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Horizontal asymtote

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    "Match functions and their horizontal asymptotes."

    2. Relevant equations
    lim f(x)
    x-> + infinity

    lim f(x)
    x-> - infinity

    3. The attempt at a solution
    http://img198.imageshack.us/img198/7678/45328107.jpg [Broken]

    im just stuck at this... im sure that everything is right!!!
    but its wrong :(

    i was thinking, arctan(x^4-x^2) = tan-1(x^4-x^2)
    i saw on a different post, that someone said, graph x^4-x^2 then take the limit and do tan(L) and thats the answer.. but it doesnt seem to work.

    the other one that may be wrong (but im sure its not) is cot-1(x).. i graphed it as 1/tan-1(x) and got 0.6 and -0.6

    please help me!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 26, 2009 #2
    So your problem was with arctan(x4 - x2)?

    What is the limit of arctan(x) as x goes to infinity? Does x4 - x2 also go to infinity as x goes to infinity?
  4. Sep 26, 2009 #3
    i dont know... im assuming so.
    thats my question, what wrong with my matching in the picture?
  5. Sep 26, 2009 #4
    F is correct. Was there another answer you thought it might be? Do you know why B would not be correct?

    Also, what did you mean where you said "but it doesnt seem to work."?
  6. Sep 26, 2009 #5
    well the system uses all of the answer i input and shows me right/wrong overall.. it doesnt show me which one is wrong.. so i have to go through each one and check it.
    i have check each one 20+ times and also graphed it on a graphing calculator and nothing seems to solve my problem because i keep on getting the same thing..
    yeah and B wouldn't be correct because it is only + infinity..
    and when plugging in -infinity, x^2 and x^4 make it positive anyways..

    how about the cot-1(x) is this one correct?
  7. Sep 26, 2009 #6
    cot-1x ≠ 1/tan-1x

    [tex]y = \cot^{-1}x[/tex]
    [tex]\cot y = x[/tex]

    [tex]\frac{1}{\cot y} = \frac{1}{x}[/tex]
    [tex]\tan y = \frac{1}{x}[/tex]
    [tex]y = \tan^{-1}\left(\frac{1}{x}\right)[/tex]
    [tex]\cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right)[/tex]

    You should be able to work with this last one now.
  8. Sep 26, 2009 #7
    ohhh really????
    so that would mean that the answer is 0!
    i thought it's 1/tan-1(x)

    so the answer would be G=0?
  9. Sep 26, 2009 #8
    Yes, looks like that's the only one wrong.
  10. Sep 26, 2009 #9
    the system shows that i already had an answer like this.. which means that something else is wrong.

    for (6x+1) / (sqrt(3x^2+1))
    i get hor.asymptote as pi and -pi..
    which means its J.. right?

    so my answers are now in this order:

    and looks like everything is right but still the system says its wrong...
    maybe that one with J is wrong?
    Last edited: Sep 26, 2009
  11. Sep 26, 2009 #10
    That one is J (why [itex]\pi[/itex] and -[itex]\pi[/itex]?). Do you know what it's horizontal asymptote is?
  12. Sep 26, 2009 #11
    yeah.. horizontal asymptote is the y value thats a line which the graph doesnt pass (the limit of it.. gets really close to it.. and goes to infinity but still doesnt touch it)

    if that one is right... then i just dont know.
    the first one is the only thing left and its %99.999 right

    the last one, when graphing it, i get 2. when solving it i get 0.. maybe i solved that one wrong?
  13. Sep 26, 2009 #12
    Which one are you talking about?
  14. Sep 26, 2009 #13
    the first matching fraction.. 6x^3+2x+1 / 2x^3-10x^2+13x+100
    but i dont think thats the problem..

    im more concerned about the last one
    f(x)=2-x + sqrt(x^2+1)

    im getting 0 when doing it algebraically, but graphing in on a calculator give me a limit of 2
  15. Sep 26, 2009 #14
    That one has a limit of 2, so you're right.

    Some, like this one, need some creative rewriting so you can find its limit.
  16. Sep 26, 2009 #15

    you meant 3 for the first one right?

    for f(x)=2-x + sqrt(x^2+1),
    what i did is multiplying both top and bottom by -2+x + sqrt(x^2+1)
    which led to:
    (4x-3) / (sqrt(x^2+1) +x -2)
    from there, i divide both top and bottom by x^2..

    (4x-3)/x^2 / (sqrt(x^2+1) +x -2)/x^2

    and i got (4/x - 3/x ) / (sqrt(1+ 1/x^2) + x -2)
    if x-> infinity..

    the top will be immediately 0 and the bottom is 1 which mean 0/1 = 0...
  17. Sep 26, 2009 #16
    3, you're right. :tongue:

    You can rewrite it a little easier so you don't make any mistakes, which I think you did

    [tex]\lim_{x\rightarrow\infty} (2 + \sqrt{x^2 + 1} - x) = 2 + \lim_{x\rightarrow\infty}(\sqrt{x^2 + 1} - x)\left(\frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} + x}\right)[/tex]
  18. Sep 26, 2009 #17
    oh man... thats what i did wrong?!!!

    now i get (1/0) / 2, which means, doesnt exist

    which means it would be "none of the above" for the final answer, am i correct?
  19. Sep 26, 2009 #18
    No, it does exist as you saw on your calculator.
  20. Sep 26, 2009 #19
    [tex]2 + \lim_{x\rightarrow\infty}(\sqrt{x^2 + 1} - x)\left(\frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} + x}\right) = 2 + \lim_{x\rightarrow\infty}\frac{x^2 + 1 - x^2}{\sqrt{x^2(1 + \frac{1}{x^2})} + x} = 2 + \lim_{x\rightarrow\infty}\frac{1}{x\sqrt{1 + \frac{1}{x^2}} + x}[/tex]

    Can you take it from there?
  21. Sep 26, 2009 #20
    i got,
    (x^2 + 1 - x^2) / (sqrt(x^2 +1) +x) =
    1/ (sqrt(x^2 +1) +x)

    now i divided both top and bottom by x

    1/x / (sqrt(x^2 +1) +x) / x

    and got:
    1/x / (sqrt(1+ 1/x^2) +1)

    >> 1/x^2 = 0 so (sqrt(1+ 0) +1) = 2

    top: 1/x = 1/infintiy = 0
    0/2 = 0

    2+0 = 2
    oh man.. so that still 2

    then what is wrong??
    we went over every single one and still its wrong..
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