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Finding vertical asymptotes of non-rational f

  1. Sep 12, 2009 #1

    honestrosewater

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    Problem
    Find the limit (if it exists). If it does not, explain why.

    [tex]
    \lim_{x \rightarrow -3^{-}} \frac{x}{\sqrt{x^{2}-9}}
    [/tex]

    (i.e. the limit from the left.)

    Definitions & theorems
    vertical asymptote definition: if f(x) approaches infinity (or negative infinity) as x approaches c from the right or the left, then the line x = c is a vertical asymptote of the graph of f.

    vertical asymptote theorem: let f and g be continuous on an open interval containing c. If f(c) != 0, g(c) = 0, and there exists an open interval containing c such that g(x) != 0 for all x != c in the interval, then the graph of the function given by h(x) = f(x)/g(x) has a vertical asymptote at x = c.

    infinite limit definition: let f be a function that is defined at every real number in some open interval containing c (except possibly at c itself). The statement

    [tex]
    \lim_{x \rightarrow c} f(x) = \infty
    [/tex]

    means that for each M > 0 there exists a d > 0 such that f(x) > M whenever 0 < |x-c| < d. [Limit for negative infinity is defined similarly.]
    To define the infinite limit from the left, replace 0< |x-c| < d by c-d<x<c. [Limit from right is defined similarly.]

    My attempt
    (I am trying to do things as the book says. Personally, I hate this book.) A graph and my intuition say that there should be a vertical asymptote at -3, but I do not see how to prove it using the definitions and theorems in the book.

    h(x) = x/sqrt(x^2 - 9) ; x^2 - 9 >= 0, sqrt(x^2 - 9) != 0
    domain(h) = (-infinity, -3) U (3, infinity)
    h(x) = f(x)/g(x) --> f(x) = x and g(x) = sqrt(x^2 - 9)
    domain(f) = R, domain(g) = (-infinity, -3] U [3, infinity)
    f(-3) != 0, g(-3) = 0
    However, g is not defined on any open interval containing -3, so I do not know how the book wants me to proceed.

    My answer would be that the limit does not exist because there is a vertical asymptote at x = -3 if I could figure out how to prove it. I will need to do so for other problems, so even if I can get away without it here, I need to know eventually.
     
  2. jcsd
  3. Sep 12, 2009 #2

    Hurkyl

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    Hrm. Your idea looks convincing, but it seems the theorems you have available aren't quite up to the task?

    Then invent a new theorem!

    Can you take the idea behind the vertical asymptote theorem and modify it so that it applies to the one-sided case?



    Alternative approach

    Can you find a new problem whose solution will tell you the answer to the original problem, but the vertical asymptote theorem does apply to the new problem?
     
  4. Sep 12, 2009 #3

    honestrosewater

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    Should I try to prove that if f is unbounded and f' is positive and increasing on an interval (a, b), then there exists a positive vertical asymptote at x = b? And similarly for the decreasing/negative case? That sounds true. No?

    But we are not supposed to know about derivatives until the next chapter. The answer in the back says "Limit does not exist. The function decreases without bound as x approaches -3 from the left". But it doesn't give any reasons. It never does, anywhere. Math is not about reasoning to them, apparently. Hah, sorry. So frustrated...

    Oh, now that I think of it, if f does have a vertical asymptote at b, f' should have no upper bound on (a, b) too, right? The slope of f should be approaching a vertical line.? So I can just say that f' is positive, increasing, and unbounded?
     
    Last edited: Sep 12, 2009
  5. Sep 12, 2009 #4

    Hurkyl

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    Yes, this sounds true. (f' increasing is an unnecessary hypothesis)

    However, I submit that in this particular case, if you can show your h is unbounded and negative over an interval whose right endpoint is -3, then it's quite likely that nearly the same argument can be used to show that the limit of h is negative infinity, so that you don't need to bother with the derivative at all!


