part ii. sequences of real numbersalmus/4389_an_part2.pdf · 2016-03-29 · part ii. sequences of...

PART II. SEQUENCES OF REAL NUMBERS

II.1. CONVERGENCE

Definition 1. A sequence is a real-valued function f whose domain is the set positive integers(N). The numbers f(1), f(2), · · · are called the terms of the sequence.

Notation Function notation vs subscript notation:

f(1) ≡ s1, f(2) ≡ s2, · · · , f(n) ≡ sn, · · · .

In discussing sequences the subscript notation is much more common than functional notation. We’lluse subscript notation throughout our treatment of analysis.

Specifying a sequence There are several ways to specify a sequence.

1. By giving the function. For example:

(a) sn =1n

or {sn} ={

1n

}. This is the sequence {1,

12,

13,

14, . . . ,

1n

, . . .}.

(b) sn =n − 1

n. This is the sequence {0,

12,

23,

34, . . . ,

n − 1n

, . . .}.

(c) sn = (−1)nn2. This is the sequence {−1, 4,−9, 16, . . . , (−1)nn2, . . .}.

2. By giving the first few terms to establish a pattern, leaving it to you to find the function. Thisis risky – it might not be easy to recognize the pattern and/or you can be misled.

(a) {sn} = {0, 1, 0, 1, 0, 1, . . .}. The pattern here is obvious; can you devise the function? It’s

sn =1 − (−1)n)

2or sn =

{0, n odd1, n even

(b) {sn} ={

2,52,103

,174

,265

, . . .

}, sn =

n2 + 1n

.

(c) {sn} = {2, 4, 8, 16,32, . . .}. What is s6? What is the function? While you might say 64and sn = 2n, the function I have in mind gives s6 = π/6:

sn = 2n + (n − 1)(n − 2)(n − 3)(n − 4)(n − 5)[

π

720− 64

120

]

3. By a recursion formula. For example:

(a) sn+1 =1

n + 1sn, s1 = 1. The first 5 terms are

{1,

12,16,

124

,1

120, . . .

}. Assuming that

the pattern continues sn =1n!

.

(b) sn+1 =12(sn + 1), s1 = 1. The first 5 terms are {1, 1, 1, 1,1, . . .}. Assuming that the

pattern continues sn = 1 for all n; {sn} is a “constant” sequence.

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Definition 2. A sequence {sn} converges to the number s if to each ε > 0 there correspondsa positive integer N such that

|sn − s| < ε for all n > N.

The number s is called the limit of the sequence.

Notation “{sn} converges to s” is denoted by

limn→∞

sn = s, or by lim sn = s, or by sn → s.

A sequence that does not converge is said to diverge.

Examples Which of the sequences given above converge and which diverge; give the limits of theconvergent sequences.

THEOREM 1. If sn → s and sn → t, then s = t. That is, the limit of a convergent sequenceis unique.

Proof: Suppose s 6= t. Assume t > s and let ε = t − s. Since sn → s, there exists a positiveinteger N1 such that |s− sn| < ε/2 for all n > N1. Since sn → t, there exists a positive integerN2 such that |t− sn| < ε/2 for all n > N2. Let N = max{N1, N2} and choose a positive integerk > N . Then

t − s = |t − s| = |t − sk + sk − s| ≤ |t − sk| + |s − sk| <ε

2+

ε

2= ε = t − s,

a contradiction. Therefore, s = t.

THEOREM 2. If {sn} converges, then {sn} is bounded.

Proof: Suppose sn → s. There exists a positive integer N such that |s− sn| < 1 for all n > N .Therefore, it follows that

|sn| = |sn − s + s| ≤ |sn − s| + |s| < 1 + |s| for all n > N.

Let M = max{|s1|, |s2|, . . . , |sN |, 1 + |s|}. Then |sn| < M for all n. Therefore {sn} isbounded.

THEOREM 3. Let {sn} and {an} be sequences and suppose that there is a positive number k

and a positive integer N such that

|sn| ≤ k an for all n > N.

If an → 0, then sn → 0.

Proof: Note first that an ≥ 0 for all n > N . Since an → 0, there exists a positive integer N1

such that |an| < ε/k. Without loss of generality, assume that N1 ≥ N . Then, for all n > N1,

|sn − 0| = |sn| ≤ k an < kε

k= ε.

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Therefore, sn → 0.

Corollary Let {sn} and {an} be sequences and let s ∈ R. Suppose that there is a positivenumber k and a positive integer N such that

|sn − s| ≤ k an for all n > N.

If an → 0, then sn → s.

Exercises 2.1

1. True – False. Justify your answer by citing a theorem, giving a proof, or giving a counter-example.

(a) If sn → s, then sn+1 → s.

(b) If sn → s and tn → s, then there is a positive integer N such that sn = tn for alln > N .

(c) Every bounded sequence converges

(d) If to each ε > 0 there is a positive integer N such that n > N implies sn < ε, thensn → 0.

(e) If sn → s, then s is an accumulation point of the set S = {s1, s2, · · · }.

2. Prove that lim3n + 1n + 2

= 3.

3. Prove that limsin n

n= 0.

4. Prove or give a counterexample:

(a) If {sn} converges, then {|sn|} converges.

(b) If {|sn|} converges, then {sn} converges.

5. Give an example of:

(a) A convergent sequence of rational numbers having an irrational limit.

(b) A convergent sequence of irrational numbers having a rational limit.

6. Give the first six terms of the sequence and then give the nth term

(a) s1 = 1, sn+1 = 12(sn + 1)

(b) s1 = 1, sn+1 = 12sn + 1

(c) s1 = 1, sn+1 = 2sn + 1

7. use induction to prove the following assertions:

(a) If s1 = 1 and sn+1 =n + 12n

sn, then sn =n

2n−1.

(b) If s1 = 1 and sn+1 = sn − 1n(n + 1)

, then sn =1n

.

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8. Let r be a real number, r 6= 0. Define a sequence {Sn} by

S1 = 1

S2 = 1 + r

S3 = 1 + r + r2

...

