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Math Help - Limit

  1. #1
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    Limit

    Hi,

    I don't know how to calculate \lim_{x\to +\infty} \frac{\sqrt{x-1}-\sqrt[6]{x}}{\sqrt[3]{x-1}-\sqrt{x}} .

    Can you help me please????
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  2. #2
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    Quote Originally Posted by bhitroofen01 View Post
    Hi,

    I don't know how to calculate \lim_{x\to +\infty} \frac{\sqrt{x-1}-\sqrt[6]{x}}{\sqrt[3]{x-1}-\sqrt{x}} .

    Can you help me please????

    try using factorisation.....
    use for both numerator n denominator
    Last edited by amul28; April 3rd 2010 at 06:11 AM. Reason: or....use L-Hospital's rule n differentiate the limit
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  3. #3
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    Hello, bhitroofen01!

    evaluate: . \lim_{x\to \infty} \frac{\sqrt{x-1}-\sqrt[6]{x}}{\sqrt[3]{x-1}-\sqrt{x}}

    We have: . \frac{(x-1)^{\frac{1}{2}} - x^{\frac{1}{6}}} {(x-1)^{\frac{1}{3}} - x^{\frac{1}{2}}}

    \text{Divide top and bottom by }x^{\frac{1}{2}}:\quad  \frac{\dfrac{(x-1)^{\frac{1}{2}}}{x^{\frac{1}{2}}} - \dfrac{x^{\frac{1}{6}}}{x^{\frac{1}{2}}}}  {\dfrac{(x-1)^{\frac{1}{3}}}{x^{\frac{1}{2}}} - \dfrac{x^{\frac{1}{2}}}{x^{\frac{1}{2}}}}

    . . . =\;\frac{\left(\dfrac{x-1}{x}\right)^{\frac{1}{2}} - \dfrac{1}{x^{\frac{1}{3}}}}{ \dfrac{(x-1)^{\frac{1}{3}}}{x^{\frac{1}{6}}\cdot x^{\frac{1}{3}}} - 1} . =\;\; \frac{\left(\dfrac{x-1}{x}\right)^{\frac{1}{2}} - \dfrac{1}{x^{\frac{1}{3}}}} {\dfrac{1}{x^{\frac{1}{6}}}\left(\dfrac{x-1}{x}\right)^{\frac{1}{3}} - 1} . =\;\; \frac{\left(1-\dfrac{1}{x}\right)^{\frac{1}{2}} - \dfrac{1}{x^{\frac{1}{3}}}}  {\dfrac{1}{x^{\frac{1}{6}}}\left(1 - \dfrac{1}{x}\right)^{\frac{1}{3}} - 1}


    Therefore: . \lim_{x\to\infty}\left[ \frac{\left(1-\dfrac{1}{x}\right)^{\frac{1}{2}} - \dfrac{1}{x^{\frac{1}{3}}}}  {\dfrac{1}{x^{\frac{1}{6}}}\left(1 - \dfrac{1}{x}\right)^{\frac{1}{3}} - 1} \right] . =\;\;\frac{(1-0)^{\frac{1}{2}} - 0}{0(1-0)^{\frac{1}{3}} - 1} \;\;=\;\;\frac{1}{-1} \;\;=\;\;-1

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