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Math Help - Numerator Degree < Denominator Degree

  1. #1
    Senior Member nycmath's Avatar
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    Numerator Degree < Denominator Degree

    If the numerator degree is less than denominator degree, what is the oblique or slant asymptote?

    T(x) = (x^3)/(x^4-1)
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  2. #2
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    Re: Numerator Degree < Denominator Degree

    This answer on StackExchange may help.
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  3. #3
    Senior Member nycmath's Avatar
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    Re: Numerator Degree < Denominator Degree

    Sorry but that link was not too helpful....
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  4. #4
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    Re: Numerator Degree < Denominator Degree

    There is NO "oblique or slant" asymptote. An "asymptote" is a line or curve that a function graph approaches as x grows large, either positive or negative. An "oblique or slant" is a non vertical or horizontal line the function graph approaches and that requires that the function be of the form f(x)= mx+ b+ F(x) where F(x) goes to 0 as x gets larger. That can only happen if the numerator has degree exactly one more than the denominator (and other curves as asymptotes if the degree of the numerator is two or more higher). As long as the degree of the denominator is greater than the degree of the numerator, the horizontal line, y= 0, is the asymptote.
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  5. #5
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    Re: Numerator Degree < Denominator Degree

    To add about the relevance of the link: As post #4 says, f(x) has an asumptote mx+b if \lim_{x\to\infty}(f(x)-mx-b)=0. The link describes how to compute the limit of a rational function f(x): one just needs to consider the ratio of the senior terms in the numerator and the denominator. Here

    \lim_{x\to\infty}\frac{x^3}{x^4-1}= \lim_{x\to\infty}\frac{x^3}{x^4}= \lim_{x\to\infty}\frac{1}{x}=0

    so the asimptote is y = 0.
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  6. #6
    Senior Member nycmath's Avatar
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    Re: Numerator Degree < Denominator Degree

    Thank you everyone. Say find the asymptotes for (4x)/(x^2-49).

    The vertical asymptotes are found by equating the denominator to zero and solving for x.

    Since the degree of the top is smaller than denominator degree, the horizontal asymptote is the line y = 0 which is
    the x-axis, right?

    There is no slant asymptote here, right?
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  7. #7
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    Re: Numerator Degree < Denominator Degree

    Yes, this is correct.
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  8. #8
    Senior Member nycmath's Avatar
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    Re: Numerator Degree < Denominator Degree

    I get it now. But if the rational function involves a radical?

    Sample: Find all asymptotes of
    (x^3)/sqrt(x^4-49)
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  9. #9
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    Re: Numerator Degree < Denominator Degree

    The idea is the same: \sqrt{x^4+a_3x^3+\dots+a_0}=(x^4+a_3x^3+\dots+a_0)  ^{1/2} behaves like (x^{4})^{1/2}=x^2 for large x. Formally, in this case we have for x > 3

    \frac{x^3}{\sqrt{x^4-49}} =\frac{x^3}{\sqrt{x^4(1-49/x^2)}} =\frac{x^3}{x^2\sqrt{1-49/x^2}}= x\frac{1}{\sqrt{1-49/x^2}}

    Now \lim_{x\to\infty}\frac{49}{x^2}=0, so \lim_{x\to\infty}\frac{1}{\sqrt{1-49/x^2}}=1. Therefore, the original expression behaves like x for large x.
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