Introduction to Analysis Problems and Solutions

Prévia do material em texto

Solutions to Introduction to Analysis
Edward D. Gaughan
March 3, 2013
Notation
Symbol Meaning
Ac X \A, where X is the universal set (assumed to be R unless otherwise stated)
f ∼ g f tends asymptotically to g
N Natural numbers (symbolized as J in the book)
LHS Left-hand side
RHS Right-hand side
incr. Increasing
decr. Decreasing
cts. Continuous
diff. Differentiable
intgbl. Integrable
lg(x) log2(x)
dxe If if a < x ≤ a+ 1 where a is a positive integer, then, dxe = a+ 1.
an → A lim
n→∞an = A.
lim an lim
n→∞an
IVT Intermediate Value Theorem
H l’Hospital’s Rule
MVT Mean Value Theorem
IFT Inverse Funcion Theorem
Mi Mi = supx∈(xi−1,xi) f(x). (Used in tandem with a function f and partition P ).
mi Similar to Mi, but replace sup with inf.
R(x; [a, b]) The set of Riemann-integrable functions on the interval [a, b].
Shortened to R(x) if interval is apparent.
U(P, f) If P is a partition of [a, b], then U(P, f) =
∑n
i=0Mi(xi − xi−1).
L(P, f) Similar to U(P, f), but replace Mi with mi
µ(P ) sup
k≤n
|xk − xk+1| for partition P0 and refinements {Pn}. Called the mesh of P .
S(P, f) S(P, f) =
∑n
i=1 f(ti)(xi − xi−1), where ti ∈ (xi−1, xi). (Riemann sum)∑
an Lower bound unimportant, upper bound ∞. (Only intereseted in convergence.)∑
n=n0
an Upper bound assumed to be ∞.
cgt Convergent.
a
±�
= b b− � < a < b+ �
1
Contents
0 Prerequisites 4
0.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.2 Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . 6
0.3 Mathematical Induction and Recursion . . . . . . . . . . . . . . . 7
0.4 Equivalent and Countable Sets . . . . . . . . . . . . . . . . . . . 9
0.5 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1 Sequences 13
1.1 Sequences and Convergence . . . . . . . . . . . . . . . . . . . . . 13
1.2 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.3 Arithmetic Operations on Sequences . . . . . . . . . . . . . . . . 17
1.4 Subsequences and Monotone Sequences . . . . . . . . . . . . . . 19
Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2 Limits of Functions 23
2.1 Definition of the Limit of a Function . . . . . . . . . . . . . . . . 23
2.2 Limits of Functions and Sequences . . . . . . . . . . . . . . . . . 24
2.3 Algebra of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.4 Limits of Monotone Functions . . . . . . . . . . . . . . . . . . . . 27
Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3 Continuity 30
3.1 Continuity of a Function at a Point . . . . . . . . . . . . . . . . . 30
3.2 Algebra of Continuous Functions . . . . . . . . . . . . . . . . . . 32
3.3 Uniform Continuity: Open, Closed, and Compact Sets . . . . . . 34
3.4 Properties of Continuous Functions . . . . . . . . . . . . . . . . . 38
Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4 Differentiation 41
4.1 The Derivative of a Function . . . . . . . . . . . . . . . . . . . . 41
4.2 The Algebra of Derivatives . . . . . . . . . . . . . . . . . . . . . 44
4.3 Rolle’s Theorem and the Mean Value Theorem . . . . . . . . . . 45
4.4 L’Hospital’s Rule and the Inverse-Function Theorem . . . . . . . 48
Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2
CONTENTS 3
5 The Riemann Integral 52
5.1 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . 52
5.2 Classes of Integrable Functions . . . . . . . . . . . . . . . . . . . 53
5.3 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
5.4 The Fundamental Theorem of Integral Calculus . . . . . . . . . . 56
5.5 Algebra of Integrable Functions . . . . . . . . . . . . . . . . . . . 56
5.6 Derivatives of Integrals . . . . . . . . . . . . . . . . . . . . . . . . 58
5.7 Mean-Value and Change-of-Variable Theorems . . . . . . . . . . 59
Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
6 Infinite Series 63
6.1 Convergence of Infinite Series . . . . . . . . . . . . . . . . . . . . 63
6.2 Absolute Convergence and the Comparison Test . . . . . . . . . 65
6.3 Ratio and Root Tests . . . . . . . . . . . . . . . . . . . . . . . . 66
6.4 Conditional Convergence . . . . . . . . . . . . . . . . . . . . . . . 69
6.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
Chapter 0
Prerequisites
0.1 Sets
1. List the elements of each of the following sets:
(a) N∩[0, 6)
(b) Z ∩ (−6, 2]
(c) {1, 2, 3, 4} ∪ {2, 3, 4, 5}
(d) {1, 2, 3, 4} ∩ {2, 3, 4, 5}
(a) {1, 2, 3, 4, 5}
(b) {−5,−4,−3,−2,−1, 0, 1, 2}
(c) {1, 2, 3, 4, 5}
(d) {2, 3, 4}
2. Write each of the following in interval notation:
(a) (0, 2) ∩ (1/2, 1)
(b) [−1, 5] ∪ [2, 7]
(a) (1/2, 1)
(b) [−1, 7]
3. Prove (vi) of Theorem 0.2. (That is, prove that for sets A, B, and C,
A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C)).
