Introduction
A sequence can be built from simpler sequences using multiplication and division. Knowing the limits of those simpler sequences often lets us find the limit of the new sequence without starting from scratch.
These calculations have an important qualification: the quotient law requires the denominator sequence to approach a nonzero limit. The product law has no such restriction. Understanding both the rules and their conditions helps prevent errors when expressions contain quantities that become small.
Definitions and Limit Laws
A real sequence \( (a_n) \) converges to \(A\) if, for every \(\varepsilon>0\), there is an \(N\in\mathbb{N}\) such that \(n\geq N\) implies \(|a_n-A|<\varepsilon\). We write \(a_n\to A\) as \(n\to\infty\), or \(\lim_{n\to\infty}a_n=A\).
The product rule says that the limit of a product is the product of the limits. The quotient rule says that the limit of a quotient is the quotient of the limits, provided the denominator limit is not zero. The condition \(B\ne 0\) is essential: it ensures the quotient is defined eventually and lets us control how close its denominator is to zero.
Why the Product Law Works
The product law rests on two facts. First, every convergent sequence is bounded. Second, the difference between the products can be split into terms involving the differences \(a_n-A\) and \(b_n-B\).
To see the boundedness fact, suppose \(a_n\to A\). Taking \(\varepsilon=1\) in the definition gives an index \(N_0\) such that, for \(n\geq N_0\), \(|a_n-A|<1\). The triangle inequality then gives \(|a_n|<|A|+1\) for those \(n\). There are only finitely many earlier terms, so together they and the tail have a finite bound \(M\) for \(|a_n|\).
Now suppose \(a_n\to A\) and \(b_n\to B\). Let \(M\) be a positive bound for \(|a_n|\), so \(|a_n|\leq M\) for every \(n\). For any \(\varepsilon>0\), choose indices large enough that
Such indices exist by the two convergence assumptions. For \(n\) beyond both indices, add and subtract \(a_nB\) and apply the triangle inequality:
The last inequality holds because \(|B|/(|B|+1)<1\). This proves, directly from the definition, that \(a_nb_n\to AB\). The boundedness step matters: convergence of \(b_n\) alone makes \(b_n-B\) small, but the factor \(|a_n|\) must also be controlled.
Why the Quotient Law Requires a Nonzero Limit
Suppose \(b_n\to B\) with \(B\ne 0\). Eventually, \(b_n\) stays away from zero. In fact, the definition of convergence gives an index after which \(|b_n-B|<|B|/2\). The reverse triangle inequality then gives
Thus \(b_n\ne 0\) eventually, and its reciprocal is bounded there by \(2/|B|\). For the quotient, subtract the proposed limit and use the lower bound on \(|b_n|\):
Given \(\varepsilon>0\), convergence lets us choose \(n\) large enough that the first term on the final line is less than \(\varepsilon/2\). If \(A\ne0\), choose \(n\) large enough also to make the second term less than \(\varepsilon/2\). If \(A=0\), that second term is zero. In either case, the whole difference is less than \(\varepsilon\), proving the quotient law.
Worked Examples
Worked Example: The Product of Two Convergent Sequences
Find the limit of \(u_nv_n\), where \(u_n=1+\frac{1}{n}\) and \(v_n=2-\frac{1}{n}\), for \(n\in\mathbb{N}\).
First, \(|u_n-1|=1/n\to0\), and \(|v_n-2|=1/n\to0\). For example, for any \(\varepsilon>0\), choosing \(N>1/\varepsilon\) makes both differences less than \(\varepsilon\) whenever \(n\geq N\). Therefore \(u_n\to1\) and \(v_n\to2\). The product law gives \(u_nv_n\to1\cdot2=2\).
Check by computing the terms and comparing with \(2\):
Since \(1/n\to0\), the difference from \(2\) tends to zero. This confirms the limit.
Worked Example: A Quotient with a Nonzero Denominator Limit
Find the limit of \(\frac{2+1/n}{3-1/n}\) for \(n\in\mathbb{N}\).
The numerator tends to \(2\), and the denominator tends to \(3\), which is nonzero. Also, for every \(n\geq1\), \(3-1/n\geq2\), so the denominator is nonzero. The quotient law therefore gives
To check directly, subtract \(2/3\). The numerator after bringing the terms over a common denominator is
Hence
The final bound tends to zero, confirming that the quotient tends to \(2/3\).
Worked Example: Combining Product and Quotient Laws
Find the limit of \(w_n=\frac{(1+1/n)(2-1/n)}{4+1/n}\).
The two factors in the numerator tend to \(1\) and \(2\), so the product tends to \(2\). The denominator tends to \(4\ne0\). Applying the quotient law gives
To check, the numerator is \(2+1/n-1/n^2\). Subtracting half the denominator from it gives
Since \(4+1/n\geq4\), it follows that
The last inequality uses \(1/n^2\leq1/n\) for \(n\geq1\). The bound tends to zero, so the direct check confirms the limit.
A Common Mistake: Dividing Limits When the Denominator Limit Is Zero
A tempting but invalid calculation is to use the quotient rule whenever the numerator and denominator both converge, even when the denominator limit is zero. The theorem gives no conclusion in that case.
For example, let the numerator be \(a_n=1/n\). If the denominator is also \(b_n=1/n\), then both sequences tend to zero, but \(a_n/b_n=1\) for every \(n\), so the quotient tends to \(1\). If instead the denominator is \(c_n=1/n^2\), it too tends to zero, while
which does not converge to a finite real number. Indeed, for any real bound \(K\), choosing an integer \(n>K\) gives \(n>K\). Thus the same numerator limit and denominator limit can lead to different quotient behavior.
Using the Laws Reliably
When applying these laws, identify the component sequences and their limits before combining them. For a product, no nonzero condition is needed. For a quotient, check that the denominator limit is not zero. Then verify that the original quotient is defined for all sufficiently large indices, as guaranteed by the theorem when its condition holds.
These laws are useful because they let a complicated sequence inherit its limit from simpler convergent components. They also make the boundary of a calculation precise: knowing the limits of the numerator and denominator is not enough if the denominator limit is zero. A direct estimate of the difference from the proposed limit, as in the examples, is a useful way to check the result.
Check Your Understanding
Use the limit laws and their conditions to answer these questions.
- If \(a_n\to-2\) and \(b_n\to5\), what is the limit of \(a_nb_n\)?
- If \(a_n\to6\) and \(b_n\to-3\), what is the limit of \(a_n/b_n\), and why is the quotient law applicable?
- Why does the product-law proof need a bound on one of the convergent sequences?
- Can the quotient law determine the limit of \(a_n/b_n\) when \(a_n\to0\) and \(b_n\to0\)? Explain.
- If \(a_n\to A\) and \(b_n\to B\ne0\), why is \(b_n\ne0\) for all sufficiently large \(n\)?
