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How do I find the horizontal asymptotes of the following equation?

$$ \frac{1}{\sqrt{x^2-2x}-x} $$

I'm having most trouble with the algebra behind it as I'm not so used to this field of math just yet. An extra detailed way of solving it would be of great appreciation.

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    hint: rationalize the denominator – 2017-02-06

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Hint

$$y=\frac{1}{\sqrt{x^2-2x}-x}=\frac{1}{\sqrt{x^2-2x}-x}\times \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x}=\frac{\sqrt{x^2-2x}+x}{x^2-2x-x^2}$$ $$y=-\frac 12 \frac{\sqrt{x^2-2x}+x}x$$

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    I can't seem to get through these kind of calculations, today, sorry. I know that I'm supposed to find lim x -> ∞ and x -> -∞ of these but I don't know how to proceed. Could you possibly show me? Thank you very much for the answer by the way! – 2017-02-06
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    @André. Just one more hint : $\sqrt{1-\frac 2x}\approx 1-\frac 1x$ when $x$ is large. – 2017-02-06