(⊆) Let x ∈ A∪ (B ∩C). Then, x ∈ A or (x ∈ B and x ∈ C). If x ∈ A, then
x ∈ (A ∪B) ∩ (A ∪ C) since A ⊆ (A ∪B) ∩ (A ∪ C). If x /∈ A, then x ∈ B and
x ∈ C. Thus, x ∈ A∪B and x ∈ A∪C. Thus, [A∪(B∩C)] ⊆ [(A∪B)∩(A∪C)].
(⊇) Let x ∈ (A ∪ B) ∩ (A ∪ C). Then, (x ∈ A or x ∈ B) and (x ∈ A or
x ∈ C). Suppose x /∈ A. Then x ∈ B and x ∈ C. Thus, x ∈ A ∪ (B ∩C). Now,
suppose x ∈ A. Then x ∈ A ∪ (B ∩ C).
4
CHAPTER 0. PREREQUISITES 5
4. Prove (ii) of Theorem 0.3. (That is, prove that for sets A, B, and C,
A \ (B ∪ C) = (A \B) ∩ (A \ C)).
(⊆) Let x ∈ A \ (B ∪ C). Then, x ∈ A and x /∈ (B ∪ C). Then, x /∈ B and
x /∈ C. Thus, x ∈ (A \B) ∩ (A \ C).
(⊇) Let x ∈ (A \ B) ∩ (A \ C). Then, (x ∈ A and x /∈ B) and (x ∈ A and
x /∈ C). Then, x is in neither B nor C, so x /∈ (B ∪C). Thus, x ∈ A \ (B ∪C).
5. Prove that for all sets A, B, and C, A ∩B ⊂ A ⊂ A ∪ C.
If x ∈ A ∩B, then x ∈ A. If x ∈ A, then x ∈ A ∪ C.
6. If A ⊂ B, prove that (C \ B) ⊂ (C \ A). Either prove the converse is
true, or give a counterexample.
If x ∈ C \B, then x ∈ C and x /∈ B. Since A ⊂ B, x /∈ A. Thus, x ∈ C \A.
The converse is false. Let A = {1, 2, 3, 4, 5}, B = {6, 7, 8, 9, 10} and C = {6, 7}.
Then, C \A = {6, 7} but C \B = ∅. Thus, (C \B) ⊂ (C \A), but A * B.
7. Under what conditions does A \ (A \B) = B?
We see A \ (A \ B) = A ∩ (A \ B)c = A ∩ (A ∩ Bc)c Q4= A ∩ (Ac ∪ B) Q3=
(A∩Ac)∪ (A∩B) = ∅∪ (A∩B) = A∩B. Thus, A \ (A \B) = B if A∩B = B
which happens when B ⊆ A.
8. Show that (A \B) ∪ (B \A) = (A ∪B) \ (A ∩B).
We see that (A∪B) \ (A∩B) = (A∪B)∩ (A∩B)c = (A∪B)∩ (Ac ∪Bc) =
[(A∪B)∩Ac]∪ [(A∪B)∩Bc] = [(A∩Ac)∪ (B ∩Ac)]∪ [(A∩Bc)∪ (B ∩Bc)] =
(B ∩Ac) ∪ (A ∩Bc) = (A \B) ∪ (B \A).
9. Look up Russell’s paradox and write a brief summary of how it relates to
Section 0.1.
