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Math Help - help finding the limit..

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    35

    help finding the limit..

    i need to find the limit as x -> negative infinity for the function

    (sqrt(x^2-2x))/(2x+7)
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  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Multiply by \frac{\tfrac{1}{x}}{\tfrac{1}{x}} to get

    \frac{\sqrt{x^2-2x}}{2x+7} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}}

    = \frac{\sqrt{\frac{x^2-2x}{x^2}}}{2 + \frac{7}{x}}

    = \frac{\sqrt{1-\frac{2}{x}}}{2 + \frac{7}{x}}
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