Limit Laws for Quotients

Dividing Sequences with Limits

The Limit of a Product shows that limits can be multiplied term by term. Division requires an additional condition: the denominator sequence must approach a nonzero number. If the limiting denominator is zero, its terms may become arbitrarily small, so taking reciprocals can magnify even a small change in the denominator.

The key fact is that if \(b_n\to M\) and \(M\neq0\), then all sufficiently late terms \(b_n\) stay a definite distance from zero. This makes their reciprocals well-defined and allows us to control how close \(1/b_n\) is to \(1/M\). We first prove this reciprocal law, then use the Limit of a Product to obtain the quotient law.

Theorem (Limit of a Reciprocal): Suppose \(b_n\to M\), where \(M\neq0\), and suppose \(b_n\neq0\) for every \(n\). Then

\frac{1}{b_n}\longrightarrow\frac{1}{M}.

Proof. Let \(\varepsilon>0\). Since \(b_n\to M\), the epsilon–N definition gives an index \(N_1\) such that

|b_n-M|<\min\left\{\frac{|M|}{2},\frac{\varepsilon |M|^2}{2}\right\}

whenever \(n\geq N_1\). The first part of this bound ensures that the denominator is bounded away from zero. Indeed, by the triangle inequality,

|M|\leq |M-b_n|+|b_n|,

so, for \(n\geq N_1\),

|b_n|\geq |M|-|M-b_n|>\frac{|M|}{2}.

The reciprocal error can now be estimated directly. For every \(n\geq N_1\),

\left|\frac{1}{b_n}-\frac{1}{M}\right| =\frac{|M-b_n|}{|b_n||M|} <\frac{2|M-b_n|}{|M|^2} <\varepsilon.

The last inequality follows from \(|M-b_n|<\varepsilon |M|^2/2\). Thus for every \(\varepsilon>0\), the reciprocal terms are within \(\varepsilon\) of \(1/M\) for all sufficiently large \(n\). Therefore \(1/b_n\to1/M\). \(\square\)

This proof uses two different consequences of \(b_n\to M\). Closeness to \(M\) first prevents \(b_n\) from getting too close to zero; once that lower bound is available, the identity for the reciprocal error gives the desired estimate. The assumption \(M\neq0\) is essential to both steps.

The Limit of a Quotient

We can now divide one convergent sequence by another, provided the denominator sequence approaches a nonzero limit. In the theorem below, requiring \(b_n\neq0\) for every \(n\) makes the quotient sequence defined at every index.

Theorem (Limit of a Quotient): Suppose \(a_n\to L\) and \(b_n\to M\), where \(M\neq0\), and suppose \(b_n\neq0\) for every \(n\). Then

\frac{a_n}{b_n}\longrightarrow\frac{L}{M}.

Proof. By the Limit of a Reciprocal, \(1/b_n\to1/M\). The Limit of a Product, established in the previous tutorial, applies to \(a_n\) and \(1/b_n\). Since \(a_n/b_n=a_n(1/b_n)\), it gives

\frac{a_n}{b_n} =a_n\frac{1}{b_n} \longrightarrow L\frac{1}{M} =\frac{L}{M}.

This proves the quotient law. \(\square\)

The quotient law also applies if \(b_n\) is nonzero only for all sufficiently large \(n\). In that case, \(a_n/b_n\) is defined on a tail of the sequence. Assign any real values to the finitely many earlier terms; the resulting full sequence has the same limit, by the theorem Finite Changes Preserve Convergence. The limit statement is about sufficiently late terms, so those initial assignments do not change it.

Worked Examples with Quotient Limits

Worked Example: A Quotient with a Positive Limiting Denominator

Consider the sequences

a_n=4-\frac{1}{n+1}, \qquad b_n=2+\frac{3}{n+1}.

Since \(1/(n+1)\to0\), the Limit of a Difference and the Limit of a Sum give \(a_n\to4\) and \(b_n\to2\). The denominator is nonzero for every \(n\), because \(2+3/(n+1)>0\). The quotient law therefore gives

\frac{a_n}{b_n}\longrightarrow\frac{4}{2}=2.

The first two terms illustrate the quotient itself. At \(n=0\), \(a_0=3\) and \(b_0=5\), so \(a_0/b_0=3/5\); substituting \(n=0\) into the defining quotient also gives \((4-1)/(2+3)=3/5\). At \(n=1\), \(a_1=7/2\) and \(b_1=7/2\), so the quotient is \(1\); direct substitution gives \((4-1/2)/(2+3/2)= (7/2)/(7/2)=1\). These early values need not equal the limit \(2\).

