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Math Help - limit question

  1. #1
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    limit question

    i know the lim as x approaches 1 of \frac{\sqrt{x+3}-\sqrt{3x+1}}{x-1}=-\frac{1}{2}

    however i don't know the steps to get there. Can someone help he? Thanks
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  2. #2
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    \frac{\sqrt{x+3}-\sqrt{3x+1}}{x-1} \cdot \frac{\sqrt{x+3}+\sqrt{3x+1}}{\sqrt{x+3}+\sqrt{3x+  1}} =

    \frac{(x+3) - (3x+1)}{(x-1)[\sqrt{x+3}+\sqrt{3x+1}]} =<br />

    \frac{2(1 - x)}{(x-1)[\sqrt{x+3}+\sqrt{3x+1}]} =<br />

    \frac{-2}{\sqrt{x+3}+\sqrt{3x+1}}

    now let x \to 1
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  3. #3
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    thanks man...u da man
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