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Math Help - [SOLVED] Find the Limit

  1. #1
    Raj
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    [SOLVED] Find the Limit



    Having trouble arranging the equation so I can find the limit. I tried getting a lcd for the top part, multiplying by the reciprocal and ended with a mess. help appreciated.
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  2. #2
    MHF Contributor
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    Hi

    Let f(x) = \frac{1}{\sqrt{x}}

    Your limit is \lim_{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}

    It is therefore equal to f'(4)
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  3. #3
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    Hello, Raj!

    \lim_{x\to4}\frac{\frac{1}{\sqrt{x}} - \frac{1}{2}}{x-4}

    Multiply top and bottom by 2\sqrt{x}

    . . \frac{2\sqrt{x}\left(\frac{1}{\sqrt{x}} - \frac{1}{2}\right)}{2\sqrt{x}(x - 4)} \;=\;\frac{2-\sqrt{x}}{2\sqrt{x}(x - 4)}


    Multiply top and bottom by (2+\sqrt{x})

    . . \frac{2-\sqrt{x}}{2\sqrt{x})x-4)}\cdot\frac{2+\sqrt{x}}{2+\sqrt{x}} \;=\;\frac{4-x}{2\sqrt{x}(x-4)(2+\sqrt{x})}   \;=\;\frac{-(x-4)}{2\sqrt{x}(x-4)(2+\sqrt{x})} . =\;\frac{-1}{2\sqrt{x}(2 + \sqrt{x})}


    Then: . \lim_{x\to4}\left[\frac{-1}{2\sqrt{x}(2+\sqrt{x})}\right] \;=\;\frac{-1}{2\sqrt{4}(2 + \sqrt{4})} \;=\;\frac{-1}{2\cdot2\cdot4} \;=\;-\frac{1}{16}

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