Curved sketching strange things happen

  • #1
rock.freak667
Homework Helper
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Main Question or Discussion Point

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)?
 

Answers and Replies

  • #2
352
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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.
 
  • #3
Vid
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There's a vertical asymptote at x=10. For x > 10, the x axis is a horizontal asymptote.
 
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  • #4
rock.freak667
Homework Helper
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There's a vertical asymptote at x=10. For x > 10, the x axis is a horizontal asymptote.
(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.
 
  • #5
737
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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.
 
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