Introduction
A rational function is a quotient of polynomials. Its behavior for very large positive or negative inputs can often be determined without evaluating the quotient at particular inputs: the decisive information is the degrees of the two polynomials and, when those degrees match, their leading coefficients.
This tutorial develops a systematic method for finding these limits. The comparison works at both \(+\infty\) and \(-\infty\), but the signs of infinite limits can differ at the two ends. In particular, when the numerator degree exceeds the denominator degree, the parity of their degree difference matters at \(-\infty\).
Degrees and Leading Coefficients
Let \(P\) and \(Q\) be nonzero polynomials, and suppose the rational function is \(f(x)=P(x)/Q(x)\). The degree of a nonzero polynomial is the highest exponent with a nonzero coefficient. Its leading coefficient is the coefficient of that highest-power term.
Write the polynomials as
where \(a\ne 0\), \(b\ne 0\), \(n=\deg P\), and \(m=\deg Q\). The values \(a\) and \(b\) are the leading coefficients. Since \(Q\) has leading term \(b x^m\), it is nonzero for all sufficiently large positive and negative \(x\). Thus \(P(x)/Q(x)\) is defined sufficiently far out in either direction, even if the denominator has zeros at some finite inputs.
The Degree-Comparison Rule
There are three cases. If the numerator degree is smaller, the quotient tends to zero. If the degrees are equal, the quotient tends to the ratio of leading coefficients. If the numerator degree is greater, the magnitude of the quotient grows without bound; its direction is determined by the signs of the leading-coefficient ratio and the relevant power of \(x\).
- If \(n<m\), \(\lim_{x\to+\infty}P(x)/Q(x)=0\) and \(\lim_{x\to-\infty}P(x)/Q(x)=0\).
- If \(n=m\), both limits equal \(a/b\).
- If \(n>m\), let \(d=n-m\). At \(+\infty\), the limit is \(+\infty\) if \(a/b>0\) and \(-\infty\) if \(a/b<0\). At \(-\infty\), the limit is \(+\infty\) if \((a/b)(-1)^d>0\) and \(-\infty\) if \((a/b)(-1)^d<0\).
The first two cases give finite limits. In the third case, the degree difference \(d\) is a positive integer. At negative inputs, \(x^d\) is negative when \(d\) is odd and positive when \(d\) is even. That sign change accounts for why the two infinite limits may point in different directions.
Why the Rule Works
The leading term controls a polynomial at either end: dividing the polynomial by its highest power of \(x\) leaves its leading coefficient plus terms that tend to zero. Applying this fact to both polynomials makes their degree comparison explicit.
Write every term below the leading term in \(P\) as \(c_kx^k\), where \(0\le k<n\), and every term below the leading term in \(Q\) as \(e_jx^j\), where \(0\le j<m\). Then
Every exponent in the denominators of these sums is positive, so each sum tends to zero as \(x\to+\infty\) and as \(x\to-\infty\). Consequently, sufficiently far out, the quotient can be written as
The fraction on the right tends to \(a/b\), which is nonzero. If \(n=m\), the factor \(x^{n-m}\) is \(1\), so the limit is \(a/b\). If \(n<m\), the factor is \(1/x^{m-n}\), which tends to zero at either end, while the other factor tends to the finite number \(a/b\). The product therefore tends to zero.
If \(n>m\), the factor is \(x^d\), where \(d=n-m>0\), and its magnitude grows without bound. The remaining factor tends to the nonzero value \(a/b\), so it eventually has the sign of \(a/b\). At \(+\infty\), \(x^d\) is positive. At \(-\infty\), its sign is \((-1)^d\). This proves the stated directions of divergence.
Worked Examples
Worked Example: The Numerator Has Smaller Degree
Evaluate the limits of
The numerator has degree \(2\) and leading coefficient \(3\). The denominator has degree \(4\) and leading coefficient \(2\). Since \(2<4\), both limits should be zero. To check the comparison directly, divide every term in the numerator and denominator by \(x^4\):
As \(x\to+\infty\), \(3/x^2\), \(-5/x^3\), and \(1/x^4\) each tend to zero. The denominator tends to \(2\), not zero. Therefore the quotient tends to \(0/2=0\). As \(x\to-\infty\), the same reciprocal powers tend to zero, and the denominator again tends to \(2\), so the quotient also tends to \(0/2=0\).
Check: The normalized numerator includes the leading term \(3x^2/x^4=3/x^2\), as well as both lower-degree terms. All three terms tend to zero, while the normalized denominator tends to its leading coefficient \(2\). This confirms both conclusions.
