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Factoring Involving Radicals

  1. Feb 22, 2007 #1
    1. The problem statement, all variables and given/known data
    This question is concerning a limit question. I have no problems finding limits but i need to be able to factor this equation. I need to be able to some how get rid of the (x-1)


    3. The attempt at a solution
    I ust keep ending up in a loop geting a bigger and bigger radical.
    An out of curiosity can 0 be divided by 0?
  2. jcsd
  3. Feb 23, 2007 #2

    Gib Z

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    Homework Helper

    Nope. Give us the actual limit, theres an easier way.
  4. Feb 23, 2007 #3


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    Staff Emeritus
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    No, you cannot divide by 0- not even 0! You should know, from a more general formula, that [itex]y^6- 1= (y- 1)(y^5+ y^4+ y^3+ y^2+ y+ 1)[/itex]. What happens if y= x^{1/6}?
  5. Feb 23, 2007 #4
    Then all thats left is to create a new limit as y approaches something. In this case its still 1 but you would need to take the 6th root of the original limit point

    For example, if you wanted to do something like the limit as x approaches 8 of:

    [itex]\displaystyle{\frac{-x^2+ 4x^\frac{4}{3}+ x^\frac{7}{8}+ x^\frac{5}{6}- 4x^\frac{1}{2}- 4x^\frac{1}{6}}{x^\frac{2}{3}- 4}}[/itex]

    You would have to again use y=x^(1/6), but take the new limit as y approaches 8^(1/6) or radical 2 of:
    [tex]\displaystyle{\frac{-y^12+ 4y^8+ y^7+ y^5- 4y^3- 4y}{y^4- 4}}[/tex]
    Last edited: Feb 23, 2007
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