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Math Help - factoring squareroots in limits

  1. #1
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    factoring squareroots in limits

    i have been stuck on this problem for awhile and would love some help.

    lim ((1/sqrt(x))-(1/sqrt(x^2+x))
    x-->0+

    thanks ahead of time

    -Dan
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  2. #2
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    \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2+x}} = \frac{\sqrt{x^2+x} - \sqrt{x}}{\sqrt{x} \sqrt{x^2+x}}  \cdot \frac{\sqrt{x^2+x} + \sqrt{x}}{\sqrt{x^2+x} + \sqrt{x}}

    \frac{x^2}{\sqrt{x} \cdot \sqrt{x} \cdot \sqrt{x+1} \cdot \sqrt{x} \cdot (\sqrt{x+1} + 1)}
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  3. #3
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    Quote Originally Posted by casesam View Post
    i have been stuck on this problem for awhile and would love some help.

    lim ((1/sqrt(x))-(1/sqrt(x^2+x))
    x-->0+

    thanks ahead of time

    -Dan
    Consider:


    \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2 + x}} = \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x} \sqrt{x + 1}} = \frac{\sqrt{x + 1}}{\sqrt{x} \sqrt{x + 1}} - \frac{1}{\sqrt{x} \sqrt{x + 1}}


    = \frac{\sqrt{x + 1} - 1}{\sqrt{x} \sqrt{x + 1}} = \frac{(\sqrt{x + 1} - 1)}{\sqrt{x} \sqrt{x + 1}} \times \frac{(\sqrt{x + 1} + 1)}{(\sqrt{x + 1} + 1)} = \frac{(x + 1) - 1}{(\sqrt{x} \sqrt{x + 1} (\sqrt{x + 1} + 1)}

    = \frac{x}{(\sqrt{x} \sqrt{x + 1} (\sqrt{x + 1} + 1)} = \frac{\sqrt{x}}{\sqrt{x + 1} (\sqrt{x + 1} + 1)}.


    Therefore ......
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  4. #4
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    Hello, Dan!

    Wow . . . This is an ugly one!


    \lim_{x\to0^+} \left(\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2+x}}\right)
    The second fraction is: . \frac{1}{\sqrt{x(x+1)}} \;=\;\frac{1}{\sqrt{x}}\cdot\frac{1}{\sqrt{x+1}}

    So we have: . \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x}}\cdot\frac{1}{\sqrt{x+1}} \;=\;\frac{1}{\sqrt{x}}\left(1 - \frac{1}{\sqrt{x+1}}\right) \;=\;\frac{1}{\sqrt{x}}\left(\frac{\sqrt{x+1} - 1}{\sqrt{x+1}}\right)


    Multiply top and bottom by (\sqrt{x+1} + 1)

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

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


    Hence: . \lim_{x\to0^+}\left[\frac{\sqrt{x}}{\sqrt{x+1}(\sqrt{x+1} + 1)}\right] \;=\;\frac{\sqrt{0}}{\sqrt{1}(\sqrt{1} + 1)} \;=\;\frac{0}{2} \;=\;0

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  5. #5
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    Rationalizing the numerator. Gets 'em every time!

    -Dan
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