Worked Example: Taking the Reciprocal of a Convergent Sequence

Let

c_n=3-\frac{2}{n+1}.

Because \(2/(n+1)\to0\), we have \(c_n\to3\), a nonzero limit. Also, \(c_n\geq1\) for every \(n\), since \(2/(n+1)\leq2\); hence \(c_n\) is never zero. The Limit of a Reciprocal yields

\frac{1}{c_n}\longrightarrow\frac{1}{3}.

For example, \(c_0=1\), so \(1/c_0=1\). At \(n=1\), \(c_1=2\), so \(1/c_1=1/2\). These agree with the defining expression: \(1/(3-2)=1\) at \(n=0\), and \(1/(3-1)=1/2\) at \(n=1\). The reciprocal terms approach \(1/3\), even though the first terms are not especially close to it.

Worked Example: A Quotient When Both Sequences Approach Zero

Define

u_n=\frac{3}{n+1}, \qquad v_n=\frac{6}{n+1}.

Both sequences tend to zero, and \(v_n\neq0\) for every \(n\). Nevertheless, the quotient law does not apply: its denominator limit would be \(M=0\). In this example the quotient can be calculated exactly:

\frac{u_n}{v_n} =\frac{3/(n+1)}{6/(n+1)} =\frac{3}{6} =\frac{1}{2}

for every \(n\). At \(n=0\), the quotient is \(3/6=1/2\); at \(n=2\), it is \(1\) divided by \(2=1/2\). Thus the quotient converges to \(1/2\), but this conclusion follows from the exact cancellation, not from the quotient law. A zero limiting denominator means the law is unavailable; it does not by itself determine whether the quotient converges.

Worked Example: A Quotient That Does Not Approach a Finite Limit

Let \(p_n=1\) and \(q_n=1/(n+1)\). The denominator is positive at every index, but \(p_n\to1\) and \(q_n\to0\). The quotient is

\frac{p_n}{q_n} =\frac{1}{1/(n+1)} =n+1.

At \(n=0\), the quotient is \(1/(1)=1\); at \(n=1\), it is \(1/(1/2)=2\). In general it is the arithmetic sequence \(n+1\), with nonzero common difference \(1\). By the Convergence Criterion for Arithmetic Sequences, it does not converge to a real number. This example shows why a denominator that tends to zero cannot simply be treated as though it had a nonzero limiting value.

Why the Nonzero-Limit Condition Matters

The condition \(M\neq0\) is not a technical convenience. It guarantees that \(b_n\) is eventually separated from zero. If \(b_n\to M\neq0\), then eventually \(|b_n|>|M|/2\), as the reciprocal proof showed. This lower bound controls the denominator in the error formula

\left|\frac{1}{b_n}-\frac{1}{M}\right| =\frac{|b_n-M|}{|b_n||M|}.

If \(M=0\), no positive lower bound on \(|b_n|\) follows from convergence. The quotients in the last two examples show that very different outcomes are possible: one quotient is constantly \(1/2\), while the other is \(n+1\) and has no finite limit. The behavior depends on how the numerator and denominator approach zero, not just on their individual limits.

A related issue is whether the quotient is defined at every index. If \(b_n\to M\neq0\), then the estimate in the reciprocal proof guarantees that \(b_n\neq0\) for all sufficiently large \(n\), whether or not some earlier terms vanish. A quotient with a denominator equal to zero at an early index is not defined there, but its tail is defined; convergence can be discussed after assigning arbitrary values at the finitely many missing indices. In applications, it is useful to distinguish this eventual definition from a quotient sequence defined at every index.

When applying a quotient law, check both parts of its hypothesis: the denominator sequence must converge, and its limit must be nonzero. Then the quotient limit is obtained by dividing the limits. If the denominator limit is zero, analyze the quotient directly or use another argument; the quotient law gives no conclusion.

Check Your Understanding

Use the reciprocal and quotient limit laws, and pay attention to the limiting denominator.

If \(a_n\to-6\) and \(b_n\to3\), with \(b_n\neq0\) for every \(n\), what is the limit of \(a_n/b_n\)?
Why does \(b_n\to M\neq0\) imply that \(|b_n|\) is bounded below by a positive number for all sufficiently large \(n\)?
If \(a_n\to0\) and \(b_n\to0\), can the quotient law determine the limit of \(a_n/b_n\)? Explain.
Suppose \(b_n\to5\), but \(b_0=0\). What can be said about the quotient \(a_n/b_n\) for sufficiently large \(n\), and how can convergence be treated if the quotient is not defined at \(n=0\)?
Give an example of two sequences that both tend to zero but whose quotient is constantly equal to a nonzero number.

What You'll Learn