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Limit Comparison and Equivalents · Tutorial 642 of 998

Leading-Order Equivalents From Factored Expressions

Extract the dominant powers and constants from polynomial factors, and use a ratio limit to confirm the resulting equivalent.

Intermediate 9 min read

What You'll Learn

  • Define asymptotic equivalence using a ratio limit.
  • Extract powers of the index from polynomial factors.
  • Combine factor degrees and leading coefficients in products and quotients.
  • Verify a proposed leading-order equivalent by computing a ratio limit.
  • Identify when denominator factors are nonzero and when the result is useful for positive sequences.

Reading the Leading Behavior of Factored Expressions

A sequence may be written as a product or quotient of several polynomial factors. Its exact value can be complicated, but for large indices each factor is governed by its highest-degree term. Extracting those dominant powers and coefficients often reveals a much simpler expression that captures the sequence's leading behavior.

For instance, a factor such as \(4n-3\) has the same leading behavior as \(4n\): the constant shift matters to individual values, but becomes small relative to \(n\) as \(n\) grows. In a product or quotient, however, the leading coefficients matter as well as the powers. Keeping track of both is essential.

Definition: Two sequences \(a_n\) and \(b_n\) are asymptotically equivalent, written \(a_n\sim b_n\), if \(b_n\ne 0\) for all sufficiently large \(n\) and $$ \lim_{n\to\infty}\frac{a_n}{b_n}=1. $$

An equivalent is not generally equal to the original sequence. It is a simpler expression whose ratio to that sequence approaches \(1\). In this tutorial, the goal is to find such an expression when the sequence is given as a finite product or quotient of polynomial factors.

Extracting a Dominant Power from One Factor

Let \(P(n)=a_dn^d+a_{d-1}n^{d-1}+\cdots+a_0\) be a polynomial of degree \(d\), with leading coefficient \(a_d\ne 0\). Factor \(a_dn^d\) from the expression. The remaining terms inside the parentheses contain powers of \(1/n\):

$$ P(n)=a_dn^d\left(1+\frac{a_{d-1}}{a_d n}+\cdots+\frac{a_0}{a_d n^d}\right). $$

Each fraction inside the parentheses tends to \(0\) as \(n\to\infty\). Thus the parenthetical factor tends to \(1\), and

$$ \frac{P(n)}{a_dn^d} = 1+\frac{a_{d-1}}{a_d n}+\cdots+\frac{a_0}{a_d n^d} \longrightarrow 1. $$

Therefore \(P(n)\sim a_dn^d\). For a linear factor \(an+b\), this is the familiar extraction \(an+b=n(a+b/n)\), with \(a+b/n\to a\). The same idea applies to higher-degree factors: pull out the highest power of \(n\), and the remaining factor tends to the leading coefficient.

The Rule for Products and Quotients

Suppose a sequence is a finite product of polynomial factors divided by another finite product of polynomial factors. Degrees add when factors are multiplied and subtract when factors are divided. Leading coefficients multiply in the numerator and denominator. These two bookkeeping rules give the power and constant in the equivalent.

Theorem: Let $$ x_n=\lambda\frac{P_1(n)\cdots P_r(n)}{Q_1(n)\cdots Q_s(n)}, $$ where \(\lambda\ne 0\), each \(P_i\) and \(Q_j\) is a nonzero polynomial, and the denominator is nonzero for all sufficiently large \(n\). Write \(d_i\) and \(e_j\) for the degrees of \(P_i\) and \(Q_j\), and \(a_i\) and \(b_j\) for their leading coefficients. Then $$ x_n\sim \left(\lambda\frac{a_1\cdots a_r}{b_1\cdots b_s}\right) n^{\,d_1+\cdots+d_r-e_1-\cdots-e_s}. $$

The expression on the right includes both the net power of \(n\) and the full leading constant. The condition on the denominator ensures that the quotient is defined eventually; it holds, for example, if each denominator polynomial has no zeros at sufficiently large integer indices.

Why the Rule Works

For each numerator polynomial, write \(P_i(n)=a_i n^{d_i}(1+u_i(n))\), where \(u_i(n)\to 0\). Similarly, write \(Q_j(n)=b_j n^{e_j}(1+v_j(n))\), where \(v_j(n)\to 0\). These factorizations follow by pulling out each polynomial's leading term.

$$ x_n = \left(\lambda\frac{a_1\cdots a_r}{b_1\cdots b_s}\right) n^{\,\sum_i d_i-\sum_j e_j} \frac{\prod_{i=1}^r(1+u_i(n))} {\prod_{j=1}^s(1+v_j(n))}. $$

There are finitely many factors. Each \(1+u_i(n)\) and \(1+v_j(n)\) tends to \(1\), so their products tend to \(1\). In particular, each factor in the denominator is nonzero for sufficiently large \(n\). Divide \(x_n\) by the proposed equivalent: the constants and powers cancel, leaving exactly the quotient of parenthetical factors, which tends to \(1\). This proves the theorem by the definition of asymptotic equivalence.

Worked Examples

Worked Example: A Quotient with Equal Total Degrees

Find a leading-order equivalent for \(a_n=\dfrac{(2n+1)(3n-4)}{(5n+2)(n+7)}\).

