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Rational Functions' Asymptotes

  1. Feb 14, 2008 #1
    First Question

    If: f(x) = (x^2+1)/x

    Then: f(x) = x + (1/x)

    From my understanding, x would be the oblique/slant asymptote. Why is that?

    Second Question

    Why and how can horizontal asymptotes be crossed?
     
  2. jcsd
  3. Feb 14, 2008 #2
    well, just glancing at it i could tell that this function has a vertical asymptote at x=0, because lim(x-->0+)f(x)=+infinity
    and lim(x-->0-)f(x)=- infinity

    there might be also some oblique asymptotes, let me see in a few sec.
     
  4. Feb 14, 2008 #3
    yeah, and y=x is an oblique asymptote also.
    y=kx+b, is the general formula for a line. k is its slope, b- is y intercept

    k=lim(x-->infinity)[f(x)/x]=1, so k=1, then

    b=lim(x->infinity)[f(x)-kx]=0, so y=x is an oblique asymptote, there are no v. asymptotes.
     
  5. Feb 14, 2008 #4

    symbolipoint

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    Notice, according to your division result, that f(x) approaches x as x approaches negative or positive infinity, but never actually becomes x.

    Another way to say this,
    [tex] \[
    \mathop {\lim }\limits_{x \to \pm \infty } \,f(x) = x
    \]
    [/tex]
     
  6. Feb 14, 2008 #5

    symbolipoint

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    Actually, you may be more comfortable reading [tex] \[
    \mathop {\lim }\limits_{x \to \pm \infty } \,(x + {\textstyle{1 \over x}}) = x
    \]
    [/tex]
     
  7. Feb 14, 2008 #6
    What the heck!!!!!!!!!
    Are u sure?????????
     
  8. Feb 15, 2008 #7

    Gib Z

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    He would be more comfortable reading [tex]\lim_{x\to \infty} \frac{ x+ \frac{1}{x}}{x} = 1[/tex], or in Landau notation, [tex]x+\frac{1}{x}[/tex] ~ [tex]x[/tex], but [tex] \[
    \mathop {\lim }\limits_{x \to \pm \infty } \,(x + {\textstyle{1 \over x}}) = x
    \]
    [/tex]

    is wrong.
     
  9. Feb 15, 2008 #8
    o_o ...I'm sorry, I'm not all too familiar with limits. -_-;
     
  10. Feb 15, 2008 #9
    Well, it is kind of hard to deal with asymptotes if you are not familiar with limits. because saying in other words, asymptotes are merely some straight lines that are said to touch the graph of a function at a point at infinity, so to find them you will always have to deal with limits.
     
  11. Feb 15, 2008 #10

    symbolipoint

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    Syukai's original function was (x^2 + 1)/x, and he correctly obtained the division result of (x + (1/x)). That makes the slant asymtote of y=x to be obvious. As x tends to positive or negative infinity, the function approaches the line y=x
     
  12. Feb 16, 2008 #11

    Gib Z

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    Yes. But to say a limit of a function of x, as x approaches infinity, is equal to another function of x, is not correct.
     
  13. Feb 16, 2008 #12

    symbolipoint

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    You should have given a solid explanation, such as can be found in the middle of the website, http://en.wikipedia.org/wiki/Asymptote , which gives some understandable steps, and includes the very same example as in the start of this thread. Apparantly, we find trouble with finding a limit of a function when the function does not approach a specific value. On the other hand, when the difference between a function and another function approaches zero as the independant variable goes to plus or minus infinity, then we have a specific value of a particular difference.

    Now this is clearer. The intuitive approach shown in some book such as PreCalculus of Larson, Hostetler & Edwards is good, but did not rigorously use statements of limits. A LIMIT IS A SPECIFIC VALUE. Is that correct?
     
  14. Feb 16, 2008 #13
    The limit of a real one-variable function as the argument of the function approaches some real number is either an element of the range of the function, an element of the closure of the range of the function, or does not exist (where behaviors like increasing without bound or decreasing without bound are usually denoted by the symbols [itex]\infty[/itex] or [itex]-\infty[/itex], respectively). This is only by a rigorous definition of the limit; one example of which is that L is the limit of a real one-variable function f(x) as x approaches x0 iff every one-dimensional disc around L contains the image f(D-{x0}) of some punctured one-dimensional disc D around x0 (this is slightly more geometrically appealing than the introductory definition directly using the metric). You can adjust the definition obviously for limits as x increases/decreases without bound. This doesn't really make sense if L is a variable.
    Gib Z just put your statement into more rigorous terms, which does fit into limit notation.
     
    Last edited: Feb 16, 2008
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