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A rational function, $\frac{p(x)}{q(x)}$ has an oblique asymptote only when the degree of $p(x)=$ degree of $q(x) -1$.

  • What "causes" the "slant" of the asymptote? Most asymptotes are caused by a function approaching an undefined value - I assume this is the same, but why (unlike others) would these asymptotes be slanted?

  • Why does this only work with a difference of 1 between degrees?

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4 Answers 4

up vote 1 down vote accepted

It works with arbitrary degrees, it's just that with higher degrees the asymptotes won't be straight lines but parabolas etc.

Consider an example: $f(x)=\frac{4x^3+1}{2x^2}=2x+\frac{1}{2x^2}$, the latter part goes towards zero as $x$ becomes large. You can do that with every such function - do polynomial long division and write out the residue. This will always look like $f(x)=\frac{p(x)}{q(x)}=cx+ d + \frac{r(x)}{q(x)}$ where $c,d$ are constant and $deg\ r < deg\ q$, so the latter fraction approaches zero as $|x|\rightarrow\infty$.

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Thanks for the help! –  user26649 Jul 17 '12 at 2:09

Write $p(x) = (a + bx)*q(x) + r(x)$; dividing we have $$ {p(x)\over q(x)} = a + bx + {r(x)\over q(x)},$$ where $\deg(r) < \deg(q)$. The last term decays to zero as $|x|\to\infty$ and your oblique asymptote will be the line $y = a + bx$.

You can have a difference larger than 1 of degrees. In this case, your graph will have the graph of a polynomial as an asymptote at $\pm\infty$.

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Let $p(x) = \displaystyle\sum_{k=0}^{n+1} a_kx^k$ and $q(x) = \displaystyle\sum_{k=0}^n b_k x^k.$

$\displaystyle\lim_{x\to \infty } \frac{p(x)}{q(x)} = \displaystyle\lim_{x\to \infty } [(\displaystyle\sum_{k=0} a_k x^{k-n})/(\displaystyle\sum_{k=0}^n b_k x^{k-n})].$ $k<n$ for all but one summand in the denominator and all but two in the numerator, so the rational function is asymptotically $\frac{a_{n+1}}{b_n} x + \frac{a_n}{b_n}.$

For cases of higher degree $p,$ the same exact method will give you the coefficients of a polynomial asymptotically similar to $p/q.$

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We usually say a funciton has an oblique asymptote $y=ax+b$ if

$$\lim_{x\to \infty}\frac{f(x)}{x}=a \text{ ; and} $$ $$\lim_{x\to \infty}{f(x)}-{ax}=b $$

This is basically saying, for $x$ large enough, the function resembles the line $y=ax+b$. This is a special case of what we call "asymptotic behaviour" of a function. Particularily, given two polynomial functions $p$ and $q$, the quotient

$$f(x)=\frac{p(x)}{q(x)}$$

will behave asymptotically like a line $y=ax+b$ if $\deg p =\deg q+1$. Why is this? Let's examine the polynomials

$$q(x)=a_nx^n+a_{n-1}x^{n-1}\cdots+a_0 \text{ ; } a_n\neq0$$

$$p(x)=b_{n+1}x^{n+1}+b_{n}x^{n}\cdots+b_0 \text{ ; } b_{n+1}\neq0$$

We have that their quotient is

$$f(x)=\frac{{p(x)}}{{q\left( x \right)}} = \frac{{{b_{n + 1}}{x^{n + 1}} + {b_n}{x^n} \cdots + {b_0}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}}$$

And now, if we divide by $x$ and find the limit for $x\to \infty$, we find

$$\mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\frac{{{b_{n + 1}}{x^{n + 1}} + {b_n}{x^n} \cdots + {b_0}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{b_{n + 1}}{x^n} + {b_n}{x^{n - 1}} \cdots + \frac{{{b_0}}}{x}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}}$$

Note that now, except for the insignificant $\dfrac{b_0}{x}$ we have a quotient of polynomials of equal degree, precisely because upon dividing by $x$ the degrees were equated. So this limit will be constant, in fact we have

$$\mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{{b_{n + 1}}{x^n} + {b_n}{x^{n - 1}} \cdots + \frac{{{b_0}}}{x}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}} = \frac{{{b_{n + 1}}}}{{{a_n}}}$$

Calculating the second part of the limit, we get

$$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) - \frac{{{b_{n + 1}}}}{{{a_n}}}x = \mathop {\lim }\limits_{x \to \infty } \frac{{{b_{n + 1}}{x^{n + 1}} + {b_n}{x^n} \cdots + {b_0}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}} - \frac{{{b_{n + 1}}}}{{{a_n}}}x$$

because of the factor of $\dfrac{{{b_{n + 1}}}}{{{a_n}}}$ the $x^{n+1}$ will cancel you (the algebra of that is a bit messy) but you'll end up again with a quotient of polynomials of equal degree:

$$\mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{b_n} - \frac{{{b_{n + 1}}}}{{{a_n}}}{a_{n - 1}}} \right){x^n} + \cdots + {b_0}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}} = \frac{{{b_n}}}{{{a_n}}} - \frac{{{b_{n + 1}}}}{{{a_n}}}\frac{{{a_{n - 1}}}}{{{a_n}}}$$

so the asymptote you'll get is

$$y = \frac{{{b_{n + 1}}}}{{{a_n}}}x + \frac{{{b_n}}}{{{a_n}}} - \frac{{{b_{n + 1}}}}{{{a_n}}}\frac{{{a_{n - 1}}}}{{{a_n}}}$$

And why do we get a straight line only when the difference is $1$? That because we divide by $x^1$ (which lowers the degree of the denominator by $1$). However, if the degrees differs by say $r$, the rational function will behave asymptotically like a polynomial of degree $r$

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