Russell’s paradox points out a flaw in naive set theory. If we examine the
set A of all sets, then A ∈ A. But then, there must be another set containing
A. This is the paradox. Since the theory in this section are based on naive set
theory (the treatment of “set” as an undefined term), it shows that the theory
is not complete.
10. Describe each of the following sets as the empty set, as R, or in interval
notation, as appropriate:
(a)
∞⋂
n=1
(− 1n , 1n)
CHAPTER 0. PREREQUISITES 6
(b)
∞⋃
n=1
(−n, n)
(c)
∞⋂
n=1
(− 1n , 1 + 1n)
(d)
∞⋃
n=1
(− 1n , 2 + 1n)
(a) {0}
(b) R
(c) (0, 1]
(d) (−1, 3)
11. Prove (ii) of Theorem 0.4. (That is, prove that S\(∩λ∈ΛAλ) = ∪λ∈Λ(S\
Aλ).)
(⊆) Let x ∈ S \ (∩λ∈ΛAλ). Then, x ∈ S and x /∈ Aλ for all λ ∈ Λ. Then,
x /∈ ∪λ∈ΛAλ. Thus, x ∈ ∪λ∈Λ(S \Aλ).
(⊇) Let x ∈ ∪λ∈Λ(S \ Aλ) = ∪λ∈Λ(S ∩ Acλ). Then, for each λ, x ∈ S and
x /∈ Aλ. Thus, x ∈ S. Since x /∈ Aλ for all λ, we see that x /∈ ∩λ∈ΛAλ. Thus,
x ∈ S \ (∩λ∈ΛSλ).
12. Use DeMorgan’s Laws to give a different and simpler description of the
following sets:
(a) R \
∞⋂
n=1
(− 1n , 1n)
(b)
∞⋃
n=1
(R \ [ 1n , 2 + 1n])
(a) R \∞⋂
n=1
(− 1n , 1n) = ⋃{(−∞,− 1n] ∪ [ 1n ,∞)} = R.
(b)
∞⋃
n=1
(R \ [ 1n , 2 + 1n]) = R \⋂[ 1n , 2 + 1n] = R \ (0, 2] = (−∞, 0] ∪ (2,∞).
0.2 Relations and Functions
13. Define f : N → N by f(n) = 2n − 1 for each n ∈ N. What is im f? Is
f 1-1? Is f onto? If f has an inverse, find the domain of the inverse and give
a formula for f−1(n).
We see that im f = {odd natural numbers}. We see that f is 1-1. Indeed,
f(n) = f(m)⇔ 2n−1 = 2m−1⇔ n = m. Since im f 6= N, f is not onto. Since
f is 1-1 and im f = {odd natural numbers}, f−1 : {odd natural numbers} → N
exists. To find it, set y = f−1(n) and write n = 2y − 1⇔ f−1(n) = n+12 .
14. What is the domain of f(x) = xx+2? What is im f? Is f injective? If
so, find the inverse.
CHAPTER 0. PREREQUISITES 7
We see thatdom f = R \ {−2}. Since xx+2 = x+2−2x+2 = x+2x+2 − 2x+2 = 1− 2x+2 ,
we see that im f = (1,∞)∪ (−∞, 1). We see that f is injective, so we find that
f−1(x) = 2x−1 − 2.
For Exercises 15-17, let A = {1, 2, 3, 4, 5}, B = {2, 3, 4, 5, 6, 7}, and C =
{a, b, c, d, e}.
15. Give an example of f : A→ B that is not 1-1.
Let f(n) = n, if 2 ≤ n ≤ 5, and f(1) = 6 and f(1) = 7. (Note that f need
not be a function.)
16. Give an example of f : A → B that has an inverse, and show the
inverse.
Let f(n) = n, if 2 ≤ n ≤ 5, and f(1) = 6. Then f−1 : B \{7} → A is defined
by f−1(n) = n, if 2 ≤ n ≤ 5 and f−1(6) = 1.
17. Give an example of f : A→ B, g : B → C such that g ◦ f is 1-1, but g
is not 1-1.
Define f = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} and g = {(2, a), (3, b), (4, c), (5, d), (6, e), (7, e)}.
Then, g ◦ f ={(1, a), (2, b), (3, c), (4, d), (5, e)} which is 1-1. But, 6 7→ e and
7 7→ e, so g is not 1-1.
*18. If f : A→ B is 1-1 and im f = B, prove that (f−1 ◦ f)(a) = a for all
a ∈ A and (f ◦ f−1)(b) = b for each b ∈ B.