Worked Example: Equal Degrees
Evaluate the limits of
Both polynomials have degree \(3\). Their leading coefficients are \(-6\) and \(4\), so the predicted limit is
To verify it, divide every term in the numerator and denominator by \(x^3\):
At \(+\infty\), each reciprocal-power term tends to zero. The numerator tends to \(-6\), and the denominator tends to \(4\), so the quotient tends to \(-6/4=-3/2\). At \(-\infty\), the reciprocal powers also tend to zero; although some are negative before taking the limit, their limits are still zero. The numerator and denominator again tend to \(-6\) and \(4\), respectively.
Check: In the normalized expression, the denominator tends to \(4\ne0\), so taking the quotient of the two limits is valid. The quotient of those limits is exactly \(-3/2\), as predicted by the leading coefficients.
Worked Example: Larger Numerator Degree and Different Signs
Evaluate the limits of
The numerator has degree \(5\) and leading coefficient \(2\). The denominator has degree \(2\) and leading coefficient \(-1\). Their degree difference is \(d=5-2=3\), which is odd, and the ratio of leading coefficients is \(2/(-1)=-2\). Thus \(h(x)\) behaves like \(-2x^3\): it tends to \(-\infty\) at the positive end and \(+\infty\) at the negative end.
For a direct verification, factor the highest power from each polynomial:
The fraction multiplying \(x^3\) tends to \(2/(-1)=-2\), which is negative and nonzero. As \(x\to+\infty\), \(x^3\to+\infty\), so the product tends to \(-\infty\). As \(x\to-\infty\), \(x^3\to-\infty\); multiplying by a factor that tends to \(-2\) gives \(+\infty\).
Check: At \(x=10\), the numerator is \(2(10^5)-10^2+4=199904\), and the denominator is \(-10^2+3(10)-5=-75\), giving \(h(10)=199904/(-75)<0\). At \(x=-10\), the numerator is \(2(-10)^5-(-10)^2+4=-200096\), and the denominator is \(-(-10)^2+3(-10)-5=-135\), giving \(h(-10)=(-200096)/(-135)>0\). These finite values do not prove the limits, but their signs agree with the result; the factorization above establishes the limits.
A Common Mistake: Forgetting the Sign at Negative Infinity
A frequent error is to identify the leading-term quotient and then assume that its sign is the same at both ends. For the function \(h\) above, that shortcut gives \(-2x^3\). It is correct to use this to identify the controlling behavior, but incorrect to conclude that the function tends to \(-\infty\) at both ends.
The correction is to check the sign of the degree-difference power \(x^d\) at each end. Here \(d=3\) is odd, so \(x^3\) changes sign between the ends. More generally, an even degree difference leaves \(x^d\) positive at both ends, while an odd degree difference makes it negative at \(-\infty\) and positive at \(+\infty\). The leading-coefficient ratio must still be included in either case.
A Reliable Procedure
For a rational function \(P(x)/Q(x)\), use this sequence of steps. The polynomials must be nonzero, and the denominator has a nonzero leading coefficient by definition.
Record \(n=\deg P\), \(m=\deg Q\), and the leading coefficients \(a\) and \(b\).
If \(n<m\), the limit is zero at both ends. If \(n=m\), the limit is \(a/b\) at both ends.
Set \(d=n-m\). The quotient grows without bound in magnitude.
At \(+\infty\), use the sign of \(a/b\). At \(-\infty\), use the sign of \((a/b)(-1)^d\).
Why the Comparison Is Useful
Degree comparison provides a quick way to distinguish three different types of end behavior: decay to zero, approach to a finite nonzero value, and unbounded growth. It also prevents a common inefficiency: expanding or manipulating every term when only the highest powers are needed to identify the limit.
The method is not a rule for substituting infinity into a rational expression. Infinity is not a real number that can be substituted for \(x\). Instead, the method compares terms as \(x\) becomes arbitrarily large in magnitude. The normalized expressions in the examples show why lower powers become negligible relative to the highest power, and why the denominator’s normalized expression has a nonzero limit.
At negative infinity, do not discard signs when using powers of \(x\). A reciprocal power such as \(1/x^3\) approaches zero from below at \(-\infty\), while \(1/x^2\) approaches zero from above. Both still have limit zero, but a positive power such as \(x^3\) grows in magnitude with a sign that depends on the direction of approach. Keeping the degree difference visible makes this distinction straightforward.
Check Your Understanding
For each question, identify the relevant degrees and leading coefficients before deciding on the limit.
- What are the limits at \(+\infty\) and \(-\infty\) of \(\frac{5x^2+1}{-3x^4+x}\)?
- What are the limits at both ends of \(\frac{7x^3-2}{-2x^3+4x}\)?
- For \(\frac{-4x^6+x}{2x^3-1}\), determine the limits at \(+\infty\) and \(-\infty\). Explain the role of the degree difference.
- If the numerator degree exceeds the denominator degree by an even number, does the degree-difference power have the same sign at both ends? What additional sign information is needed?
- Why does dividing the numerator and denominator by the denominator’s highest power show that a lower-degree numerator contributes its leading term as well as its lower-degree terms to the limit calculation?
