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Why does this expression not have a vertical asymptote?

  1. Jul 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the vertical asymptote, if there is one, of this rational function.

    [tex]\frac{\sqrt{16x^2 + 3x + 6} -5}{x-1}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    This was actually a Calculus problem, I had to find the limit at infinity. I was able to do that easily; an extra question asked what the vertical asymptote was. I see an obvious V.A. at [tex]x = 1[/tex] but the answer was that there isn't one. What am I doing wrong? I feel as I'm just overlooking the obvious. Canceling out doesn't change the fact that a vertical asymptote exists, correct? Plus I don't really see a way to simplify the function, other than multiplying by the conjugate, but that wasn't the point of the problem. I think there's a more obvious way of getting the answer.
     
    Last edited: Jul 10, 2011
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  3. Jul 10, 2011 #2

    phinds

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    what I see right off is that as x approaches 1, the value approaches 5, not infinity, so I don't see an asymtote there, but I can't give you a rigorous proof.
     
  4. Jul 10, 2011 #3

    SammyS

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    Wrong.
    What did you get after the "cancelling out" occurred?
     
  5. Jul 10, 2011 #4
    Multiplying by the conjugate didn't affect anything. I don't think simplifying it is the correct way to get the answer.

    I realize now that I was wrong about the canceling out part. For some reason I was confusing discontinuities with vertical asymptotes.

    Just to clarify, I was always told that if you have the function [tex]\frac{x^2-1}{x-1}[/tex], the correct graph would include the removable discontinuity at x = 1 even though the function can be simplified. Is that true?
     
  6. Jul 10, 2011 #5

    Hurkyl

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    Yes and no.

    It's probable that, for what you actually mean by the symbols, the answer is "yes". (and the answer is certainly yes for whoever told you that)


    Sometimes, however, people use the same symbols for different purposes. The two most important variations are:
    1. They are working with polynomials rather than polynomial functions. (The distinction is rather subtle) If you take the quotient first, and only then convert into a function, the result doesn't have any removable singularities.
    2. Sometimes people use / to mean "divide and then remove all removable singularities" rather than just "divide"


    Another important variation is to simply treat functions that differ only at a finite number of points* as if they were the same function. Of course, in this context, it doesn't make any difference whether or not you include a removable singularity at some point, because both graphs are treated as if they were the same.

    *: actually, if they differ on a set of measure zero. You're not expected to know what that means -- you just need to know what I wrote is a special case.
     
    Last edited: Jul 10, 2011
  7. Jul 11, 2011 #6

    SammyS

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    Actually, multiplying the numerator & denominator by the conjugate of the numerator does "cancel out" the x - 1 in the denominator.
    In the context of pre-calculus and/or Calc. I, I would say, Yes, the correct graph has a removable discontinuity at x = 1.
     
  8. Jul 11, 2011 #7

    NascentOxygen

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    The expression isn't a function, or were you told to consider only the +ve root of 25? When x=1, the numerator has two values, -10 and 0

    EDIT. changed -25 to -10
     
    Last edited: Jul 11, 2011
  9. Jul 11, 2011 #8

    SammyS

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    When x = 1, the numerator has only one value, 0.

    [itex]\sqrt{16x^2 + 3x + 6} -5[/itex] is indeed a function of x.
     
  10. Jul 12, 2011 #9
    Thanks for all of the clarification. Unfortunately my high school doesn't teach calculus so I've been doing some self studying. I get the major concepts but get hung up on details like this.
     
  11. Jul 12, 2011 #10

    NascentOxygen

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    What you have is really two questions in one. The equation you present is not a function, because each value of x gives rise to two values for y.

    If you confine your interest to the positive value of the square root, then at x=0 you have y = 0/0 and hence undefined. Without calculus, to see what it does near x you can take x closer and closer to 1 and recalculate y for each. Say, x=0.999, 0.9999, 0.999999 etc. Then, on the other side of 1, say, x=1.001, 1.0001, 1.000001, etc. Construct a table, and you'll clearly see what y is doing in the region of x=1. There is no asymptote, just a discontinuity in an almost level curve.

    Turning attention to the other solution, the negative value of the square root, then you have the classic asymptote* at x=1. You discover this by, again, taking values of x increasingly close to 1, first from one side, then from the other. The denominator changes sign at x=1, but our numerator here is steadfastly negative.

    * http://www.jtaylor1142001.net/calcjat/DEFINITN/CalcGlos.html

    Good luck with your self-study of calculus. It's an exciting topic. Imagine how Newton (and Leibniz) would have stayed up late into the night developing their topic by candlelight, and wondering how the mathematical world would receive such a momentous advance.
     
  12. Jul 12, 2011 #11

    SammyS

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    When we write: [itex]\sqrt{25}\,,[/itex] for instance, this is accepted to mean the principal square root of 25, which is the number 5, not -5, even though it's true that (-5)2 = 25 .

    As long as the context is real valued functions with real valued arguments there should be no ambigutiy about this.
     
  13. Jul 12, 2011 #12

    NascentOxygen

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    May I ask who is this we who allows for only the positive root? I shall suggest that struggling students consider moving to that region or planet, as it will make their study of mathematics so much simpler. And if perchance, the local language is Gallic, then having Pi defined as 3.0000000000000 will be an added bonus!
     
  14. Jul 13, 2011 #13

    ehild

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    Read, for example: http://en.wikipedia.org/wiki/Square_root. The notation √a means the non-negative root of the equation x^2=a. The negative root is -√a.
    f(x)=√x is a function, so √x can not be double valued.

    ehild
     
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