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Math Help - Horizontal Asymptotes

  1. #1
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    Horizontal Asymptotes

    Let f be the function given by f(x) =
    <br /> <br />
\frac {x}{sqrt(x^2 - 4)}<br />

    write an equation for each horizontal asymptote.

    How do I find these without graphing it on my calc?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by ny_chow View Post
    Let f be the function given by f(x) =
    <br /> <br />
\frac {x}{sqrt(x^2 - 4)}<br />

    write an equation for each horizontal asymptote.

    How do I find these without graphing it on my calc?
    Horizontal tangents occur when the derivative is eqal to zero.

    So we need to take the derivative

    f'(x)=\frac{\sqrt{x^2-4}-x\frac{1}{2}\cdot (x^2-4)^{-\frac{1}{2}}2x}{x^2-4}

    Factoring out a (x^2-4)^{-\frac{1}{2}} from the numerator we get

    (x^2-4)^{-\frac{1}{2}}\left( \frac{(x^2-4)-x^2}{x^2-4}\right)=\frac{-4}{(x^2-4)^{\frac{3}{2}}}

    Since this is never equal to zero the function has no horizontal tangents
    Last edited by TheEmptySet; May 2nd 2009 at 07:39 PM. Reason: did the wrong problem but I will leave it :(
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    sorry I answered the wrong question

    I found the horiztonal tangents (or lack there of)
    To find the horizontal asymptote take the limit as x goes to infinity and negative infinity

    \lim_{x \to \infty}\frac{x}{\sqrt{x^2}-4}=\frac{1}{\sqrt{1-\frac{4}{x^2}}}=1


    \lim_{x \to -\infty}\frac{x}{\sqrt{x^2}-4}=\frac{1}{\sqrt{1-\frac{4}{x^2}}}=-1

    so they are y= \pm 1
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