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Asymptotes to Absolute Values

  1. Sep 27, 2009 #1
    Hi! I've got a problem that's got me a bit.. well, in the end I guess the proper word is sceptic, because it feels a bit like I've made the answer up, which is why I'm taking it up here at all. Anyway:

    1. The problem statement, all variables and given/known data
    Determine the domain for f(x) and the asymptotes to the curve y=f(x) for

    [tex]f(x) = \frac{|x^3 + x^2|}{x^2 -1}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I started with the domain, which was simple enough - it's not defined for -1 or 1. The question I've got here, however, is how I write this in the best way. I mean, I can say that the function is defined in the intervals x>1, -1<x<1 and x<-1, but we've also learned another way of writing:

    If a function is defined from -∞ to +∞ the domain is Df = ]-∞,∞[ , but can I write this when the interval is broken like above? I mean, it looks bizarre; Df = ]-∞,-1[ , Df = ]-1,1[ , Df = ]1,∞[ .. ?


    Anyway, for the asymptotes:

    I checked the vertical asymptotes first, which wasn't too hard:

    [tex]\lim_{x\to 1} \frac{|x^3 + x^2|}{x^2 -1} = \frac{2}{0} = \infty[/tex]

    So, 1 is a vertical asymptote.

    [tex]\lim_{x\to -1} \frac{|x^3 + x^2|}{x^2 -1} = \frac{0}{0} [/tex]

    -1 isn't an asymptote, is what I've figured from my reading.

    Anyway, now we come to the real.. sceptism. We've learnt slant asymptotes too. And well.. it's.. odd.

    First of all, I came to the conclusion that when x → +∞, x3 + x2 is defintitely positve so:

    [tex]\lim_{x\to \infty} \frac{|x^3 + x^2|}{x^2 -1} -(ax+b) = \lim_{x\to \infty} \frac{x^3 + x^2}{x^2 -1} -(ax+b) = \lim_{x\to \infty} \frac{x^2(x + 1)}{x^2(1- \frac{1}{x^2})} -(ax+b) = \lim_{x\to \infty} \frac{x + 1}{1- \frac{1}{x^2}} -(ax+b)[/tex]

    And here I start coming up with an answer, and I really feel like I've imagined it.

    To find the asymptote I want the whole thing to go to zero and well, my reasoning went something like this:

    I first thought that when x → ∞, (1/x2) → 0, so then I thought, well, that sort of leaves me with x + 1 - (ax+b), which goes to zero when it's x+1 - (x+1), which means that it has an asymptote in a=1, b=1. Which is what I feel might be rather imagined. (so, asymptote is y = x + 1.)

    So I started searching around and found a webpage where it said that you can determine a slant asymptote by doing long division in this type of situation, so I tried that to

    (x3 + x2)/(x2-1)

    which in turn gave me:

    f(x) = (x+1) + (x+1)/(x2-1), which also means that the asymptote is y=x+1..

    And I did the same thing for x → -∞, saying that when this happens x3 + x2 is most definitely negative, so the absolute value of it is -(x3 + x2) and by this got the asymptote (using both aforementioned methods) y = -x - 1.

    First: I did not learn the long division method in school, not sure if my teacher will accept it and since I learned it from the internet, is it really a trustworthy method?

    Second: If I've done it wrong, help?!

    Third: Sorry if this is a confusing post--it was definitely confusing to write!
     
  2. jcsd
  3. Sep 27, 2009 #2
    You could say Df = {x|x ≠ -1, 1} or Df = (-∞, -1) U (-1, 1) U (1, ∞) with the union.


    Long division is often the easiest way to find a slant asymptote, but you can get the same result by factoring, rewriting, and canceling:

    [tex]\frac{x^3 + x^2}{x^2 - 1} = \frac{x^2(x + 1)}{(x - 1)(x + 1)} = \frac{x^2 - 1 + 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} + \frac{1}{x - 1} = x + 1 + \frac{1}{x - 1}[/tex]

    Your method seems just as good and shorter, and you found the two slant asymptotes.
     
  4. Sep 28, 2009 #3
    OK, thank you! :D

    I'm trying to find some more information on it, but is it really true that if x→-1 the expression goes to (0/0) that it's not an asymptote? I might have to justify it to my teacher and I'd like to be able too, really..
     
  5. Sep 28, 2009 #4
    0/0 could approach a value or diverge to +/- infinity.

    x/x approaches 1 as x goes to 0, but x/x2 has a vertical asymptote at x=0. Both are 0/0 if you let x=0.
     
  6. Sep 28, 2009 #5
    But.. how do I find out? I am so confused. Cos I can't see whether it goes to +/- infinity or a value.

    I can see it in your examples, sort of, but how can I look at it in my case?
     
  7. Sep 28, 2009 #6
    You don't always know; that's when you have to try canceling whatever is making the denominator 0.
    In the case of x/x, it turned out to be like the constant function 1. Sometimes you cancel until you get to a point where you won't be able to cancel a part that makes the denominator 0. With x/x2, one x in the numerator and one in the denominator canceled, but you're still left with 1/x, and you can't do anything else with that x; there is a vertical asymptote at x=0.

    0/0 is just one case of what is called an indeterminate form. You can read more about them http://en.wikipedia.org/wiki/Indeterminate_form" [Broken]. Often they approach a number, and you'll learn about L'Hôpital's rule in Calc II, also on that page, which usually helps you find that number it approaches.
     
    Last edited by a moderator: May 4, 2017
  8. Sep 29, 2009 #7
    Ah, I get it now.. Thank you!
     
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