Sn = 1 + r + r2 + · · ·+ rn−1

...

(a) Suppose r = 1. What is Sn for Sn = 1, 2, 3, . . . ?

(b) Suppose r 6= 1. Find a formula for Sn.

9. Set an =1

n(n + 1), n = 1, 2, 3, . . ., and form the sequence

S1 = a1

S2 = a1 + a2

S3 = a1 + a2 + a3

...

Sn = a1 + a2 + a3 + · · ·+ an

...

Find a formula for Sn.

II.2. LIMIT THEOREMS

THEOREM 4. Suppose sn → s and tn → t. Then:

1. sn + tn → s + t.

2. sn − tn → s − t.

3. sntn → st.

Special case: ksn → ks for any number k.

4. sn/tn → s/t provided t 6= 0 and tn 6= 0 for all n.

THEOREM 5. Suppose sn → s and tn → t. If sn ≤ tn for all n, then s ≤ t.

Proof: Suppose s > t. Let ε =s − t

2. Since sn → s, there exists a positive integer N1 such that

|sn − s| < ε for all n > N1. This implies that s − ε < sn < s + ε for all n > N1. Similarly, thereexists a positive integer N2 such that t− ε < tn < t + ε for all n > N2. Let N = max {N1, N2}.Then, for all n > N , we have

tn < t + ε = t +s − t

2=

s + t

2= s − ε < sn

16

which contradicts the assumption sn ≤ tn for all n.

Corollary Suppose tn → t. If tn ≥ 0 for all n, then t ≥ 0.

Infinite Limits

Definition 3. A sequence {sn} diverges to +∞ (sn → +∞) if to each real number M thereis a positive integer N such that sn > M for all n > N . {sn} diverges to −∞ (sn → −∞)if to each real number M there is a positive integer N such that sn < M for all n > N .

THEOREM 6. Suppose that {sn} and {tn} are sequences such that sn ≤ tn for all n.

1. If sn → +∞, then tn → +∞.

2. If tn → −∞, then sn → −∞.

THEOREM 7. Let {sn} be a sequence of positive numbers. Then sn → +∞ if and only if1/sn → 0.

Proof: Suppose sn → ∞. Let ε > 0 and set M = 1/ε. Then there exists a positive integer N

such that sn > M for all n > N . Since sn > 0,

1/sn < 1/M = ε for all n > N

which implies 1/sn → 0.

Now suppose that 1/sn → 0. Choose any positive number M and let ε = 1/M . Then thereexists a positive integer N such that

0 <1sn

< ε =1M

for all n > N. that is,1sn

<1M

.

Since sn > 0 for all n, 1/sn < 1/M for all n > N implies sn > M for all n > N . Therefore,sn → ∞.

Exercises 2.2

1. Prove or give a counterexample.

(a) If sn → s and sn > 0 for all n, then s > 0.

(b) If {sn} and {tn} are divergent sequences, then {sn + tn} is divergent.

(c) If {sn} and {tn} are divergent sequences, then {sntn} is divergent.

(d) If {sn} and {sn + tn} are convergent sequences, then {tn} is convergent.

(e) If {sn} and {sntn} are convergent sequences, then {tn} is convergent.

(f) If {sn} is not bounded above, then {sn} diverges to +∞.

2. Determine the convergence or divergence of {sn}. Find any limits that exist.

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(a) sn =3 − 2n

1 + n(b) sn =

(−1)n

n + 2

(c) sn =(−1)nn

2n − 1(d) sn =

23n

32n

(e) sn =n2 − 2n + 1

(f) sn =1 + n + n2

1 + 3n

3. Prove the following:

(a) limn→∞

(√n2 + 1 − n

)= 0.

(b) limn→∞

(√n2 + n − n

)=

12.

4. Prove Theorem 4.

5. Prove Theorem 6.

6. Let {sn}, {tn}, and {un} be sequnces such that sn ≤ tn ≤ un for all n. Prove that ifsn → L and un → L, then tn → L.

II.3. MONOTONE SEQUENCES AND CAUCHY SEQUENCES

Monotone Sequences

Definition 4. A sequence {sn} is increasing if sn ≤ sn+1 for all n; {sn} is decreasing ifsn ≥ sn+1 for all n. A sequence is monotone if it is increasing or if it is decreasing.

Examples

(a) 1, 12 , 1

3 , 14 , . . . , 1

n , . . . is a decreasing sequence.

(b) 2, 4, 8, 16, . . . , 2n, . . . is an increasing sequence.

(c) 1, 1, 3, 3, 5, 5, . . . , 2n − 1, 2n − 1, . . . is an increasing sequence.

(d) 1, 12 , 3, 1

4 , 5, . . . is not monotonic.

Some methods for showing monotonicity:

(a) To show that a sequence is increasing, show thatsn+1

sn≥ 1 for all n. For decreasing, show

sn+1

sn≤ 1 for all n.

The sequence sn =n

n + 1is increasing: Since

sn+1

sn=

(n + 1)/(n + 2)n/(n + 1)

=n + 1n + 2

· n + 1n

=n2 + 2n + 1

n2 + 2n> 1

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(b) By induction. For example, let {sn} be the sequence defined recursively by

sn+1 = 1 +√

sn, s1 = 1.

We show that {sn} is increasing. Let S be the set of positive integers for which sk+1 ≥ sk.Since s2 = 1 +

√1 = 2 > 1, 1 ∈ S. Assume that k ∈ S; that is, that sk+1 ≥ sk. Consider

sk+2:sk+2 = 1 +

√sk+1 ≥ 1 +

√sk = sk+1.

Therefore, sk+1 ∈ S and {sn} is increasing.

THEOREM 8. A monotone sequence is convergent if and only if it is bounded.

Proof: Let {sn} be a monotone sequence.

If {sn} is convergent, then it is bounded (Theorem 2).

