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Math Help - Horizontal Asymptotes

  1. #1
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    Horizontal Asymptotes

    f(x)= 2x/(x^2 + 2)^(1/2).

    How do I find the negative infinity limit part of the problem? I found one Horizontal asymptote at x =2, how do I find the other?
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  2. #2
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    follow these rules

    This procedure leads to the following general result:

    * 1 * If the degree of the denominator exceeds that of the numerator, y = 0 is a horizontal asymptote;
    * 2 * If the leading coefficient of the numerator is a and the leading coefficient of the denominator is b and their degrees are equal then is a horizontal asymptote;

    * 3 * If the degree of the numerator exceeds that of the denominator, there is no horizontal asymptote.


    http://www.math.csusb.edu/math110/src/rationals/hasym.html
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  3. #3
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    How do I find the negative limit though?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by unluckykc View Post
    f(x)= 2x/(x^2 + 2)^(1/2).

    How do I find the negative infinity limit part of the problem? I found one Horizontal asymptote at x =2, how do I find the other?
    as x goes to infinity, the 2 in the denominator essentially does not matter. thus we have that:

    \lim_{x \to \pm \infty} \frac {2x}{\sqrt{x^2 + 2}} = \lim_{x \to \pm \infty} \frac {2x}{\sqrt{x^2}} =  \lim_{x \to \pm \infty} \frac {2x}{|x|} = \left \{ \begin{array}{lr} 2 & \mbox{ as } x \to \infty \\ -2 & \mbox { as } x \to - \infty \end{array} \right.
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