1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Asymptote problem

  1. Jul 10, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{2x-3}{x-1}[/tex] find the asymptotes of this equation.

    3. The attempt at a solution

    We already know that x=1 is a vertical asymptote from calculating the domain.

    the denominator of x-1=0 when x=1.

    However I'm not sure what it is they mean when later they go onto say..

    [tex]f(x)=\frac{2x-3}{x-1}=\frac{2-\frac{3}{x}}{1-\frac{1}{x}}=\frac{2-0}{1-0}=2 as x\rightarrow \infty[/tex]

    What value of x are they using here?

    Another more complicated one.

    1. The problem statement, all variables and given/known data

    I don't know what to do to achieve the same say for this:-

    [tex]\frac {5(12x+65)}{x^2-25}}[/tex]

    What exactly are they doing when they insert the value of x? It doesn't appear they have the same value for x as they do in the numerator as the denominator? :uhh:

    3. The attempt at a solution

    This for a graph sketching excercise, by finding certain values you can make a sketch, would I be correct in assuming that there would be a horizontal asymptote in the equation above at y=60, and of course x=-5 and 5 Or am I missing something here?

    [tex]\frac{60x+\frac{325}{x}}{x^2-\frac{25}{x^2}}=\frac{60}{1}[/tex]

    I think I'm missing something here.
     
    Last edited: Jul 10, 2007
  2. jcsd
  3. Jul 10, 2007 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For the first one they let x tend to infinity and thus the denominator going to infinity makes the fraction tend to zero. That means that there is a non-vertical asymptote at f(x) = 2.

    If the numerator is of greater order than the denominator then you'd do a polynomial division and the asymptote would be the leading straight line term.
     
    Last edited: Jul 10, 2007
  4. Jul 10, 2007 #3

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Above, they are just finding the limit of the function as x tends to infinity. This is a standard method of finding limits of quotients of polynomials: divide through by the highest power of x and then use the fact that 1/x tends to zero as x tends to infinity.

    I'm not sure what you're doing here.
     
  5. Jul 10, 2007 #4

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The second example you'll have to divide the numerator and denominator by x2 then investigate as x tends to plus or minus infinity.
     
  6. Jul 10, 2007 #5
    That makes a lot more sense, they don't explain what they have done at all really particularly well, I'll take a look bearing that in mind.

    Neither was I that was kind of the point, :smile: thanks all though.

    [tex]\frac{\frac{60x}{x^2}+\frac{325}{x^2}}{\frac{x^2}{x^2}-\frac{25}{x^2}}\rightarrow\frac{\frac{60}{x}-0}{1-0}[/tex]

    What would I do now?

    I mean I can see [itex]\frac {x}{x^2}[/itex] tends to zero? [itex]x\rightarrow\infty[/itex]

    I get it with the simple example but what do I do when I end up with x/x^2?

    EDITED: to try the second one, still don't understand what I'm going to end up with here? In the case of 60/x then as x tends to infinity x becomes 0? But then this means the answer is 0? Is that right?
     
    Last edited: Jul 10, 2007
  7. Jul 10, 2007 #6

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You will notice that sometimes horizontal asymptotes don't work within the region of vertical asymptotes and the origin and often only apply as x tends to infinity in the positive and negative direction.
     
  8. Jul 10, 2007 #7
    Sorry I edited the previous post^^, am I looking a this correctly?

    I understand now what they mean but I'm still not sure what the answer would be, is it 60/x? Or do I then go further?

    I looked at the graph on Mathcad, there appears to be an asymptote at -5,5, I don't think there exactly is a horizontal asymptote at y=0? There appears to be one at ~y=-4 although I could be imagining it.
     
    Last edited: Jul 10, 2007
  9. Jul 10, 2007 #8

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It would of course tend to 0/1 which is just 0. Therefore there is a horizontal asymptote at y=0. As I mentioned before its sometimes not very good around the origin which you can see on your graph and thus there won't be one at y = -4.4. You will of course still get an accurate graph because your investigation of local minimums and maximums will tell you whats going on there.
     
  10. Jul 10, 2007 #9
    Great that's fantastic Kurdt and Cristo, I did a bit of work on this yesterday for the first time, and couldn't really grasp at the time what they were driving at, but had I taken more note I would have seen it, anyway live and learn :smile:
     
    Last edited: Jul 10, 2007
  11. Jul 10, 2007 #10

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Most text books I've seen never explain this concept properly. They just kind of introduce it and never explain that a horizontal asymptote sometimes only works for very large or very small x. And they almost always never explain the method used.

    My favourite undergraduate introductory text (Guide 2 Mathematical Methods) which I think is almost perfect in every other way doesn't do this either and it frustrates me immensly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Asymptote problem
  1. Asymptotic expansions (Replies: 7)

Loading...