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Math Help - Need help over a problem with horizontal assymptote

  1. #1
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    Need help over a problem with horizontal assymptote

    Hey,
    Basically I have attempted a solution to my problem, it did not get the correct answer and I cannot figure out how to get the correct answer. Looking for where I am making my mistake and how to correct it!

    Question
    Find the limit
     \lim_{x\to -\infty} (x + \sqrt{x^2 + 2x})

    Attempted solution:
    *pretend that  \lim_{x\to -\infty} is in front, I am making a mess trying to coordinate this with that and it's just easier to show without..

     (x + \sqrt{x^2 + 2x})* \frac{(x - \sqrt{x^2 + 2x})}{(x - \sqrt{x^2 + 2x})} ***Multiplied by conjugate
     \frac{-2x}{x - \sqrt{x^2 + 2x}}
    The highest value of x is "x" so I divide that through the numerator and denominator and get:
     \frac{-2}{1 - \sqrt{1}}
    Which is giving me 0 on the bottom. I do not know how to fix this and was hoping someone could give me a hand here.

    Thanks alot!
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  2. #2
    MHF Contributor chisigma's Avatar
    Joined
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    near Piacenza (Italy)
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    Take into account that is...

    \displaystyle \lim_{x \rightarrow - \infty} x + \sqrt{x^{2} + 2 x} = \lim_{x \rightarrow + \infty} \sqrt{x^{2} - 2 x} - x

    Kind regards

    \chi \sigma
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