Introduction
A rational function is a quotient of polynomials. Its graph can have several different features near finite values of the input, but its behavior far to the left or right is often determined by a compact comparison: the degrees of its numerator and denominator.
This comparison answers a specific question: does the function approach a constant as \(x\) becomes very large in magnitude? If so, that constant gives a horizontal asymptote. The answer depends first on the degrees, and, when the degrees are equal, on the leading coefficients.
Horizontal Asymptotes and Degree
A polynomial is nonzero if it is not identically zero. For a nonzero polynomial, its degree is the largest exponent of \(x\) with a nonzero coefficient. The coefficient of that highest-power term is its leading coefficient.
Suppose \(P\) and \(Q\) are polynomials and \(Q\) is not the zero polynomial. The rational function \(f(x)=P(x)/Q(x)\) is defined at inputs for which \(Q(x)\ne 0\). Its behavior as \(x\) tends toward positive or negative infinity can be described by limits.
For a rational function with a nonzero numerator, let \(m=\deg P\), \(n=\deg Q\), \(a\) be the leading coefficient of \(P\), and \(b\) the leading coefficient of \(Q\). The main classification is as follows.
- If \(m<n\), then \(y=0\) is a horizontal asymptote at both ends.
- If \(m=n\), then \(y=a/b\) is a horizontal asymptote at both ends.
- If \(m>n\), then \(f\) has no horizontal asymptote at either end.
If \(P\) is the zero polynomial, then \(f(x)=0\) wherever it is defined, so \(y=0\) is a horizontal asymptote at both ends. The degree of the zero polynomial is not used in the comparison above.
Why Comparing Degrees Works
The key idea is that a polynomial's highest-power term determines its end behavior. For example, in \(4x^3-2x+7\), the terms \(-2x\) and \(7\) become small relative to \(4x^3\) as \(|x|\) grows. We can make this precise by factoring out the highest power of \(x\) from each polynomial.
Write \(P(x)=a_mx^m+\cdots+a_0\) and \(Q(x)=b_nx^n+\cdots+b_0\), where \(a_m=a\ne0\) and \(b_n=b\ne0\). Factoring out the leading powers gives
where \(u(x)\) is a sum of terms involving negative powers of \(x\), and \(v(x)\) is also a sum of terms involving negative powers of \(x\). Each such term tends to zero as \(x\to+\infty\) and as \(x\to-\infty\). Thus \(u(x)\to0\) and \(v(x)\to0\) at both ends.
In particular, \(1+v(x)\to1\), so \(1+v(x)\ne0\) for all sufficiently large \(|x|\). Therefore the quotient is defined near either end, and we can write
The final fraction tends to \(1\) at both ends. The behavior is therefore determined by \(x^{m-n}\), with the leading-coefficient ratio \(a/b\) as a constant factor.
If \(m<n\), then \(m-n<0\), so \(x^{m-n}\to0\) at both ends. Multiplying by a factor that tends to \(a/b\) still gives a limit of \(0\).
If \(m=n\), then \(x^{m-n}=1\). The quotient tends to \(a/b\) at both ends.
If \(m>n\), then \(k=m-n\) is a positive integer. The factor \(x^k\) grows without bound in magnitude. Since the remaining factor tends to the nonzero number \(a/b\), the rational function's magnitude grows without bound, so it cannot approach a finite horizontal level.
This proves the degree comparison. It also shows why the same classification holds at positive and negative infinity: the lower-degree terms tend to zero after factoring at either end. In the larger-numerator-degree case, the sign of the function may differ between the two ends, but its magnitude does not remain bounded.
Worked Examples
Worked Example: Numerator Degree Smaller
Find the horizontal asymptote of \(f(x)=\dfrac{3x+2}{x^2-4}\).
The numerator has degree \(1\), and the denominator has degree \(2\). Since \(1<2\), the theorem predicts a horizontal asymptote \(y=0\). Verify this by dividing numerator and denominator by \(x^2\):
As \(x\to+\infty\), \(3/x\), \(2/x^2\), and \(4/x^2\) all tend to zero. The numerator tends to \(0\), and the denominator tends to \(1\), so the quotient tends to \(0/1=0\). The same terms tend to zero as \(x\to-\infty\), giving the same limit there. Thus the result is checked: \(y=0\) is a horizontal asymptote at both ends.
Worked Example: Equal Degrees
Find the horizontal asymptotes of \(g(x)=\dfrac{6x^2-5x+1}{3x^2+4}\).
Both polynomials have degree \(2\). Their leading coefficients are \(6\) and \(3\), so the predicted horizontal asymptote is \(y=6/3=2\). To verify, divide every term in the numerator and denominator by \(x^2\):
At either end, \(5/x\), \(1/x^2\), and \(4/x^2\) tend to zero. The numerator tends to \(6\) and the denominator tends to \(3\), so
Both limits confirm that \(y=2\) is the horizontal asymptote at both ends.
Worked Example: Numerator Degree Larger
Determine whether \(h(x)=\dfrac{2x^3-x+4}{x^2+1}\) has a horizontal asymptote.
The numerator has degree \(3\) and the denominator has degree \(2\). Since \(3>2\), the degree comparison predicts no horizontal asymptote. To check the end behavior directly, divide the numerator and denominator by \(x^2\):
Dividing this expression by \(x\) gives
Because \(h(x)/x\to2\), \(h(x)\) grows in magnitude like \(2x\). In particular, \(h(x)\to+\infty\) as \(x\to+\infty\), while \(h(x)\to-\infty\) as \(x\to-\infty\). Neither end has a finite limit, confirming that there is no horizontal asymptote.
A Common Mistake: Comparing Coefficients Before Degrees
Leading coefficients determine the horizontal asymptote only when the numerator and denominator have equal degrees. Using their ratio in other cases can give a wrong answer.
For instance, consider \(r(x)=\dfrac{5x^2+1}{2x^3+3}\). A mistaken calculation might take the ratio of the leading coefficients, \(5/2\), and claim \(y=5/2\) is the horizontal asymptote. But the numerator degree is \(2\), while the denominator degree is \(3\), so the degrees are not equal. Divide through by \(x^3\) instead:
The correct horizontal asymptote is \(y=0\). The ratio \(5/2\) is relevant only when the degrees match.
Using the Rule Carefully
For a rational function, identify the degree of each polynomial and then compare them. Do not compare individual terms that are not leading terms, and do not assume that the constant terms determine the end behavior.
The comparison concerns the function where it is defined. A denominator can be zero at some finite inputs, but a polynomial with nonzero leading coefficient is nonzero for all sufficiently large \(|x|\), as the factorization in the proof shows. Thus finite domain restrictions do not prevent us from studying either end.
Also, the classification is about horizontal asymptotes only. If the numerator degree is larger, the theorem says there is no horizontal asymptote; it does not identify another type of asymptote. The degree comparison is enough to answer the horizontal-asymptote question without making any further claim about the graph.
Check Your Understanding
For each question, decide which degree case applies before finding or ruling out a horizontal asymptote.
- What is the horizontal asymptote of \(\dfrac{4x^2+7}{2x^2-x+1}\)?
- What is the horizontal asymptote of \(\dfrac{x+5}{3x^3-2}\)?
- Does \(\dfrac{7x^4+1}{x^3+6x}\) have a horizontal asymptote? Explain using the degrees.
- Why is the ratio of leading coefficients not the correct answer for every rational function?
- If the numerator is the zero polynomial, what is the function's horizontal asymptote wherever the function is defined?
