until you do not get an indeterminate form anymore.


    3. The attempt at a solution

    lim x→∞ of ((x^3+2x-∏)^(1/3))-x

    lim x→∞ (((x^3+2x-∏)^(1/3))-1)/(x^3)/(1/x)

    lim x→∞ (1/3(1+2/x^2-∏/x^3)^(2/3))/(-1/x^2)

    I do not know how to continue from here. I think I'm missing something in my 3rd step (multiplying by the derivative of the inside function?) Any help would be awesome!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 11, 2012 #2

    Mark44

    Staff: Mentor

    This is not accurate. L'Hopital's Rule doesn't apply to just any old indeterminate form - only the forms [0/0] or [±∞/∞].

    What you have above is the indeterminate form [∞ - ∞], so L'H doesn't apply.

    What I would do is to multiply by 1 in the form of something over itself. The "something" would be whatever is needed to come up with the difference of cubes.
     
  4. Apr 11, 2012 #3
    Oh okay I understand that. Thanks! I think I've got the right answer now
     
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