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Homework Help: Finding asymptotes of |x|/x

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data
    List all the asymptotes of
    f(x) = |x| / x

    2. Relevant equations

    3. The attempt at a solution
    This is a problem on a limits test in a Calculus AB class. I tried vertical asymptotes and horizontal asymptotes by setting the numerator and denominator equal to 0 but only getting x=0


    Unfortunately that wasn't the answer to the question, but I plugged the equation in the calculator and it did appear to have a vertical asymptote at x=0 and two horizontal asymptotes at x=1 and x=-1
    However I don't know how to find that algebraically.
    I did some research and saw an example where they found the limit of a function as it went to positive and negative infinity in order to find the asymptotes. I tried that:

    lim |x|/x = 1

    lim |x|/x = -1

    At the end of this I was left with
    x=0 (from my very first attempt)

    1. Is setting finding the limit of the function as it goes to infinity a solid way of finding asymptotes?
    2. a. Is my final answer correct?
    b. If not, Could someone explain the mistakes/false-reasoning I made or lead me in the right direction?
  2. jcsd
  3. Jan 21, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper

    It doesn't have a vertical asymptote at x=0. What are the limits from the right and left? And you are right about the behavior at infinity, but doesn't that make the asymptote lines y=1 and y=(-1) (not x=1 and x=(-1))?
  4. Jan 21, 2014 #3
    There is no vertical asymptote at x=0 because
    the limit as x goes to 0 from the left is -1
    the limit as x goes to 0 from the right is 1

    but there is two horizontal asymptotes at
    because of the limit as x goes to ∞ and -∞?

    Does everything look good now?
  5. Jan 21, 2014 #4


    User Avatar
    Science Advisor
    Homework Helper

    Yes. Vertical asymptotes are where a one sided limit goes to infinity. Doesn't happen here.
  6. Jan 21, 2014 #5


    Staff: Mentor

    I don't think I would trust that calculator.

    The function here is simple enough that you can sketch a graph in about the same time it would take to enter it into the calculator.

    f(x) = 1 if x > 0, and
    f(x) = -1 if x < 0
    It is undefined at x = 0 (jump discontinuity).
    The graph consists of two horizontal half-lines.
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