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Vertical and Horizontal Asymptotes

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data
    What are the vertical and horizontal asymptotes for:
    2x2/ SQRT(x4-81)?

    2. Relevant equations

    3. The attempt at a solution
    For one VA, I got x=3. Are there any more? I am troubled with this square root on the bottom.
  2. jcsd
  3. Sep 14, 2009 #2


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    Practically, what the square root on the bottom does is two things:
    1) It limits where the function can be defined. If x4>81, then the function is not defined. It also has the potential to make asymptotic behavior different... an example is something like

    [tex] \frac{x-1}{ \sqrt{(x^4-1)(x-1)}}[/tex]

    At first glance it appears to have an asymptote at x=1 and x=-1, but the x-1 at the top cancels with the x-1 in the denominator (you can see this by factoring the denominator) in the limit as x approaches 1, so no asymptotic behavior occurs.

    What's required for a zero in the denominator to cause a vertical asymptote is for there to be no canceling zeroes, or not enough canceling zeroes at least, in the numerator. Having a square root can effect this as it effectively "halves" the power of the zero in the denominator.

    You definitely have at least one more vertical asymptote. For what x is x4-81=0? You should be able to fully solve this (either by factoring or by noticing you can set x2=y and solving y2-81=0 first)
  4. Sep 15, 2009 #3
    You got x4-81=x4-34=(x2-32)(x2+32)

    Can you determine the other values for x for which the denominator is 0 ?
    Last edited: Sep 15, 2009
  5. Sep 15, 2009 #4


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    For very large x (but negative since x cannot be larger than [itex]\sqrt[4]{81}[/itex]) we can ignore the "81" in comparison with the "x4". Then [itex]2x^2/\sqrt{x^4- 81}[/itex] is approximately [itex]2x^2/\sqrt[4]{x^4}[/itex].
  6. Sep 15, 2009 #5


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    Oh? So x must be less than [itex]\sqrt[4]{81}[/itex]? I didn't know we were searching for complex solutions as well :tongue2: Please consider revising this sentence.

    [itex]x^4[/itex] is positive for all real x, not just negative x.

    Therefore, the domain is only defined where [tex]\sqrt{x^4-81}> 0[/tex]

    If you know how to solve this, it leaves x<-3, x>3

    Just to fix the denominator up slightly: for large [itex]\pm x[/itex] the function tends towards [tex]y=\frac{2x^2}{\sqrt{x^4}}[/tex] so you can simplify this and this will give you the horizontal asymptote :smile:
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