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Thread: How do you find horizontal asymptotes?

  1. #1
    s3a
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    How do you find horizontal asymptotes?

    How do I find the horizontal asymptotes in this (or any) limit? I attached the question as well as my work but I am lost with horizontal asymptotes so my work might be meaningless and I would appreciate any detailed explanation.

    Thanks in advance!
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    Quote Originally Posted by s3a View Post
    How do I find the horizontal asymptotes in this (or any) limit? I attached the question as well as my work but I am lost with horizontal asymptotes so my work might be meaningless and I would appreciate any detailed explanation.

    Thanks in advance!
    The horizontal asymptotes of this function will be the vertical asymptotes of the inverse function.

    So, the inverse function is defined by

    $\displaystyle x = \frac{17y}{(y^4 + 1)^{\frac{1}{4}}}$

    $\displaystyle x^4 = \frac{83521y^4}{y^4 + 1}$

    $\displaystyle x^4 = 83521 - \frac{83521}{y^4 + 1}$

    $\displaystyle x^4 - 83521 = - \frac{83521}{y^4 + 1}$

    $\displaystyle \frac{1}{x^4 - 83521} = -\frac{y^4 + 1}{83521}$

    $\displaystyle -\frac{83521}{x^4 - 83521} = y^4 + 1$

    $\displaystyle -\frac{83521}{x^4 - 83521} - 1 = y^4$

    $\displaystyle \left(-\frac{83521}{x^4 - 83521} - 1\right)^{\frac{1}{4}} = y$.

    So the vertical asymptotes of the inverse function are $\displaystyle x = -17$ and $\displaystyle x = 17$.


    Thus, the horizontal asymptotes of the original function are $\displaystyle y = -17$ and $\displaystyle y = 17$.
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  3. #3
    s3a
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    What am I doing wrong? (Work attached)

    (Ignore the black handwriting)
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  4. #4
    s3a
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    Oh I get it! x^4 = +/- x^(1/4) that's why x = 17 and x = -17 as well.
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