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One-Sided and Infinite Limits · Tutorial 244 of 1495

Slant Asymptotes By Polynomial Division

Use polynomial division to find a candidate slant asymptote, then verify it by showing the remainder divided by the denominator tends to zero.

Intermediate 9 min read

What You'll Learn

  • Define a slant asymptote using the difference between a function and a line.
  • Use polynomial division to identify a candidate line for a rational function.
  • Verify the asymptote by evaluating the remainder term at both ends of the real line.
  • Understand why a numerator degree exactly one greater than the denominator produces a linear quotient.
  • Recognize when polynomial division does not produce a slant asymptote.
  • Avoid confusing a nonzero remainder with a failure of the asymptote test.

What a Slant Asymptote Describes

A rational function can grow without bound as \(x\) becomes large in magnitude and still approach a straight line. The line describes the function’s end behavior: the vertical difference between the function and the line tends to zero.

Polynomial division makes this behavior visible. It rewrites the rational function as a polynomial plus a remainder term. When the quotient is linear and the remainder term tends to zero, that linear quotient is the slant asymptote.

Definition: A line \(y=mx+b\), with \(m\ne 0\), is a slant asymptote of \(f\) at positive infinity if \(\lim_{x\to+\infty}\bigl(f(x)-(mx+b)\bigr)=0\). It is a slant asymptote at negative infinity if the corresponding limit as \(x\to-\infty\) is zero. If both limits are zero, the same line is a slant asymptote at both ends.

The condition is about the difference between the function and the line—not whether the function itself has a finite limit. A function approaching a slant asymptote may grow without bound.

Polynomial Division Reveals the Candidate

Let \(f(x)=\frac{P(x)}{Q(x)}\), where \(P\) and \(Q\) are polynomials and \(Q\) is not the zero polynomial. Polynomial division gives polynomials \(S\) and \(R\) such that

$$ P(x)=Q(x)S(x)+R(x), \qquad \deg R<\deg Q. $$

At inputs where \(Q(x)\ne 0\), divide by \(Q(x)\) to obtain

$$ f(x)=S(x)+\frac{R(x)}{Q(x)}. $$

If the degree of \(P\) is exactly one greater than the degree of \(Q\), the quotient \(S\) is linear. It is then a candidate slant asymptote. The remaining question is whether \(\frac{R(x)}{Q(x)}\) tends to zero at the end or ends being considered.

Key takeaway: If \(\deg P=\deg Q+1\), polynomial division gives a linear quotient \(S(x)=mx+b\). If also \(\lim_{x\to+\infty}\frac{R(x)}{Q(x)}=0\), that line is a slant asymptote at positive infinity. The corresponding limit at negative infinity verifies it there.

Why the Remainder Term Tends to Zero

Suppose \(\deg Q=n\ge 1\), and write \(Q(x)=a_nx^n+\cdots+a_0\), where \(a_n\ne 0\). Because \(\deg R<n\), every term of \(R(x)/x^n\) tends to zero as \(x\to+\infty\) or \(x\to-\infty\). Dividing numerator and denominator of \(R(x)/Q(x)\) by \(x^n\) therefore gives a numerator tending to zero and a denominator tending to \(a_n\ne 0\). Thus the ratio tends to zero at both ends.

Theorem: If \(\deg P=\deg Q+1\), polynomial division gives a linear quotient \(S\) and a remainder \(R\) with \(\deg R<\deg Q\). For \(\deg Q\ge 1\), \(\frac{R(x)}{Q(x)}\to 0\) as \(x\to+\infty\) and as \(x\to-\infty\). Consequently, \(S\) is a slant asymptote at both ends. If \(Q\) is constant, division is exact and \(R=0\), so the function equals its linear quotient wherever it is defined.

This result follows directly from the definition of the asymptote. Since \(f(x)=S(x)+R(x)/Q(x)\), subtracting \(S(x)\) leaves precisely the remainder term. The degree condition ensures that the remainder has lower degree than the denominator, so that term vanishes at both ends.

Worked Examples

Worked Example: Finding a Slant Asymptote by Division

Find the slant asymptote of \(f(x)=\frac{2x^2+3x-2}{x-1}\).

The numerator has degree 2 and the denominator has degree 1, so their degree difference is 1. Divide the polynomials:

$$ \begin{aligned} 2x^2\div x&=2x,\\ 2x(x-1)&=2x^2-2x,\\ (2x^2+3x-2)-(2x^2-2x)&=5x-2. \end{aligned} $$

The next quotient term is \(5\), since \(5x\div x=5\). Multiply and subtract again:

$$ \begin{aligned} 5(x-1)&=5x-5,\\ (5x-2)-(5x-5)&=3. \end{aligned} $$

Thus the quotient is \(2x+5\) and the remainder is \(3\). The division identity checks:

$$ (x-1)(2x+5)+3 =2x^2+3x-5+3 =2x^2+3x-2. $$

Therefore, for \(x\ne1\),

$$ f(x)=2x+5+\frac{3}{x-1}. $$

Subtract the candidate line. As \(x\to+\infty\), \(\frac{3}{x-1}\to0\); as \(x\to-\infty\), it also tends to zero. Hence \(y=2x+5\) is a slant asymptote at both ends.

Check: \(f(x)-(2x+5)=\frac{3}{x-1}\), and both required limits of this difference are zero.

