76 Chapter 3. Limits
(2) Letbn = 1
3 − 1
2n+ 1
6n2. The sequence (bn)∞n=1can be represented by 0, 1
8, 5
27, 7
32, 6
25, . . . (3.2.2)
Remark
• It is not a good way to describe a sequence by listing a few terms in the sequence. For example, in (3.2.1) or (3.2.2), it may not be easy to find a formula for thenth term. Moreover, different people may obtain different formulas. It is better to describe a sequence by writing down a formula for thenth term explicitly.
• To denote a sequence, some authors use the notation{an}∞n=1instead of (an)∞n=1.
Definition A sequence (an)∞n=1is said to beconvergentif there exists a real numberLsuch that (∗) anis arbitrarily close toLifnis sufficiently large.
RemarkCondition (∗) means that we can make|an−L|as small as we want by takingnlarge enough. For the sequence (an)∞n=1wherean= 1
2n, we can makeanarbitrarily close to 0 by takingnlarge enough. For example, if we want1
2n −0<0.01, we can taken>7; if we want1
2n −0<0.001, we can taken>10 etc.
Intuitively, Condition (∗) means that if we letnincrease without bound (or letnapproach “∞”, an imag- inary point very far on the right), the value an approachesL. Geometrically, this means that the point (n,an) approaches the horizontal liney= Lasnincreases without bound.
1 2 3 4 5 6 7 8 9 10 11 12 13
Figure 3.4 RemarkFor simplicity, instead of saying Condition (∗), we will say
(∗∗) anis close toLifnis large.
In the definition of “convergent”, it is clear that ifLexists, then it is unique. We say thatLis thelimitof (an)∞n=1and we write lim
n→∞an= L.
FAQ Can we just write liman=L, omittingn→ ∞?
Answer For sequences, this will not cause ambiguity. However, for functions, we will consider (in later sections)limits at infinityas well aslimits at a point a(wherea∈R). The notations lim
x→∞f(x) and lim
x→af(x) have
different meanings.
The following rules can be proved by definition using an alternative method, called-Nmethod, to describe condition (∗). However, the-Ndefinition is outside the scope of this course. Readers may convince themselves that the rules are true using intuition.
3.2. Limits of Sequences 77
Rules for Limits of Sequences (L1) lim
n→∞k=k (wherekis a constant) (L2) lim
n→∞
1
np =0 (where pis a positive constant) (L3) lim
n→∞
1
bn =0 (wherebis a constant greater than 1) (L4) lim
n→∞(an+bn)= lim
n→∞an+ lim
n→∞bn (L5) lim
n→∞anbn = lim
n→∞an· lim
n→∞bn (L6) lim
n→∞
an bn =
n→∞lim an
n→∞lim bn provided that lim
n→∞bn,0.
Remark
• The meaning of (L1) is that if an = k for all n where k is a constant, then the sequence (an)∞n=1 is convergent and its limit isk.
• The meaning of (L4) is that if both (an)∞n=1 and (bn)∞n=1 are convergent and their limits are L and M respectively, then (an+bn)∞n=1is also convergent and its limit isL+M.
• The following is a special case of (L5). It can be obtained by puttingan=kfor allnand applying (L1).
(L5s) lim
n→∞kbn=k lim
n→∞bn
• Using (L4) and (L5s), we get (L40) lim
n→∞(an−bn)= lim
n→∞an− lim
n→∞bn
In fact, Rule (L4) is valid for sum and difference of finitely many sequences. This general result will be referred to as Rule (L4). Similarly, the result for product of finitely many sequences will be referred to as Rule (L5).
In Problem 1 in the last section, the sequence obtained can be represented by the formulaan =4+ 1
2n. Our intuition tells us that the limit of the sequence is 4. Below we use rules for limits to justify this result.
Example Find lim
n→∞ 4+ 1 2n
!
, if it exists.
ExplanationThe sequence under consideration is given byan=4+ 1
2n. The question asks for the following (1) Does the limit of (an)∞n=1exist or not (or equivalently, is the sequence convergent)?
(2) If the answer to (1) is affirmative, find the limit.
Solution lim
n→∞ 4+ 1 2n
!