    It turns out that doesn't always imply vertical asymptote -- sometimes, it can just be a vertical tangent line. For example, consider the graph of
    [tex]f(x) = \sqrt[3]{x}[/tex]
     
  6. Sep 12, 2009 #5

    honestrosewater

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    Hm, the slope of f has to be increasing because if it was constant on some (a', b), f(x) would reach b and b would be an upper bound? I can sort of see it in my head, but I haven't looked at it formally yet. I will think about the rest. Thanks. :^)
     
    Last edited: Sep 12, 2009
  7. Sep 12, 2009 #6

    honestrosewater

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    So yeah, I had to make a similar argument for another problem, and the professor seemed surprised to hear it. (He seems surprised by a lot of things that I say, which makes me worry about passing my tests.) Here goes.

    h(x) = x/sqrt(x2 - 9), so h is continuous on (-infinity, -3).

    All variables are real and universally quantified unless stated otherwise.

    [tex]
    x_1 < x_2 < -3 \ \Rightarrow\ {x_1}^2 > {x_2}^2 > 9 \ \Rightarrow\ {x_1}^2 - 9 > {x_2}^2 - 9 > 0 \ \Rightarrow\ \sqrt{{x_1}^2 - 9} > \sqrt{{x_2}^2 - 9} > 0
    [/tex]

    So, on (-infinity, -3), the denominator is positive and decreasing and the numerator is negative and increasing. So I need to see what that means for the quotient.

    [tex]
    \mbox{\bf{Conjecture 1. }} a < c < 0 \ \wedge\ 0 < d < b \ \rightarrow\ a/b \ ??\ c/d
    [/tex]

    It doesn't look like I can prove this the same way I proved (a > c > 0 & d > b > 0) --> (a/b > c/d), which was just to show that a/b - c/d was positive by introducing r and s > 0 such that a = c+r and d = b+s and using some algebraic manipulation to end up with a quotient of sums of products of positive numbers, which must itself then be positive. Will something similar work here? I will think about it. Perhaps I should remove the negative sign and deal with the magnitudes and put it back in afterwards.

    Forgetting the signs for a moment and just looking at the magnitudes (so reversing the order for a and c), what does a > c and b > d mean for a/b and c/d, thinking in a model? The numerator decreasing means that you are taking fewer pieces (of whatever, pie), but the denominator decreasing means that the pieces are getting bigger. So whether you are getting more or less pie is a function of... I don't know, something, maybe a and b, maybe just their ratio. Hm.

    Or I could try to prove

    [tex]
    \mbox{\bf{Conjecture 2. }} \forall b \exists x\ [ x < -3 \rightarrow \frac{x}{\sqrt{x^2 - 9}} < b]
    [/tex]
     
    Last edited: Sep 12, 2009
  8. Sep 18, 2009 #7

    Hurkyl

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    As you noticed, it's not obvious how ?? could be replaced with < or > -- your pie argument suggests you could find examples for both < and >.

    You could add extra conditions (e.g. a hypothesis about how ad and bc compare, or a/c and b/d), but those are really equivalent problems.

    [/quote]
    That should do it. (it's very nearly the definition of the limit from the left side going to -infinity!)

    Were you able to manage?
     
  9. Sep 20, 2009 #8

    honestrosewater

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    I had to move on since the book doesn't require a proof. I am interested in (C1) and its related problems. It will be on my mind while learning about derivatives. (If f = g/h, surely f' is determined by the relationship between g' and h', no? But I need a bigger whiteboard to see how.)

    I think I see part of (C2). There should be a general technique for this kind of problem, I would think.

    Assume

    [tex]1)\ \exists b \forall x\ [ x < -3 \ \wedge\ \frac{x}{\sqrt{x^2 - 9}} \geq b] \\
    [/tex]

    Either b < -3 or b > -3. b < -3 implies b2 > 9, which implies b2 > (b2 - 9) > 0, which implies b2 > b > sqrt(b2 - 9). Yes? But (1) implies

    [tex]2) \frac{b}{\sqrt{b^2 - 9}} \geq b
    [/tex]

    which implies b2 < sqrt(b2 - 9), a contradiction. Right?

    I don't see what problem b > -3 will pose, though. I might think about it. I'm not terribly interested in it except perhaps for a general technique.
     
  10. Sep 20, 2009 #9

    honestrosewater

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    Oh, maybe that will make b > x/sqrt(x2 - 9).
     
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