Now suppose that {sn} is a bounded, monotone sequence. In particular, suppose {sn} isincreasing. Let u = sup {sn} and let ε be a positive number. Then there exists a positive integerN such that u− ε < sN ≤ u. Since {sn} is increasing, u− ε < sn ≤ u for all n > N . Therefore,|u − sn| < ε for all n > N and sn → u.

A similar argument holds for the case {sn} decreasing.

THEOREM 9. (a) If {sn} is increasing and unbounded, then sn → +∞.

(b) If {sn} is decreasing and unbounded, then sn → −∞.

Proof: (a) Since {sn} is increasing, sn ≥ s1 for all n. Therefore, {sn} is bounded below.Since {sn} is unbounded, it is unbounded above and to each positive number M there is a positiveinteger N such that sN > M . Again, since {sn} is increasing, sn ≥ sN > M for all n > N .Therefore sn → ∞.

The proof of (b) is left as an exercise.

Cauchy Sequences

Definition 5. A sequence {sn} is a Cauchy sequence if to each ε > 0 there is a positive integerN such that

m, n > N implies |sn − sm| < ε.

THEOREM 10. Every convergent sequence is a Cauchy sequence.

Proof: Suppose sn → s. Let ε > 0. There exists a positive integer N such that |s − sn| < ε/2for all n > N . Let n, m > N . Then

|sm − sn| = |sm − s + s − sn| ≤ |sm − s| + |s − sn| <ε

2+

ε

2= ε.

Therefore {sn} is a Cauchy sequence.

THEOREM 11. Every Cauchy sequence is bounded.

19

Proof: Let {sn} be a Cauchy sequence. There exists a positive integer N such that |sn−sm| < 1whenever n, m > M . Therefore

|sn| = |sn − sN+1 + sN+1| ≤ |sn − sN+1| + |sN+1| < 1 + |sN+1| for all n > N.

Now let M = max {|s1|, |s2|, . . . , |sN |, 1 + |sN+1|}. Then |sn| ≤ M for all n.

THEOREM 12. A sequence {sn} is convergent if and only if it is a Cauchy sequence.

Exercises 2.3


(a) If a monotone sequence is bounded, then it is convergent.

(b) If a bounded sequence is monotone, then it is convergent.

(c) If a convergent sequence is monotone, then it is bounded.

(d) If a convergent sequence is bounded, then it is monotone.

2. Give an example of a sequence having the given properties.

(a) Cauchy, but not monotone.

(b) Monotone, but not Cauchy.

(c) Bounded, but not Cauchy.

3. Show that the sequence {sn} defined by s1 = 1 and sn+1 = 14 (sn + 5) is monotone and

bounded. Find the limit.

4. Show that the sequence {sn} defined by s1 = 2 and sn+1 =√

2sn + 1 is monotone andbounded. Find the limit.

5. Show that the sequence {sn} defined by s1 = 1 and sn+1 =√

sn + 6 is monotone andbounded. Find the limit.

6. Prove that a bounded decreasing sequence converges to its greatest lower bound.

7. Prove Theorem 9 (b).

II.4. SUBSEQUENCES

Definition 6. Given a sequence {sn}. Let {nk} be a sequence of positive integers such thatn1 < n2 < n3 < · · · . The sequence {snk} is called a subsequence of {sn}.

Examples

THEOREM 13. If {sn} converges to s, then every subsequence {snk of {sn} also convergesto s.

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Corollary If {sn} has a subsequence {tn} that converges to α and a subsequence {un} thatconverges to β with α 6= β, then {sn} does not converge.

THEOREM 14. Every bounded sequence has a convergent subsequence.

THEOREM 15. Every unbounded sequence has a monotone subsequence that diverges either to+∞ or to −∞.

Limit Superior and Limit Inferior

Definition 7. Let {sn} be a bounded sequence. A number α is a subsequential limit of {sn}if there is a subsequence {snk} of {sn} such that snk → α.

Examples

Let {sn} be a bounded sequence. Let

S = {α : α is a subsequential limit of {sn}.

Then:

1. S 6= ∅.

2. S is a bounded set.

Definition 8. Let {sn} be a bounded sequence and let S be its set of subsequential limits. Thelimit superior of {sn} (denoted by lim sup sn) is

lim sup sn = sup S.

The limit inferior of {sn} (denoted by lim inf sn) is

lim inf sn = inf S.

Examples

Clearly, lim inf sn ≤ lim sup sn.

Definition 9. Let {sn} be a bounded sequence. {sn} oscillates if lim inf sn < lim sup sn.

Exercises 2.4


(a) A sequence {sn} converges to s if and only if every subsequence of {sn} converges tos.

21

(b) Every bounded sequence is convergent.

(c) Let {sn} be a bounded sequence. If {sn} oscillates, then the set S of subsequentiallimits of {sn} has at least two points.

(d) Every sequence has a convergent subsequence.

(e) {sn} converges to s if and only if lim inf sn = lim sup sn = s.


(a) Every oscillating sequence has a convergent subsequence.

(b) Every oscillating sequence diverges.

(c) Every divergent sequence oscillates.

(d) Every bounded sequence has a Cauchy subsequence.

(e) Every monotone sequence has a bounded subsequence.

22

PART III. FUNCTIONS: LIMITS AND CONTINUITY

III.1. LIMITS OF FUNCTIONS

This chapter is concerned with functions f : D → R where D is a nonempty subset ofR. That is, we will be considering real-valued functions of a real variable. The set D iscalled the domain of f .

Definition 1. Let f : D → R and let c be an accumulation point of D. A number L

is the limit of f at c if to each ε > 0 there exists a δ > 0 such that

|f(x)− L| < ε whenever x ∈ D and 0 < |x − c| < δ.

This definition can be stated equivalently as follows:

Definition. Let f : D → R and let c be an accumulation point of D. A number L isthe limit of f at c if to each neighborhood V of L there exists a deleted neighborhoodU of c such that f(U ∩ D) ⊆ V .

Notation limx→c

f(x) = L.

