Why Compare Degrees?
A rational function is a quotient of polynomials. When \(x\) becomes very large in magnitude, the terms with the greatest powers of \(x\) determine the function’s end behavior. Comparing those powers gives a systematic way to find a limit at positive or negative infinity.
The comparison must account for both the degrees and the leading coefficients. If the numerator degree is smaller, the quotient approaches zero. If the degrees match, the quotient approaches the ratio of the leading coefficients. If the numerator degree is larger, the quotient grows without bound in magnitude; its sign can differ at positive and negative infinity.
The Degree-Comparison Rule
Let \(P(x)\) and \(Q(x)\) be nonzero polynomials, with degrees \(m\) and \(n\), and leading coefficients \(a_m\) and \(b_n\), respectively. The rational function \(P(x)/Q(x)\) is defined for all sufficiently large positive and negative \(x\): the leading term of \(Q(x)\) ensures that \(Q(x)\) is nonzero once \(|x|\) is large enough.
- If \(m<n\), then \(\displaystyle\lim_{x\to\infty}f(x)=0\) and \(\displaystyle\lim_{x\to-\infty}f(x)=0\).
- If \(m=n\), then both limits equal \(a_m/b_n\).
- If \(m>n\), put \(d=m-n\). The quotient grows without bound in magnitude. At positive infinity its sign is the sign of \(a_m/b_n\). At negative infinity its sign is the sign of \((a_m/b_n)(-1)^d\).
Here, a limit of \(+\infty\) means that the function’s values eventually exceed every fixed positive bound. A limit of \(-\infty\) means that they eventually fall below every fixed negative bound. These are statements about unbounded behavior, not finite real-number values.
Why the Rule Works
Write each polynomial as its leading term times a factor. For example, the lower-degree terms of \(P\), after factoring out \(x^m\), become powers of \(1/x\):
Every term after \(1\) in the parentheses approaches zero as \(x\) approaches either positive or negative infinity. The same reasoning applies to \(Q(x)\). Therefore, for sufficiently large \(|x|\),
The last factor approaches \(1\). The remaining power \(x^{m-n}\) determines whether the limit is zero, finite, or unbounded. When the exponent is positive, its sign at negative infinity depends on whether that exponent is even or odd.
Worked Examples
Worked Example: Numerator Degree Less Than Denominator Degree
Evaluate both limits of \(f(x)=\frac{3x^2-5x+1}{2x^4+x-7}\).
The numerator has degree \(2\) and leading coefficient \(3\). The denominator has degree \(4\) and leading coefficient \(2\). Since \(2<4\), the rule predicts zero at both ends. Factoring out \(x^2\) from the numerator and \(x^4\) from the denominator gives
As \(x\) approaches either infinity, the second factor approaches \(3/2\), while \(1/x^2\) approaches zero. Thus
Check: the denominator factor in parentheses approaches \(2\), so it stays away from zero for sufficiently large \(|x|\). The other factor approaches the finite value \(3/2\), and multiplication by \(1/x^2\) makes the entire expression approach zero at both ends.
Worked Example: Equal Degrees
Evaluate both limits of \(g(x)=\frac{-6x^3+4x-5}{3x^3-2x^2+9}\).
Both polynomials have degree \(3\), so the limit at each end is the ratio of their leading coefficients, \(-6/3=-2\). Dividing every term by \(x^3\) makes this comparison explicit:
As \(x\to\infty\), every displayed reciprocal-power term approaches zero, so the quotient approaches \(-6/3=-2\). The same terms approach zero as \(x\to-\infty\). Therefore,
Check: the denominator after division approaches \(3\), which is nonzero, and the numerator approaches \(-6\). Their quotient is \(-6/3=-2\), confirming the result at both ends.
Worked Example: A Larger Numerator Degree with an Even Difference
Evaluate both limits of \(h(x)=\frac{4x^4-x+2}{-2x^2+3x+1}\).
