Horizontal Asymptote of f(x)=x^3+1/x-1: Is it y=x^2?

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The discussion centers on the horizontal asymptote of the rational function f(x) = (x^3 + 1) / (x - 1). Participants clarify that the function does not have a horizontal asymptote but rather a slant asymptote, specifically y = x^2 + x + 1, as the degree of the numerator (3) is one higher than that of the denominator (1). The consensus is that horizontal asymptotes occur only when the degrees of the numerator and denominator are equal, while slant asymptotes arise when the numerator's degree exceeds the denominator's by one. Thus, the textbook's claim of a horizontal asymptote at y = x^2 is incorrect.

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When the power of the leading coefficient of the numerator of a rational function is a lot greater than the power of the leading coefficient of the denominator , ie

f(x)=\frac{x^3+1}{x-1}

The horizontal asymtote is y=x^2 according to the book . Is that true ? Is there any proof for this . I only know when its one power difference , that would be oblique asymtote .
 
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I think that is right , look at the highest power on top and the highest power on the bottom and divide them .
 
I really doubt that your textbook says horizontal asymptote. A horizontal asymptote is a straight horizontal line and only occurs in a rational function when the numerator and denominator have the same degree. A "slant" or "skew" asymptote, a non-horizontal line, occurs when the numerator has degree one higher than the denominator.

I've never seen a curve called an "asymptote" before but since
\frac{x^3}{x- 1}= x^2+ x+ 1+ \frac{2}{x-1}
so for large x, the graph comes close to x^2+ x+ 1 which itself will be close to x^2. (Though I would say the "asymptote" was x^2+ x+ 1, not just x^2.)
 
HallsofIvy said:
I really doubt that your textbook says horizontal asymptote. A horizontal asymptote is a straight horizontal line and only occurs in a rational function when the numerator and denominator have the same degree. A "slant" or "skew" asymptote, a non-horizontal line, occurs when the numerator has degree one higher than the denominator.

I've never seen a curve called an "asymptote" before but since
\frac{x^3}{x- 1}= x^2+ x+ 1+ \frac{2}{x-1}
so for large x, the graph comes close to x^2+ x+ 1 which itself will be close to x^2. (Though I would say the "asymptote" was x^2+ x+ 1, not just x^2.)

thank you . It's not horizontal asymtote , i just didn't know what to call it , is there a name for that kind of asymtote ?

Is it necessary to include this asymtote in my sketching ? Usually , i would just include the vertical asymtote in my sketching if there is no horizontal asym.
 
I think you might be able to call it a slant asymptote.
 

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