Examples:

(a) limx→−2

(x2 − 2x + 4) = 12.

(b) limx→2

x2 − 4x − 2

= 4.

(c) limx→3

x2 + 3x + 5x − 3

does not exist.

(d) limx→1

|x− 1|x − 1

does not exist.

Example: Let f(x) = 4x − 5. Prove that limx→3

f(x) = 7.

Proof: Let ε > 0.

|f(x)− 7| = |(4x− 5) − 7| = |4x − 12| = 4|x− 3|.

Choose δ = ε/4. Then

|f(x)− 7| = 4|x − 3| < 4ε

4= ε whenever 0 < |x − 3| < δ.

Two Obvious Limits:

23

(a) For any constant k and any number c, limx→c

k = k.

(b) For any number c, limx→c

x = c.

THEOREM 1. Let f : D → R and let c be an accumulation point of D. Thenlimx→c

f(x) = L if and only if for every sequence {sn} in D such that sn → c, sn 6= c

for all n, f(sn) → L.

Proof: Suppose that limx→c

f(x) = L. Let {sn} be a sequence in D which converges toc, sn 6= c for all n. Let ε > 0. There exists δ > 0 such that

|f(x)− L| < ε whenever 0 < |x− c| < δ (x ∈ D).

Since sn → c there exists a positive integer N such that |c − sn| < δ for all n > N .Therefore

|f(sn) − L| < ε for all n > N and f(sn) → L.

Now suppose that for every sequence {sn} in D which converges to c, f(sn) → L.Suppose that lim

x→cf(x) 6= L. Then there exists an ε > 0 such that for each δ > 0 there is

an x ∈ D with 0 < |x− c| < δ but f(x)−L| ≥ ε. In particular, for each positive integern there is an sn ∈ D such that |c− sn| < 1/n and |f(sn) − L| ≥ ε. Now, sn → c but{f(sn)} does not converge to L, a contradiction.

Corollary Let f : D → R and let c be an accumulation point of D. If limx→c

f(x) exists,then it is unique. That is, f can have only one limit at c.

THEOREM 2. Let f : D → R and let c be an accumulation point of D. If limx→c

f(x)

does not exist, then there exists a sequence {sn} in D such that sn → c, but {f(sn)}does not converge.

Proof: Suppose that limx→c

f(x) does not exist. Suppose that for every sequence {sn} in

D such that sn → c (sn 6= c), {f(sn)} converges. Let {sn} and {tn} be sequences in D

which converge to c. Then {f(sn)} and {f(tn)} are convergent sequences. Let {un} bethe sequence {s1, t1, s2, t2, . . . }. Then {un}} converges to c and {f(un)} converges tosome number L. Since {f(sn)} and {f(tn)} are subsequences of {f(un)}, f(sn) → L

and f(tn) → L. Therefore, for every sequence {sn} in D such that sn → c, sn 6= c forall n, f(sn) → L and lim

x→cf(x) = L.

Arithmetic of Limits

THEOREM 3. Let f, g : D → R and let c be an accumulation point of D. If

limx→c

f(x) = L and limx→c

g(x) = M,

then

24

1. limx→c

[f(x) + g(x)] = L + M ,

2. limx→c

[f(x)− g(x)] = L − M ,

3. limx→c

[f(x)g(x)] = LM, limx→c

[k f(x)] = kL, k constant,

4. limx→c

f(x)g(x)

=L

Mprovided M 6= 0, g(x) 6= 0.

Examples:

(a) Since limx→c

x = c, limx→c

xn = cn for every positive integer n, by (3).

(b) If p(x) = 2x3 + 3x2 − 5x + 4, then, by (1), (2) and (3),

limx→−2

p(x) = 2(−2)3 + 3(−2)2 − 5(−2) + 4 = 10 = p(−2).

(c) If R(x) =x3 − 2x2 + x − 5

x2 + 4, then, by (1) – (4),

limx→2

R(x) =23 − 2(2)2 + 2 − 5

22 + 4=

−38

= R(2).

THEOREM 4. (“Pinching Theorem”) Let f, g, h : D → R and let c be anaccumulation point of D. Suppose that f(x) ≤ g(x) ≤ h(x) for all x ∈ D, x 6= c. If

limx→c

f(x) = limx→c

h(x) = L,

then limx→c

g(x) = L.

Proof: Let ε > 0. There exists a positive number δ1 such that

|f(x)− L| < ε whenever 0 < |x− c| < δ1 (x ∈ D).

That is−ε < f(x) − L < ε whenever 0 < |x − c| < δ1.

Similarly, there exists a positive number δ2 such that

−ε < h(x) − L < ε whenever 0 < |x − c| < δ2.

Let δ = min {δ1, δ2}. Then

−ε < f(x)− L ≤ g(x)− L ≤ h(x)− L < ε whenever 0 < |x− c| < δ.

Therefore, limx→c

g(x) = L.

One-Sided Limits

25

Definition 2. Let f : D → R and let c be an accumulation point of D. A number L

is the right-hand limit of f at c if to each ε > 0 there exists a δ > 0 such that

|f(x)− L| < ε whenever x ∈ D and c < x < c + δ.

Notation: limx→c+

f(x) = L.

A number M is the left-hand limit of f at c if to each ε > 0 there exists aδ > 0 such that

|f(x)− L| < ε whenever x ∈ D and c − δ < x < c.

Notation: limx→c−

f(x) = M.

Examples

(a) limx→1−

|x− 1|x − 1

= −1; limx→1+

|x− 1|x − 1

= 1.

(b) Let f(x) =

x2 − 1, x ≤ 2

1x − 2

, x > 2; lim

x→2−f(x) = 3, lim

x→2+f(x) does not exist.

THEOREM 5. limx→c

f(x) = L if and only if each of the one-sided limits limx→c+

f(x) and

limx→c−

f(x) exists, and

limx→c+

f(x) = limx→c−

f(x) = L.