The degree difference is \(4-2=2\), which is even. The leading-coefficient ratio is \(4/(-2)=-2\). Factoring out the leading powers gives
The fraction multiplying \(x^2\) approaches \(4/(-2)=-2\). Since \(x^2\) grows without bound and is positive at both ends, the product decreases without bound at both ends:
Check: the fraction in the factored expression approaches a negative number, \(-2\), while \(x^2\) is positive and becomes arbitrarily large. Thus the product is negative with arbitrarily large magnitude at both ends.
Worked Example: A Larger Numerator Degree with an Odd Difference
Evaluate both limits of \(r(x)=\frac{x^3+2x+1}{-2x^2+x-3}\).
The degree difference is \(3-2=1\), which is odd, and the leading-coefficient ratio is \(1/(-2)=-1/2\). Factoring out the leading powers yields
The fraction multiplying \(x\) approaches \(-1/2\). At positive infinity, \(x\) is positive and grows without bound, so the product approaches \(-\infty\). At negative infinity, \(x\) is negative and its product with a factor approaching \(-1/2\) is positive and grows without bound. Hence
Check: the denominator factor in parentheses approaches \(-2\), so it is nonzero for sufficiently large \(|x|\). The remaining fraction approaches \(-1/2\), confirming that its product with \(x\) has the stated opposite signs at the two ends.
A Reliable Process
For each limit, identify the two degrees first. If the numerator degree is greater, also track the degree difference and the leading-coefficient ratio. This sequence keeps the magnitude and sign decisions separate.
Use the highest-power nonzero term in each polynomial.
Smaller numerator degree gives zero; equal degrees give a finite ratio; larger numerator degree gives unbounded magnitude.
At positive infinity, use the sign of the leading-coefficient ratio. At negative infinity, also account for whether the degree difference is odd or even.
Confirm that the remaining factor approaches the predicted leading-coefficient ratio.
Common Mistake: Ignoring Negative Infinity
A frequent error is to find the sign at positive infinity and copy it to negative infinity without checking the degree difference. For example, for \(r(x)=\frac{x^3+2x+1}{-2x^2+x-3}\), one might incorrectly say that both limits are \(-\infty\), because the leading-coefficient ratio is negative.
The degree difference is \(1\), so the remaining power \(x\) changes sign between the two ends. At positive infinity, \(-\frac12x\) is negative and unbounded; at negative infinity, it is positive and unbounded. The corrected limits are \(-\infty\) and \(+\infty\), respectively.
Another useful check is to avoid treating a quotient’s leading-coefficient ratio as the whole answer when the degrees differ. That ratio alone gives the limit only when the degrees are equal. If the numerator degree is larger, the remaining power of \(x\) cannot be discarded.
Why This Comparison Is Useful
Degree comparison turns a complicated-looking rational expression into a small number of decisions: compare two integers, divide two leading coefficients, and, when needed, check the parity of their degree difference. It also explains why lower-degree terms do not change these limits: after factoring out the leading powers, those terms become reciprocal powers that approach zero.
The same reasoning works if numerator and denominator share a polynomial factor. Cancelling a common factor does not change the degree difference or the ratio of the leading coefficients: the factor’s degree is subtracted from both degrees, and its leading coefficient cancels from the ratio. Factoring the original polynomials directly therefore leads to the same limit rule.
Check Your Understanding
For each question, identify the degrees before deciding what happens at each end.
- Evaluate the limits of \(\frac{5x+1}{x^3-4}\) as \(x\to\infty\) and \(x\to-\infty\).
- What are both limits of \(\frac{7x^2-3}{-2x^2+x+1}\)?
- Evaluate both limits of \(\frac{3x^5+x}{6x^2-1}\), and explain why the signs differ or agree.
- For a rational function with numerator degree \(6\) and denominator degree \(3\), how does the degree difference affect the sign at negative infinity?
- Why does dividing numerator and denominator by the denominator’s highest power of \(x\) help when the polynomial degrees are equal?
