3.3. THE LIMIT OF A SEQUENCE 83

Then it seems clear that limn→∞1

n2+1 = 0.In fact, this is true from the definition. Let

ε > 0 be given. Let nε ≥√

ε−1. Then if n > nε ≥√

ε−1, it follows that n2 +1 > ε−1 andso 0 < 1

n2+1 = an < ε . Thus |an −0|< ε whenever n is this large.Note the definition was of no use in finding a candidate for the limit. This had to be

produced based on other considerations. The definition is for verifying beyond any doubtthat something is the limit. It is also what must be referred to in establishing theoremswhich are good for finding limits.

Example 3.3.5 Let an = n2

Then in this case limn→∞ an does not exist.

Example 3.3.6 Let an = (−1)n .

In this case, limn→∞ (−1)n does not exist. This follows from the definition. Let ε = 1/2.If there exists a limit l, then eventually, for all n large enough, |an − l| < 1/2. However,|an −an+1|= 2 and so, 2= |an −an+1| ≤ |an − l|+ |l −an+1|< 1/2+1/2= 1 which cannothold. Therefore, there is no limit for this sequence.

Theorem 3.3.7 Suppose {an} and {bn} are sequences and that

limn→∞

an = a and limn→∞

bn = b.

Also suppose x and y are in R. Then

limn→∞

xan + ybn = xa+ yb (3.1)

limn→∞

anbn = ab (3.2)

If b ̸= 0,limn→∞

an

bn=

ab. (3.3)

Proof: The first of these claims is left for you to do. To do the second, let ε > 0 be givenand choose n1 such that if n ≥ n1 then |an −a|< 1.Then for such n, the triangle inequalityimplies

|anbn −ab| ≤ |anbn −anb|+ |anb−ab| ≤ |an| |bn −b|+ |b| |an −a|

≤ (|a|+1) |bn −b|+ |b| |an −a| .

Now let n2 be large enough that for n ≥ n2,

|bn −b|< ε

2(|a|+1), and |an −a|< ε

2(|b|+1).

Such a number exists because of the definition of limit. Therefore, let nε > max(n1,n2) .For n ≥ nε ,

|anbn −ab| ≤ (|a|+1) |bn −b|+ |b| |an −a|

< (|a|+1)ε

2(|a|+1)+ |b| ε

2(|b|+1)≤ ε.