Quotients of sequences

1. Properties of the notion of limit with division

We saw in previous pages that limits work nicely with respect to the sum of two sequences and with respect to the product of two sequences. We look at the case for division here.

Suppose ( a n ) and ( b n ) are convergent sequences. We look at the sequence a n b n . Immediately we encounter a problem we didn't see for addition or multiplication: the sequence a n b n may not be well-defined as some of the terms b n may be zero.

To rescue the situation when the sequence ( b n ) does not converge to zero, we have the following proposition.

Proposition.

Let b n m 0 . Then there is N 0 such that n   ( n N 0 | b n | > | m | / 2 ) .

Proof.

Since b n m 0 there is N 0 such that n N 0   | b n - m | < | m | / 2 . In particular n N 0   | b n | > | m | / 2 .

This means that a n b n is at least defined for all n N 0 , for some N 0 , and this is good enough to mean that the sequence a n b n is defined from some point onwards.

Theorem.

Let ( a n ) and ( b n ) be convergent sequences with limits l and m respectively, and suppose m 0 . Let the sequence ( c n ) be defined by c n = a n b n . Then c n l m as n .

Proof.

This is like the analogous result for product, except the algebra is a touch trickier.

Let sequences ( a n ) and ( b n ) and l , m be such that ε > 0   N   n   n N | a n - l | < ε and ε > 0   N   n   n N | b n - m | < ε . Assume that m 0 . We must prove that a n / b n l / m .

Since b n m 0 there is N 0 such that n N 0   | b n | > | m | / 2 .

Subproof.

Assume that l 0 .

Subproof.

Now let ε > 0 be arbitrary.

Subproof.

For the moment, assume that l 0 and let N 1 such that n   n N 1 | a n - l | < | m ε | 4 and N 2 such that n   n N 2 | b n - m | < | m 2 ε | 4 | l | .

Let N = max ( N 0 , N 1 , N 2 ) .

Subproof.

Let n be arbitrary.

Subproof.

Assume n N .

Then n N 1 and n N 2 so | a n - l | < | m ε | 4 and | b n - m | < | m 2 ε | 4 | l | .

Now we estimate a n b n - l m using the triangle inequality and the fact m 0 :

| a n b n - l m | = | a n m - l b n b n m | | a n m - l m | + | l m - l b n | | b n m | = 1 | b n | | a n - l | + | l | | b n m | | b n - m |

Each of these last two terms is less that ε / 2 : for the first, | a n - l | < | m ε | 4 < | b n | ε 2 since n N 0 , N 1 and | b n | > | m | 2 ; and for the second, | b n - m | < | m 2 ε | 4 | l | < | m | | b n | | l | ε 2 since n N 0 , N 2 .

Thus n N | a n b n - l m | < ε .

Thus n   n N | a n b n - l m | < ε .

So N   n N   | a n b n - l m | < ε .

Subproof.

In the case l = 0 , the argument is easier: choose N N 0 such that | a n | < | m ε | 2 , so for all n N we have | a n b n | < | m ε | 2 | b n | < ε as you can check.

So N   n N   | a n b n - l m | < ε holds in either case, l = 0 or l 0 .

Thus ε > 0   N   n N   | a n b n - l m | < ε , as required.

2. What happens when the sequence on the bottom converges to zero?

Division, x / y , is not defined at y = 0 , so it is meaningless to even talk about the division function at such places. For sequences ( a n ) and ( b n ) both converging to zero as n , the sequence a n b n might not make sense, as b n might be zero for infinitely many n , though in other cases this limit might exist.

Even if b n is always nonzero, the limit lim a n b n sometimes does exist and sometimes doesn't. For a case when it does exist, take a n = sin 1 n and b n = 1 n . For a case when it doesn't, take a n = 1 n and b n = 1 n 2 .

3. Continuity

The idea of pushing functions through limits in expressions involving plus, times, divides, etc., is an important one and is discussed in more abstract terms in the next web page. This idea is often called Algebra Of Limits. I dislike the name AOL a lot, and in any case this result is not algebra, but analysis, as this idea applies to all sorts of functions, not just algebraic ones. For some nice functions, like plus and times, this operation of pushing though limits works and for others it doesn't. Those functions for which it works are called continuous. The definition and more examples are the subject of the next page.