= lim
n→∞4+ lim
n→∞
1
2n Rule (L4)
= 4+0 Rules (L1) and (L3)
= 4
78 Chapter 3. Limits
RemarkBelow is the logic in the above calculation:
(1) In the first step, because the constant sequence (4)∞n=1 and the sequence1
2n
∞
n=1are convergent, we can apply Rule (L4).
(2) The limits of the two sequences are found by Rule (L1) and Rule (L3) respectively in the second step.
The sequence in the next example is the one obtained in Problem 2 in the last section.
Example Find lim
n→∞
2n3−3n2+n
6n3 , if it exists.
Solution
n→∞lim
2n3−3n2+n
6n3 = lim
n→∞
1 3 − 1
2n+ 1 6n2
!
Rewrite the expression
= lim
n→∞
1 3− lim
n→∞
1 2· 1
n
! + lim
n→∞
1 6 · 1
n2
!
Rule (L4), rewrite 2nd and 3rd terms
= 1 3 − 1
2· lim
n→∞
1 n + 1
6· lim
n→∞
1
n2 Rules (L1) and (L5s)
= 1 3 − 1
2·0+ 1
6·0 Rule (L2)
= 1
3
Example Find lim
n→∞(1+2n), if it exists.
Solution Limit does not exist. This is because we can’t find any real number Lsatisfying the condition that
2n+1 is close toLifnis large.
RemarkIf we apply rules for limits, we get
n→∞lim(2n+1) = lim
n→∞2n+ lim
n→∞1 Rule (L4)
= 2 lim
n→∞n+1 Rules (L1) and (L5s) However, we can’t proceed because lim
n→∞ndoes not exist. From this, we see that the given limit does not exist.
FAQ Can we say that lim
n→∞(1+2n) is∞?
Answer Limit of a sequence is a real number satisfying Condition (∗) given in the definition on page 76.
Because∞is not a real number, we should say that the limit does not exist.
In the next section, we will discuss the meaning of lim
x→∞f(x)=∞etc.
Example Find lim
n→∞
n+1
2n+1, if it exists.
ExplanationWe can’t use Rule (L6) because limits of the numerator and the denominator do not exist. However, we can’t conclude from this that the given limit does not exist. To find the limit, we use a trick: divide the numerator and the denominator by n.
3.2. Limits of Sequences 79
Solution lim
n→∞
n+1
2n+1 = lim
n→∞
n+1 n 2n+1
n
Divide numerator and denominator byn
=
n→∞lim 1+ 1n
n→∞lim 2+ 1n Rule (L6), rewrite numerator and denominator
=
n→∞lim 1+ lim
n→∞
1n n→∞lim 2+ lim
n→∞
1n
Rule (L4)
= 1+0
2+0 Rules (L1) and (L2)
= 1
2
RemarkWe can apply the following shortcut (called theLeading Terms Rule). The method is to throw away the constant term 1 in the numerator and the denominator(note that if n is very large, compared with n or2n, 1is very small).
n→∞lim n+1
2n+1 = lim
n→∞
n 2n
= lim
n→∞
1
2 = 1
2.
The Leading Terms Rule for limits of functions at infinity will be discussed in more details in the next section (see page 83).
Exercise 3.2
1. For each of the following, find the limit if it exists.
(a) lim
n→∞
√1
n (b) lim
n→∞(7− 5
n2) (c) lim
n→∞
3n2−4000
2n2+10000 (d) lim
n→∞
n2−12345 n+1
(e) lim
n→∞
5n2+4
2n3+3 (f) lim
n→∞
1
n +(−1)n
2. Suppose $50,000 is deposited at a bank and the annual interests rate is 2%.
(a) What amount (correct to the nearest cent) will the account have after one year if interests is (i) compounded quarterly;
(ii) compounded monthly?
(b) If interest is compoundedntimes a year, express the amountAnafter one year in terms ofn.
∗(c) Does lim
n→∞Anexist? What is the value?
∗3. For each of the following sequences (an)∞n=1, use computer to find the first 100 (or more) terms. Does
n→∞limanexists? If yes, what is the value?
(a) an= 1+ 1
n
n
(b) an= 1+ 2
n
n
(c) an=nsin1
n (angles are in radians)
80 Chapter 3. Limits
∗4. Suppose (an) is a sequence such that 0< anfor allnanda1 > a2 >a3 > · · ·. Does lim
n→∞anexist? What can you tell about the limit?