Exercises 3.1

1. Evaluate the following limits.

(a) limx→2

x2 − 4x + 3x − 1

(b) limx→1

x2 − 4x + 3x − 1

(c) limx→2

x2 − x − 6x + 2

(d) limx→−2

x2 − x − 6x + 2

(e) limx→2

x2 − x − 6(x + 2)2

(f) limx→1

√x − 1

x − 1

(g) limx→0

x√4 + x − 2

(h) limx→1+

1 − x2

|x− 1|

2. Given that f(x) = x3, evaluate the following limits.

26

(a) limx→3

f(x) − f(3)x − 3

(b) limx→3

f(x) − f(2)x − 3

(c) limx→3

f(x) − f(2)x − 2

(d) limx→1

f(x) − f(1)x − 1

3. True – False. Justify your answer by citing a theorem, giving a proof, or giving acounter-example.

(a) limx→c f(x) = L if and only if to each ε > 0, there is a δ > 0 such that

|f(x)− f(c)| < ε whenever |x− c| < δ, x ∈ D.

(b) limx→c f(x) = L if and only if for each deleted neighborhood U of c there isa neighborhood V of L such that f(U ∩ D) ⊆ V .

(c) limx→c f(x) = L if and only if for every sequence {sn} in D that convergesto c, sn 6= c for all n, the sequence {f(sn)} converges to L.

(d) limx→c f(x) = L if and only if limh→0 f(c + h) = L.

(e) If f does not have a limit at c, then there exists a sequence {sn} in D

sn 6= c for all n, such that sn → c, but {f(sn)} diverges.

(f) For any polynomial P and any real number c, limx→c

P (x) = P (c).

(g) For any polynomials P and Q, and any real number c,

limx→c

P (x)Q(x)

=P (c)Q(c)

.

4. Find a δ > 0 such that 0 < |x− 3| < δ implies |x2 − 5x + 6| < 14 .

5. Find a δ > 0 such that 0 < |x− 2| < δ implies |x2 + 2x − 8| < 110 .

6. Prove that limx→1

(4x + 3) = 7.

7. Prove that limx→3

(x2 − 2x + 3) = 6.

8. Determine whether or not the following limits exist:

(a) limx→0

∣∣∣∣sin1x

∣∣∣∣.

(b) limx→0

x sin1x

.

9. Let f : D → R and let c be an accumulation point of D. Suppose that limx→c

f(x) = L

and L > 0. Prove that there is a number δ > 0 such that f(x) > 0 for all x ∈ D

with 0 < |x− c| < δ.

10. (a) Suppose that limx→c

f(x) = 0 and limx→c [f(x)g(x)] = 1. Prove that limx→c g(x)does not exist.

27

(b) Suppose that limx→c

f(x) = L 6= 0 and limx→c [f(x)g(x)] = 1. Does limx→c g(x)exist, and if so, what is it?

III.2 CONTINUOUS FUNCTIONS

Definition 3. Let f : D → R and let c ∈ D. Then f is continuous at c if to eachε > 0 there is a δ > 0 such that

|f(x)− f(c)| < ε whenever |x − c| < δ, x ∈ D.

Let S ⊆ D. Then f is continuous on S if it is continuous at each point c ∈ S. f iscontinuous if f is continuous on D.

THEOREM 6. Characterizations of Continuity Let f : D → R and let c ∈ D.The following are equivalent:

1. f is continuous at c.

2. If {xn} is a sequence in D such that xn → c, then f(xn) → f(c).

3. To each neighborhood V of f(c), there is a neighborhood U of c such thatf(U ∩ D) ⊆ V .

Proof: See Theorem 1.

Corollary If c is an accumulation point of D, then each of the above is equivalent to

limx→c

f(x) = f(c).

THEOREM 7. Let f : D → R and let c ∈ D. Then f is discontinuous at c if andonly if there is a sequence {xn} in D such that xn → c but {f(xn)} does not convergeto f(c).

Continuity of Combinations of Functions

THEOREM 8. Arithmetic: Let f, g : D → R and let c ∈ D. If f and g arecontinuous at c, then

1. f + g is continuous at c.

2. f − g is continuous at c.

3. fg is continuous at c; kf is continuous at c for any constant k.

28

4. f/g is continuous at c provided g(c) 6= 0.

THEOREM 9. Composition: Let f : D → R and g : E → R be functions such thatf(D) ⊆ E. If f is continuous at c ∈ D and g is continuous at f(c) ∈ E, then thecomposition of g with f , g ◦ f : D → R, is continuous at c.

Proof: Let ε > 0. Since g is continuous at f(c) ∈ E there is a positive number δ1

such that |g(f(x))− g(f(c))| < ε whenever |f(x) − f(c)| < δ1, f(x) ∈ E. Since f iscontinuous at c there is a positive number δ such that |f(x) − f(c)| < δ1 whenever|x − c| < δ, x ∈ D. It now follows that

|g(f(x))− g(f(c))| < ε whenever |x − c| < δ, x ∈ D

and g ◦ f is continuous at c.

Definition 4. Let f : D → R, and let G ⊆ R. The pre-image of G, denoted by f−1(G)is the set

f−1(G) = {x ∈ D : f(x) ∈ G}.

THEOREM 10. A function f : D → R is continuous on D if and only if for each openset G in R there is an open set H in R such that H ∩ D = f−1(G).

Proof: Suppose f is continuous on D. Let G ⊆ R be an open set. If c ∈ f−1(G), thenf(c) ∈ G. Since G is open, there exists a neighborhood V of f(c) such that V ⊆ G.Therefore, there exists a neighborhood Uc of c such that f(Uc ∩ D) ⊆ V . Let

H = ∪c∈f−1(G) Uc.

H is open and H ∩ D = f−1(G).

Conversely, choose any c ∈ D, and let V be a neighborhood of f(c). Since V is anopen set, there is an open set H ⊆ R such that H ∩D = f−1(V ). Since f(c) ∈ V, c ∈ H .But H is an open set so there is a neighborhood U of c such that U ⊆ H . Now

f(U ∩ D) ⊆ f(H ∩ D) = v.

