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Homework Help: Horizontal asymptotes - approaches from above or below?

  1. Feb 22, 2009 #1
    I seem to be having a lot of difficulty finding whether for a horizontal asymptote, whether the curve approaches the asymptote from above or below.


    For example, for the problem [tex]y = \frac{6x + 1}{1 - 2x}[/tex], I know that:

    For the vertical asymptote, x = 1/2, and that [tex]\lim_{x \to \frac{1}{2}^{+}} = - \infty[/tex], while [tex]\lim_{x \to \frac{1}{2}^{-}} = \infty[/tex].


    Meanwhile, for the horizontal asymptote, there is one at y = -3. However, I do not know understand whether it approaches from above or below.

    This is how I tried to find it:
    As x --> -3 from the left, eg. x = -10, [tex]\frac{6(-3)+1}{1-2(-10)}[/tex] = [tex]\frac{-17}{21}[/tex] = -0.81
    therefore approaches from above, because -0.81 is greater than -3.

    As x --> -3 from the right, eg. x = 10, [tex]\frac{6(-3)+1}{1-2(10)}[/tex] = [tex]\frac{-17}{-19}[/tex] = 0.95
    therefore also approaches from above, because -0.81 is greater than -3.


    However, my graphing calculator says otherwise. The graph shows that, left of the vertical asymptote, the curve approaches from above; to the right, it approaches from the bottom. What am I doing wrong?
     
  2. jcsd
  3. Feb 22, 2009 #2

    gabbagabbahey

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    Why are you substituting x=-3 into the numerator and then substituting [itex]x=\pm10[/itex] into the denominator?
     
  4. Feb 22, 2009 #3
    Hmm, that's what I saw in my class notes. I assumed that the numerator's x-value remained to be -3, while the denominator was supposed to have a "test" value.

    Ok, so I'm guessing I'm supposed to be subbing [itex]x=\pm10[/itex] into both numerator and denominator?
     
  5. Feb 22, 2009 #4

    gabbagabbahey

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    Yes!:smile:

    After all, it's y that approaches -3 not x.
     
  6. Feb 22, 2009 #5
    Thanks for clarifying, gabbagabbahey!

    I don't know how I've been pondering over this simple mistake, haha... :redface:
     
  7. Feb 23, 2009 #6
    Also, another question: is what I shown above involving the calculations of [tex]\lim_{x \to - \infty}[/tex] and [tex]\lim_{x \to +\infty}[/tex]?
     
    Last edited: Feb 23, 2009
  8. Feb 23, 2009 #7
    [itex] \displaystyle \lim_{x\to\infty} \frac{6x+1}{1-2x} [/itex]

    [itex] \displaystyle \lim_{x\to\infty} \frac{6+\frac{1}{x}}{\frac{1}{x}-2} [/itex]

    Does that help?
     
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