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Math Help - Limit

  1. #1
    Senior Member Stroodle's Avatar
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    Limit

    Hi there,

    How would I evaluate

    \lim_{b\rightarrow\infty}\frac{a+b-\sqrt{a^2-ab+b^2}}{6}

    Thanks for your help.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Stroodle View Post
    Hi there,

    How would I evaluate

    \lim_{b\rightarrow\infty}\frac{a+b-\sqrt{a^2-ab+b^2}}{6}

    Thanks for your help.
    Hi Stroodle,

    \frac{a+b-\sqrt{a^2-ab+b^2}}{6}=\frac{a+b-\sqrt{\left(b-\frac{a}{2}\right)^2+\frac{3a^2}{4}}}{6}

    As b\ \rightarrow\ \infty this approaches \frac{a+b-\sqrt{\left(b-\frac{a}{2}\right)^2}}{6}

    =\frac{a+b-b+\frac{a}{2}}{6}=\frac{3a}{12}
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  3. #3
    Senior Member Stroodle's Avatar
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    Nice one. Thanks for that.
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  4. #4
    Eater of Worlds
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    Rewrite it as:

    \frac{1}{6}\lim_{b\to {\infty}}\frac{a^{2}-ab}{-b-\sqrt{a^{2}-ab+b^{2}}}+\frac{a}{6}

    Now, divide the top by b, the b in the bottom by b and the radical by \sqrt{b^{2}}

    \frac{1}{6}\lim_{b\to {\infty}}\frac{\frac{a^{2}}{b}-\frac{ab}{b}}{\frac{-b}{b}-\sqrt{\frac{a^{2}}{b^{2}}-\frac{ab}{b^{2}}+\frac{b^{2}}{b^{2}}}}+\frac{a}{6}

    Then, it whittles down to:

    \frac{1}{6}\lim_{b\to {\infty}}\frac{-a}{-2}+\frac{a}{6}

    \frac{a}{12}+\frac{a}{6}=\frac{a}{4}
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