It follows that f is continuous on D by Theorem 6.

Corollary A function f : R → R is continuous if and only if f−1(G) is open in Rwhenever G is open in R.

Exercises 3.2

1. Let f(x) =x2 + 2x− 15

x − 3. Define f at 3 so that f will be continuous at 3.

29

2. Each of the following functions is defined everywhere except at x = 1. Where possible,define f at 1 so that it becomes continuous at 1.

(a) f(x) =x2 − 1x − 1

(b) f(x) =1

x − 1

(c) f(x) =x − 1|x− 1| (d) f(x) =

(x − 1)2

|x − 1|

3. In each of the following define f at 5 so that it becomes continuous at 5.

(a) f(x) =√

x + 4 − 3x − 5

(b) f(x) =√

x + 4 − 3√x − 5

(c) f(x) =√

2x − 1 − 3x − 5

(d) f(x) =√

x2 − 7x + 16 −√

6(x − 5)

√x + 1

4. Let f(x) =

{A2x2, x < 2

(1− A)x, x ≥ 2.For what values of A is f continuous at 2?

5. Give necessary and sufficient conditions on A and B for the function

f(x) =

Ax − B, x ≤ 13x, 1 < x < 2

Bx2 − A, x ≥ 2

to be continuous at x = 1 but discontinuous at x = 2.

6. Let f : D → R and let c ∈ D. True – False. Justify your answer by citing adefinition or theorem, giving a proof, or giving a counter-example.

(a) f is continuous at c if and only if to each ε there is a δ > 0 such that

|fx)− f(c)| < ε whenever |x − c| < δ and x ∈ D.

(b) If f(D) ⊆ R is bounded, then f is continuous on D.

(c) If c is an isolated point of D, then f is continuous at c.

(d) If f is continuous at c and {xn} is a sequence in D, then xn → c wheneverf(xn) → f(c).

(e) If {xn} is a Cauchy sequence in D, then {f(xn)} is convergent.


(a) If f and f + g are continuous on D, then g is continuous on D.

(b) If f and fg are continuous on D, then g is continuous on D.

30

(c) If f and g are not continuous on D, then f + g is not continuous on D.

(d) If f and g are not continuous on D, then fg is not continuous on D.

(e) If f2 is continuous on D, then f is continuous on D.

(f) If f is continuous on D, then f(D) is a bounded set.

8. Let f : D → R.

(a) Prove that if f is continuous at c, then |f | is continuous at c.

(b) Suppose that |f | is continuous at c. Does it follow that f is continuous at c?Justify your answer.

9. Let f : D → R be continuous at c ∈ D. Prove that if f(c) > 0, then there is anα > 0 and a neighborhood U of c such that f(x) > α for all x ∈ U ∩ D.

10. Let f : D → R be continuous at c ∈ D. Prove that there exists an M > 0 and aneighborhood U of c such that |f(x)| ≤ M for all x ∈ U ∩ D.

III.3. PROPERTIES OF CONTINUOUS FUNCTIONS

Definition 5. A function f : D → R is bounded if there exists a number M such that|f(x)| ≤ M for all x ∈ D. That is, f is bounded if f(D) is a bounded subset of R.

THEOREM 11. Let f : D → R be continuous. If D is compact, then f(D) is compact.(The continuous image of a compact set is compact.)

Proof: Let G = {Gα} be an open cover of f(D). Since f is continuous, for eachopen set Gα in G there is an open set Hα such that Hα ∩ D = f−1(Gα). Also, sincef(D) ⊆ ∪Gα, it follows that

D ⊆ ∪ f−1(Gα) ⊆ ∪Hα.

Thus, the collection {Hα} is an open cover of D. Since D is compact this open coverhas a finite subcover Hα1, Hα1, . . . , Hαn . Now,

D ⊆ (Hα1 ∩ D) ∪ (Hα2 ∩ D) ∪ · · · ∪ (Hαn ∩ D)

andf(D) ⊆ Gα1 ∪ Gα2 ∪ · · · ∪ Gαn .

Therefore, the open cover G has a finite subcover and f(D) is compact.

Definition 6. Let f : D → R. f(x0) is the minimum value of f on D if f(x0) ≤ f(x)for all x ∈ D. f(x1) is the maximum value of f on D if f(x) ≤ f(x1) for allx ∈ D.

31

COROLLARY 1. If f : D → R is continuous and D is compact, then f has amaximum value and a minimum value. That is, there exist points x0, x1 ∈ D such thatf(x0) ≤ f(x) ≤ f(x1) for all x ∈ D.

COROLLARY 2. If f : D → R is continuous and D is compact, then f(D) is closedand bounded.

THEOREM 12. Let f : [a, b] → R be continuous. If f(a) and f(b) have oppositesign, then there is at least one point c ∈ (a, b) such that f(c) = 0.

Proof: Suppose that f(a) < 0 and f(b) > 0. Since f(a) < 0 we know from thecontinuity of f that there is an interval [a, δ) such that f(x) < 0 on [a, δ). (SeeExercises 3.2, #9) Let

c = sup {δ : f is negative on [a, δ)}.

Clearly c ≤ b.

We cannot have f(c) > 0 for then f(x) > 0 on some interval to the left of c, andwe know that to the left of c, f(x) < 0. This also shows that c < b.

We cannot have f(c) < 0 for then f(x) < 0 on some interval [a, t), with t > c whichcontradicts the definition of c.

It follows that f(c) = 0.

THEOREM 13. Intermediate Value Theorem Let f : [a, b] → R be continuous.Suppose that f(a) 6= f(b). If k is a number between f(a) and f(b), then there is atleast one number c ∈ (a, b) such that f(c) = k.

COROLLARY If f : D → R is continuous and I ⊆ D is an interval, then f(I) is aninterval.

THEOREM 14. Suppose that f : D → R is continuous. If I ⊆ D is a compact interval,then f(I) is a compact interval.

