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Math Help - Limit as X approaches infinity

  1. #1
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    Limit as X approaches infinity

    Hi,
    I was given some review questions for a test and one of them is
    Lim as(x->infinity) of (2x-(4x^2+2x)^1/2). I've tried multiplying by the reciprocal but it didn't seem to get me anywhere; I'm not really sure how to approach the problem in any other way. Is there something I'm missing?
    Thanks!
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  2. #2
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    Quote Originally Posted by geliot View Post
    Hi,
    I was given some review questions for a test and one of them is
    Lim as(x->infinity) of (2x-(4x^2+2x)^1/2). I've tried multiplying by the reciprocal
    Did you get \dfrac{-2x}{2x+\sqrt{4x^2+2x}}?

    You should have. Then divide through by 2x.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Did you get \dfrac{-2x}{2x+\sqrt{4x^2+2x}}?

    You should have. Then divide through by 2x.
    oh wow I'm feeling pretty dumb right now. I forgot to square the 2x >.>
    Thanks so much!
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    but sorry actually, now that I'm looking at it, how do I pull the 2x out of the radical? Looking at the answer, the limit approaches -.5, but I don't see how one would end up with a 2 on the bottom of the fraction...sorry, I'm sure i'm just missing something obvious here
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  5. #5
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    Quote Originally Posted by geliot View Post
    but sorry actually, now that I'm looking at it, how do I pull the 2x out of the radical? Looking at the answer, the limit approaches -.5, but I don't see how one would end up with a 2 on the bottom of the fraction...sorry, I'm sure i'm just missing something obvious here
    \displaystyle \frac{-2x}{2x+\sqrt{4x^2+2x}} =  \frac{-2}{2 + \frac{\sqrt{4x^2+2x}}{x}} = \frac{-2}{2 + \sqrt{\frac{4x^2+2x}{x^2}}} = ....
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