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Math Help - Need HELP with very tricky limit problem!

  1. #1
    dahbch
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    Need HELP with very tricky limit problem!

    I need to find the limit using algebra of the limit as x approaches 1 of (x^(2/3)-1) / (x-1). I know from using L'Hospital's Rule that the answer is 2/3, however, I am not allowed to use that rule in the class, so does anybody know how to solve this with algebra. I would definitely appreciate the help.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dahbch
    I need to find the limit using algebra of the limit as x approaches 1 of (x^(2/3)-1) / (x-1). I know from using L'Hospital's Rule that the answer is 2/3, however, I am not allowed to use that rule in the class, so does anybody know how to solve this with algebra. I would definitely appreciate the help.
    Hang on to your hats, folks. We're going to play!

    Take a look at the numerator. This can be thought of as the difference between the square of two numbers, x^{1/3} and 1. So
    x^{2/3} - 1 = (x^{1/3} - 1)(x^{1/3} + 1)

    In a similar fashion, the denominator is the difference between two cubes: x^{1/3} and 1. The factorization here is a^3 - b^3 = (a - b)(a^2 + ab + b^2) so
    x - 1 = (x^{1/3} - 1)(x^{2/3} + x^{1/3} + 1)

    Putting these together gives:
    \frac{x^{2/3} - 1}{x - 1} = \frac{(x^{1/3} - 1)(x^{1/3} + 1)}{(x^{1/3} - 1)(x^{2/3} + x^{1/3} + 1)} = \frac{x^{1/3} + 1}{x^{2/3} + x^{1/3} + 1}

    And now you can take your limit.

    -Dan

    BTW Thank you for this problem. I haven't had this much fun doing a limit in ages!
    Last edited by topsquark; September 1st 2006 at 05:19 PM. Reason: Addendum
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  3. #3
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    Hello, topsquark!

    Great insight . . . lovely and elegant!

    Nice work, Dan!

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban
    Hello, topsquark!

    Great insight . . . lovely and elegant!

    Nice work, Dan!

    (I'm blushing!)

    -Dan
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