Worked Example: A Linear Remainder

Find the slant asymptote of \(g(x)=\frac{3x^3-2x^2+4x-1}{x^2+1}\).

The degrees are 3 and 2, so their difference is 1. Begin the division with \(3x\):

$$ \begin{aligned} 3x(x^2+1)&=3x^3+3x,\\ (3x^3-2x^2+4x-1)-(3x^3+3x)&=-2x^2+x-1. \end{aligned} $$

The next quotient term is \(-2\). Subtract its product with the denominator:

$$ \begin{aligned} -2(x^2+1)&=-2x^2-2,\\ (-2x^2+x-1)-(-2x^2-2)&=x+1. \end{aligned} $$

The quotient is \(3x-2\) and the remainder is \(x+1\). Verify the division:

$$ (x^2+1)(3x-2)+(x+1) =3x^3-2x^2+3x-2+x+1 =3x^3-2x^2+4x-1. $$

Now write the rational function in quotient-and-remainder form:

$$ g(x)=3x-2+\frac{x+1}{x^2+1}. $$

To check the remainder term at either end, divide its numerator and denominator by \(x^2\):

$$ \frac{x+1}{x^2+1} = \frac{\frac1x+\frac1{x^2}}{1+\frac1{x^2}} \longrightarrow 0 \quad\text{as }x\to+\infty\text{ and as }x\to-\infty. $$

Thus \(y=3x-2\) is a slant asymptote at both ends.

Check: Subtracting the line \(3x-2\) from \(g(x)\) gives exactly \(\frac{x+1}{x^2+1}\), whose limits at both ends are zero.

Worked Example: A Nonzero Constant Remainder

Find the slant asymptote of \(h(x)=\frac{x^2-4x+7}{2x+2}\).

The numerator degree exceeds the denominator degree by 1. The first quotient term is \(\frac12x\). Multiplication and subtraction give

$$ \begin{aligned} \frac12x(2x+2)&=x^2+x,\\ (x^2-4x+7)-(x^2+x)&=-5x+7. \end{aligned} $$

The next quotient term is \(-\frac52\), because \(-5x\div 2x=-\frac52\). Then

$$ \begin{aligned} -\frac52(2x+2)&=-5x-5,\\ (-5x+7)-(-5x-5)&=12. \end{aligned} $$

The quotient is \(\frac12x-\frac52\) and the remainder is \(12\). Verify:

$$ (2x+2)\left(\frac12x-\frac52\right)+12 =x^2-4x-5+12 =x^2-4x+7. $$

Therefore

$$ h(x)=\frac12x-\frac52+\frac{12}{2x+2}. $$

The remainder term tends to zero at both ends because its denominator grows in magnitude without bound. So the slant asymptote is \(y=\frac12x-\frac52\).

Check: \(h(x)-(\frac12x-\frac52)=\frac{12}{2x+2}\), which tends to zero as \(x\to+\infty\) and as \(x\to-\infty\).

When to Expect a Slant Asymptote

For a rational function with nonzero leading coefficients, the degree difference is a useful guide. If the numerator degree is exactly one greater than the denominator degree, division produces a linear quotient and the lower-degree remainder term vanishes at both ends. This is the standard setting for a slant asymptote.

If the numerator degree is more than one greater, the quotient has degree at least 2, not degree 1. In that case, the function’s leading growth is faster than linear, so subtracting any line cannot leave a difference tending to zero. If the numerator degree is no greater than the denominator degree, the end behavior is bounded or approaches a finite value, so it does not approach a line with nonzero slope. Horizontal asymptotes are treated separately in the earlier material on Horizontal Asymptotes And Their Limits.

The division identity applies wherever \(Q(x)\ne0\). Values at which \(Q(x)=0\) are excluded from the domain of the rational expression, but those finite inputs do not change the limits at infinity.

Common Pitfall: The Remainder Need Not Be Zero

A nonzero remainder does not prevent a slant asymptote. The test concerns the ratio of the remainder to the denominator, not the remainder by itself.

For example, division in the first worked example leaves remainder \(3\). It would be incorrect to conclude, “The remainder is not zero, so there is no slant asymptote.” The correct calculation is

$$ \frac{3}{x-1}\longrightarrow0 \quad\text{as }x\to+\infty\text{ and as }x\to-\infty. $$

It is also not enough to report the quotient as a candidate without checking what is left over. The precise verification is to subtract the proposed line from the function and confirm that the resulting remainder term tends to zero.

Why Polynomial Division Is Useful

Polynomial division turns an expression that may be difficult to interpret directly into two parts: a polynomial describing the dominant behavior and a remainder term describing the difference. For a linear quotient, this gives both the slope and intercept of the candidate line immediately. Checking the remainder term then gives the exact limit required by the definition of a slant asymptote.

Check Your Understanding

For each question, use the quotient-and-remainder form to explain your conclusion.

  1. If the degree of the numerator is one greater than the degree of the denominator, what degree must the quotient have?
  2. What limit must the remainder divided by the denominator satisfy for the quotient line to be a slant asymptote?
  3. Why can a nonzero constant remainder still produce a slant asymptote?
  4. For \(f(x)=\frac{x^2+1}{x+1}\), divide the polynomials and identify the candidate line.
  5. What is the degree difference that typically signals a slant asymptote for a rational function?

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