Exercises 3.3

1. Show that the equation x3 − 4x + 2 = 0 has three distinct roots in [−3, 3] andlocate the roots between consecutive integers.

2. Prove that sin x + 2 cos x = x2 for some x ∈ [0, π/2].

3. Prove that there exists a positive number c such that c2 = 2.

32

4. True – False. Justify your answer by citing a theorem, giving a proof, or giving acounter-example.

(a) Suppose that f : D → R is continuous. Then there exists a point x1 ∈ D suchthat f(x) ≤ f(x1) for all x ∈ D.

(b) If D ⊆ R is bounded and f : D → R is continuous, then f(D) is bounded.

(c) Let f : [a, b] → R be continuous and suppose that f(a) ≤ k ≤ f(b). Thenthere exists a point c ∈ [a, b] such that f(c) = k.

(d) Let f : (a, b) → R be continuous. Then there is a point x1 ∈ (a, b) such thatf(x) ≤ f(x1) for all x ∈ (a, b).

(e) If f : D → R is continuous and bounded on D, then f has a maximum valueand a minimum value on D.

5. Let f : D → R be continuous. For each of the following, prove or give a counterex-ample.

(a) If D is open, then f(D) is open.

(b) If D is closed, then f(D) is closed.

(c) If D is not open, then f(D) is not open.

(d) If D is not closed, then f(D) is not closed.

(e) If D is not compact, then f(D) is not compact.

(f) If D is not bounded, then f(D) is not bounded.

(g) If D is an interval, then f(D) is an interval.

(h) If D is an interval and f(D) ⊆ Q (the rational numbers), then f is constant.

6. Prove that every polynomial of odd degree has at least one real root.

7. Prove Theorem 13.

8. Prove Theorem 14.

9. Suppose that f : [a, b] → [a, b] is continuous. Prove that there is at least one pointc ∈ [a, b] such that f(c) = c. (Such a point is called a fixed point of f .)

10. Suppose that f, g : [a, b] → R are continuous, and suppose that f(a) ≤ g(a), f(b) ≥g(b). Prove that there is at least one point c ∈ [a, b] such that f(c) = g(c).

33

III.4. THE DERIVATIVE

DEFINITION 1. Let I be an interval, let f : I → R, and let c ∈ I. f isdifferentiable at c if

limx→c

f(x) − f(c)x − c

= m

exists. f is differentiable on I if it is differentiable at each point of I.

Notation: If f is differentiable at c, then the limit m is called the derivative of f

at c and is denoted by f ′(c).

An equivalent definition of differentiability is:

Definition. f is differentiable at c if

limh→0

f(c + h) − f(c)h

= m

exists.

Two basic derivatives

(a) Let f(x) ≡ k, x ∈ R, k constant. For any c ∈ R, f ′(c) = 0

limx→c

f(x) − f(c)x − c

= limx→c

k − k

x − c= lim

x→c0 = 0.

(b) Let f(x) = x, x ∈ R. For any c ∈ R, f ′(c) = 1

limx→c

f(x) − f(c)x − c

= limx→c

x − c

x − c= lim

x→c1 = 1.

Examples:

(a) Let f(x) = x2 + 3x− 1 on R. Then for any c ∈ R, we have

limx→c

f(x) − f(c)x − c

= limx→c

x2 + 3x − 1− (c2 + 3c − 1)x − c

= limx→c

x2 − c2 + 3(x − c)x − c

= limx→c

(x − c)(x + c + 3x − c

= limx→c

(x + c + 3) = 2c + 3

Thus f ′(c) = 2c + 3.

34

(b) Same function using the alternative definition.

limh→0

f(c + h) − f(c)h

= limh→0

(c + h)2 + 3(c + h)− 1 − (c2 + 3c− 1)x − c

= limh→0

c2 + 2ch + h2 + 3c + 3h − c2 − 3c

h= lim

h→0

2ch + h2 + 3h

h

= limh→0

(2c + h + 3) = 2c + 3

Note: The alternative definition is usually easier to use when calculating the deriva-tive of a given function because it’s usually easier to expand an expression than it isto factor. For example:

(c) Let f(x) = sin x on R, and let c ∈ R. Then

limx→c

f(x) − f(c)x − c

= limx→c

sin x − sin c

x − c=????

On the other hand

limh→0

f(c + h) − f(c)h

= limh→0

sin (c + h) − sin c

h

= limh→0

sin c cos h + cos c sin h − sin c

h

= limh→0

sin c[cos h − 1] + cos c sin h

h

= limh→0

sin c

[cos h − 1

h

]+ lim

h→0cos c

[sin h

h

]= cos c.

Therefore f ′(c) = cos c. Here we used the important trigonometric limits:

limθ→0

sin θ

θ= 1 and lim

θ→0

cos θ − 1θ

= 0.

(c) Let f(x) =√

x, x ≥ 0 and let c > 0.

limh→0

f(c + h) − f(c)h

= limh→0

√c + h −

√c

h= lim

h→0

√c + h −

√c

h

√c + h +

√c√

c + h +√

c

= limh→0

h

h(√

c + h +√

c) = lim

h→0

1(√c + h +

√c) =

12√

c

Thus, f ′(c) =1

2√

c.

NOTE: In each of the examples we started with a function f and “derived” a newfunction f ′ which is called the derivative of f . If we start with a function of x, then it

35

is standard to denote the derivative as a function of x. For example, if f(x) = x2 +3x− 1,then f ′(x) = 2x + 3; if f(x) = sin x, then f ′(x) = cos x; if f(x) =

√x, then

f ′(x) = 1/2√

x

Example: A function that fails to be differentiable at a point c.

Set

f(x) =

{x2 + 1, x ≤ 1

3− x, x > 1

You can verify that f is continuous for all x; in particular, f continuous at x = 1. Weshow that f is not differentiable at 1.

For h < 0,

f(1 + h) − f(1)h

=(1 + h)2 + 1− (2)

h=

2h + h2

h= 2 + h

andlim

h→0−

f(1 + h) − f(1)h

= limh→0−

(2 + h) = 2.