There are two linear factors in the numerator and two in the denominator, so both total degrees are \(2\). The leading-coefficient quotient is \((2\cdot3)/(5\cdot1)=6/5\). To verify this directly, extract \(n\) from each factor:

$$ a_n = \frac{n(2+1/n)\,n(3-4/n)} {n(5+2/n)\,n(1+7/n)} = \frac{(2+1/n)(3-4/n)} {(5+2/n)(1+7/n)}. $$

Thus \(a_n\to (2\cdot3)/(5\cdot1)=6/5\). For the claimed equivalent \(6/5\), compute the ratio:

$$ \frac{a_n}{6/5} = \frac{5(2+1/n)(3-4/n)} {6(5+2/n)(1+7/n)} \longrightarrow \frac{5\cdot2\cdot3}{6\cdot5\cdot1} =1. $$

Therefore \(a_n\sim 6/5\). The ratio limit verifies the equivalent, including its constant.

Worked Example: A Quotient with One Net Power of \(n\)

Find a leading-order equivalent for \(b_n=\dfrac{(3n-2)^2(2n+1)}{(5n+4)(n-3)}\).

The numerator has total degree \(2+1=3\), and the denominator has total degree \(1+1=2\), so the net power is \(n^{3-2}=n\). The leading-coefficient quotient is \((3^2\cdot2)/(5\cdot1)=18/5\). Extract the powers:

$$ b_n = \frac{n^2(3-2/n)^2\,n(2+1/n)} {n(5+4/n)\,n(1-3/n)} = n\frac{(3-2/n)^2(2+1/n)} {(5+4/n)(1-3/n)}. $$

To check the proposed equivalent \((18/5)n\), divide by it:

$$ \frac{b_n}{(18/5)n} = \frac{5(3-2/n)^2(2+1/n)} {18(5+4/n)(1-3/n)} \longrightarrow \frac{5\cdot9\cdot2}{18\cdot5\cdot1} =1. $$

The denominator factor \(n-3\) is nonzero for every integer \(n>3\), so the sequence is defined for all sufficiently large indices. The ratio limit confirms \(b_n\sim (18/5)n\).

Worked Example: Quadratic Factors in a Quotient

Find a leading-order equivalent for \(c_n=\dfrac{(2n^2+3n+1)(3n-2)}{(n^2-4)(4n+5)}\).

The numerator and denominator both have total degree \(3\). Their leading coefficients are \(2\cdot3=6\) and \(1\cdot4=4\), respectively. The predicted equivalent is therefore the constant \(6/4=3/2\). Extract \(n^2\) from each quadratic factor and \(n\) from each linear factor:

$$ c_n = \frac{n^2(2+3/n+1/n^2)\,n(3-2/n)} {n^2(1-4/n^2)\,n(4+5/n)} = \frac{(2+3/n+1/n^2)(3-2/n)} {(1-4/n^2)(4+5/n)}. $$

Now divide by the proposed equivalent and take the limit:

$$ \frac{c_n}{3/2} = \frac{2(2+3/n+1/n^2)(3-2/n)} {3(1-4/n^2)(4+5/n)} \longrightarrow \frac{2\cdot2\cdot3}{3\cdot1\cdot4} =1. $$

The factors \(n^2-4\) and \(4n+5\) are nonzero for all sufficiently large \(n\). Hence the quotient is eventually defined, and \(c_n\sim 3/2\).

A Common Mistake: Losing the Leading Constant

When extracting dominant terms, it is easy to track the powers correctly but drop one of the coefficients. For example, consider \(d_n=\dfrac{(2n+1)(3n-4)}{5n+2}\). A mistaken calculation might keep the numerator's leading term \(6n^2\), note that the denominator has degree \(1\), and conclude \(d_n\sim 6n\). That conclusion has omitted the denominator's leading coefficient \(5\).

Extracting \(n\) from each linear factor gives the correct expression:

$$ d_n = \frac{n^2(2+1/n)(3-4/n)} {n(5+2/n)} = n\frac{(2+1/n)(3-4/n)}{5+2/n}. $$

The correct candidate is \((6/5)n\), not \(6n\). Check the ratio:

$$ \frac{d_n}{(6/5)n} = \frac{5(2+1/n)(3-4/n)} {6(5+2/n)} \longrightarrow \frac{5\cdot2\cdot3}{6\cdot5} =1. $$

By contrast, \(d_n/(6n)\to 1/5\), not \(1\), so \(6n\) is not asymptotically equivalent to \(d_n\). A ratio limit is a useful final check precisely because it catches a missing constant.

Using the Equivalent Carefully

For a quotient built from polynomial factors, a reliable calculation has three parts: add and subtract the factor degrees to find the net power; multiply the leading coefficients, keeping their signs, to find the constant; and verify that the ratio to the proposed equivalent tends to \(1\). The order matters less than checking both the power and the coefficient.

An equivalent describes behavior for sufficiently large indices. It does not assert equality at each index, and a factor may even vanish at a few small indices without affecting an eventual ratio limit. The denominator must, however, be nonzero for all sufficiently large indices so that the sequence and ratio are defined there. If the expression is being used as a comparison for positive sequences, also check that the original sequence is positive eventually; asymptotic equivalence alone does not guarantee positivity.

This method turns a complicated factored term into a simple power of \(n\) times a constant. The ratio check ensures that the simplification preserves the leading behavior rather than merely resembling it.

Check Your Understanding

For each question, track both the net power and the leading constant. Where requested, confirm the proposed equivalent using a ratio limit.

  1. Find a leading-order equivalent for \(\dfrac{(4n+1)(n-2)}{(3n+5)}\). What are its net power and leading constant?
  2. Find a leading-order equivalent for \(\dfrac{(2n^2+n)(5n-1)}{(4n^2+3)(n+2)}\).
  3. Why must the leading coefficients be included when finding the constant in an equivalent?
  4. What condition on the denominator is needed for a quotient sequence to be defined eventually?
  5. For \(\dfrac{(n+1)^2}{2n+3}\), verify the candidate equivalent \(n/2\) by computing the ratio limit.

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