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Math Help - Need help on limits......

  1. #1
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    Need help on limits......

    HELP!

    \lim_{x\to 1} \frac{\sqrt[3]{x} - 1}{x-1} = ????
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    HELP!

    \lim_{x\to 1} \frac{\sqrt[3]{x} - 1}{x-1} = ????
    By direct substitution we have that the limit looks like \frac{0}{0} so we may use L'Hopital's rule:

    \lim_{x \to 1} \frac{\sqrt[3]{x} - 1}{x-1} = \lim_{x \to 1} \frac{ \frac{1}{3} x^{-2/3} }{1} = \frac{1}{3} 1^{-2/3} = \frac{1}{3}

    If that doesn't work for you, (this is gonna be weird!) note that we may write x - 1 as the difference between two cubes:
    x - 1 = \left ( x^{1/3} \right )^3 - 1^3

    And
    \left ( x^{1/3} \right )^3 - 1^3 = \left ( x^{1/3} - 1 \right ) \left ( x^{2/3} + x^{1/3} + 1 \right )

    So
    \lim_{x \to 1} \frac{\sqrt[3]{x} - 1}{x-1} = \lim_{x \to 1} \frac{x^{1/3} - 1}{\left ( x^{1/3} - 1 \right ) \left ( x^{2/3} + x^{1/3} + 1 \right )}

    = \lim_{x \to 1} \frac{1}{x^{2/3} + x^{1/3} + 1}

    = \frac{1}{1^{2/3} + 1^{1/3} + 1} = \frac{1}{3}

    -Dan
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    HELP!

    \lim_{x\to 1} \frac{\sqrt[3]{x} - 1}{x-1} = ????
    ---
    Rather, you can factor,
    x-1=(\sqrt[3]{x})^3-1^3=(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)

    Thus,
    \lim_{x\to 1}\frac{\sqrt[3]{x}-1}{(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}
    Thus,
    \lim_{x\to 1}\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}
    Which is,
    \frac{1}{3}
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