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Curved sketching strange things happen

  1. May 8, 2008 #1

    rock.freak667

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    Homework Helper

    I was told to sketch this graph:

    [tex]y=\frac{x-1}{x^2-100}[/tex]

    when x=0,y=.01


    [tex]y=\frac{\frac{1}{x}-\frac{1}{x^2}}{1-\frac{100}{x^2}}[/tex]

    as [itex]x \rightarrow \infty;y \rightarrow 0 \Rightarrow[/itex] y=0 (i.e. x axis is a horizontal asymptote to the curve)

    BUT

    when y=0; x-1=0 so that x=1. So a point on the curve is (1,0)

    How can the x-axis be a horizontal asymptote to the curve yet the curve passes through (1,0)?
     
  2. jcsd
  3. May 8, 2008 #2
    What do you mean, how can it? It just is.

    The x-axis is an asymptote for [tex]sin(x)/x[/tex], and this is equal to zero an infinite number of times.
     
  4. May 8, 2008 #3

    Vid

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    There's a vertical asymptote at x=10. For x > 10, the x axis is a horizontal asymptote.
     
    Last edited: May 8, 2008
  5. May 8, 2008 #4

    rock.freak667

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    (fixed)


    OH...that clears it right up for me! Thank you. I was under the impression that the x-axis would be a horizontal asymptote for the entire graph no matter what.
     
  6. May 8, 2008 #5
    It is a horizontal asymptote for the entire graph. The notion of horizontal asymptote just means what the function f(x) approaches something as x approaches positive or negative infinity. It has nothing to do with f(1), and as you have found, it is quite possible to cross over an asymptote. See uman's post for a good example of a such a function.
     
    Last edited: May 8, 2008
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