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Math Help - Grade 12 calculas limits question

  1. #1
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    Grade 12 calculas limits question

    lim x->1



    solve algebraeically...ive tried it a bunch of times and always seem to head nowhere
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  2. #2
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    Re: Grade 12 calculas limits question

    Quote Originally Posted by koolaid123 View Post
    lim x->1

    solve algebraeically...ive tried it a bunch of times and always seem to head nowhere
    Simple algebra will show that:
    \frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}.
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  3. #3
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    Re: Grade 12 calculas limits question

    Hello, koolaid123!

    \lim_{x\to1}\frac{\sqrt[3]{x} - 1}{\sqrt{x}-1}

    \text{Multiply by }\frac{\sqrt{x}+1}{\sqrt{x}+1}\,\text{ and }\,\frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}

    . . \lim_{x\to1}\frac{\sqrt[3]{x} - 1}{\sqrt{x}-1}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} \cdot \frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}

    . . =\;\lim_{x\to1} \frac{(\sqrt[3]{x} - 1)(\sqrt[3]{x^2} + \sqrt[3]{x} + 1)}{(\sqrt{x}-1)(\sqrt{x}+1)} \cdot\frac{\sqrt{x}+1}{\sqrt[3]{x^2} + \sqrt[3]{x}+1}

    . . =\;\lim_{x\to1}\frac{x-1}{x-1}\cdot\frac{\sqrt{x}+1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}

    . . =\;\lim_{x\to1}\frac{\sqrt{x}+1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1} \;=\;\frac{1+1}{1+1+1} \;=\;\frac{2}{3}
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