For h > 0,f(1 + h) − f(1)

h=

3 − (1 + h) − (2)h

=−h

h= −1

andlim

h→0+

f(1 + h) − f(1)h

= limh→0+

(−1) = −1.

Therefore,

limh→0

f(1 + h) − f(1)h

does not exist.

THEOREM 15. If f : I → R is differentiable at c ∈ I, then f is continuous at c.

Proof: For x ∈ I, x 6= c, we have

f(x) = (x − c)f(x) − f(c)

x − c+ f(c).

Since f is differentiable at c,

limx→c

f(x) − f(c)x − c

= f ′(c)

exists. Therefore,

limx→c

f(x) =[limx→c

(x− c)]

limx→c

[f(x) − f(c)

x − c

]+ lim

x→cf(c) = 0 · f ′(c) + f(c) = f(c)

By the Corollary to Theorem 6, f is continuous at c.

Differentiability of Combinations of Functions

36

THEOREM 16. Arithmetic: Let f, g : I → R and let c ∈ I. If f and g aredifferentiable at c, then

(a) f + g is differentiable at c and

(f + g)′(c) = f ′(c) + g′(c).

(b) f − g is differentiable at c and

(f − g)′(c) = f ′(c)− g′(c).

(c) fg is differentiable at c and

(fg)′(c) = f(c)g′(c) + g(c)f ′(c).

For any constant k, kf is differentiable at c and (kf)′(c) = kf ′(c).

(d) If g(c) 6= 0, then f/g is differentiable at c and(

f

g

)′(c) =

g(c)f ′(c)− f(c)g′(c)g2(c)

.

Proof: (c)

f(x)g(x)− f(c)g(c)x − c

=f(x)g(x)− f(x)g(c) + f(x)g(c)− f(c)g(c)

x − c

= f(x)g(x)− g(c)

x − c+ g(c)

f(x)− f(c)x − c

.

Since f is continuous at c, limx→c

f(x) = f(c). Therefore, since f and g are continuousat c,

limx→c

f(x)g(x)− f(c)g(c)x − c

= f(c)g′(c) + g(c)f ′(c).

(d) We show first that [1

g(c)

]′=

−g′(c)g2(c)

.

Since g is continuous at c and g(c) 6= 0, there is an interval I containing c suchthat g(c) 6= 0 on I . Now

1g(x)

− 1g(c)

x − c=

1g(x)g(c)

g(c)− g(x)x − c

= − 1g(x)g(c)

g(x)− g(c)x − c

.

Since g is continuous at c, limx toc

g(x) = g(c). Therefore

limx→c

1g(x)

− 1g(c)

x − c= − 1

g2(c)g′(c) =

−g′(c)g2(c)

37

(d) now follows by differentiating the product f(x)1

g(x)using (c).

Example: If f(x) = xn, n an integer, then f ′(x) = nxn−1.

Proof: Assume first that n is a positive integer, and use induction. Let S be the setof positive integers for which the statement holds. Then 1 ∈ S since if f(x) = x, thenf ′(x) = 1 = 1 x0. Now assume that the positive integer k ∈ S and set f(x) = xk+1. Since

xk+1 = xk x

we have, by the product rule,

f ′(x) = xk 1 + x k xk−1 = (k + 1)xk

and so k + 1 ∈ S and the statement holds for all positive integers n.

If n is a negative integer, then, for x 6= 0,

f(x) = xn =1

x−n

where −n is a positive integer. By the quotient rule,

f ′(x) =x−n(0)− (−n)x−n − 1

(x−n)2=

n x−(n+1)

x−2n= nxn−1.

Finally, if f(x) = x0 ≡ 1, then f ′(x) = 0 = 01x. There is slight difficulty with x = 0

in this case; 00 is a so-called indeterminate form.

THEOREM 17. (The Chain Rule) Suppose that f : I → R and g : J → R, andsuppose that g(J) ⊂ I. If g is differentiable at c ∈ I and f is differentiable atg(c) inJ, then f(g) is differentiable at c and

(f [g(c)])′ = f ′[g(c)] g′(c).

Pseudo-proof:

f [g(x)]− f(g(c)]x − c

=f [g(x)]− f [g(c)]

g(x)− g(c)f [g(x)]− f(g(c)]

x − c.

Set u = g(x) and a = g(c). Then, as x → x, u → a since g is continuous at c. Thus

limx→c

f [g(x)]− f(g(c)]x − c

= limu→a

f(u) − f(a)u − a

limx→c

f [g(x)]− f(g(c)]x − c

= f ′(a)g′(c) = f ′[g(c)]g′(c).

The problem with this proof is that while we know x−c 6= 0, we don’t know that u−a 6= 0;that is, we don’t know that g(x) 6= g(c). This proof can be modified to take care of thatcontingency.

38

Exercises 3.4

1. Use either of the definitions of the derivative to find the derivative of each of thefollowing functions.

(a) f(x) =1x

.

(b) f(x) =√

x.

(c) f(x) =1√x

.

(d) f(x) = x1/3.

(e) f(x) = cos x.

2. Determine the values of x for which the given function is differentiable and find thederivative.

(a) f(x) = |x− 3|.

(b) f(x) = |x2 − 1|.

(c) f(x) = x |x|.

3. Set f(x) =

{x2, if x ≥ 00, x < 0

(a) Sketch the graph of f and show that f is differentiable at 0.

(b) Find f ′ and sketch the graph of f ′.

(c) Is f ′ differentiable at 0?

4. Set f(x) =

{x sin (1/x), if x 6= 0

0, x = 0Determine whether or not f is differentiable

at 0.

5. Set g(x) =

{x2 sin (1/x), if x 6= 0

0, x = 0

(a) Calculate the derivative of g at any number c 6= 0.

(b) Use the definition to show that g is differentiable at 0 and find g′(0).

(c) Is g′ continuous at 0?

39

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