For each (x, y) ∈ f , assign (y, x) ∈ f−1. Since f is 1-1, this assignment is
also 1-1. Since im f = B, f is also onto. Now, let a ∈ A and (a, b) ∈ f . Then,
(b, a) ∈ f−1. Thus, (f−1 ◦ f)(a) = f−1(b) = a. That (f ◦ f−1)(b) = b follows
from a similar argument.
0.3 Mathematical Induction and Recursion
19. Prove that for all n ∈ N, 1 + 2 + · · ·+ n = n(n+1)2 .
For n = 1, 1 = 1(2)2 = 1. Suppose that the statement is true for all n ≤ N .
Examine 1 + 2 + · · ·+N + (N + 1) = N(N+1)2 + (N + 1) = (N+1)(N+2)2 .
20. Prove that for all n ∈ N, 1 + 3 + 5 + · · ·+ (2n− 1) = n2.
CHAPTER 0. PREREQUISITES 8
For n = 1, 1 = 12. Suppose the statement is true for all n ≤ N . Examine
1 + 3 + 5 + · · ·+ (2N − 1) + (2N + 1) = N2 + 2N + 1 = (N + 1)2.
21. Prove that n3 + 5n is divisible by 6 for each n ∈ N.
For n = 1, 12 + 5(1) = 6 which is divisible by 6. Suppose the statement
is true for all n ≤ N and N3 + 5N = 6k. Then, (N + 1)3 + 5(N + 1) =
N3 + 3N2 + 8N + 6 = (N3 + 5N) + (3N2 + 3N) + 6 = 6(k + 1) + 3N(N + 1).
Thus, we need only examine the term 3N(N + 1) to prove divisibility. Suppose
N is even. Then, 3N has 6 as a factor and 6 | 3N(N +1). Now, suppose that N
is odd. Then, N +1 is even and 3(N +1) is divisible by 6. Thus, 6 | 3N(N +1).
Thus, the statement is proven.
22. Prove that n2 < 2n for all n ∈ N, n ≥ 5.
For n = 5, 25 < 32. Suppose the statement is true for all n ≤ N . Then,
(N + 1)2 = N2 + 2N + 1 < 2N + 2(2N−1) = 2N + 2N = 2(2N ) + 2 = 2N+1.
23. Prove the second principle of mathematical induction.
Let P (n) be a statement,P (1), P (2), ..., P (N) are true and P (k)⇒ P (k+ 1)
for all k ∈ N and k ≥ m. Suppose there is some K such that P (K) is false.
Then, since P (k) ⇒ P (k + 1), by contraposition, ¬P (k + 1) ⇒ ¬P (k). By the
second hypothesis, we must then have the following sequence of implications:
¬P (k)⇒ ¬P (k − 1)⇒ ¬P (k − 2)⇒ · · · ⇒ ¬P (N). Contradiction.
24. Define f : N → N by f(1) = 1, f(2) = 2, f(3) = 3 and f(n) =
f(n− 1) + f(n− 2) + f(n− 3) for n ≥ 4. Prove that f(n) ≤ 2n for all n ∈ N.
We see that f(4) = 6 ≤ 24 = 16. Suppose the statement is true for all
n ≤ N . Then, f(N + 1) = f(N − 2) + f(N − 1) + f(N) ≤ 2N−2 + 2N−1 + 2N =
2N+1
23 +
2N+1
22 +
2N+1
2 =
1
82
N+1 + 142
N+1 + 122
N+1 = 782
N+1 < 2N+1.
25. Define f : N → N by f(1) = 2 and, for n ≥ 2, f(n) = √3 + f(n− 1).
Prove that f(n) < 2.4 for all n ∈ N. You may want to use your calculator for
this exercise.
For n = 2, f(2) =
√
3 + 2 ≤ 2.4. Suppose the statement is true for all
n ≤ N . Then, f(N + 1) = √3 + f(N) ≤ √3 + 2.4 ≈ 2.32 < 2.4.
26. Define f : N → N by f(1) = 2, f(2) = −8, and, for n ≥ 3, f(n) =
8f(n − 1) − 15f(n − 2) + 6 · 2n. Prove that, for all n ∈ N, f(n) = −5 · 3n +
5n−1 + 2n+3.
CHAPTER 0. PREREQUISITES 9
For f(3) = 8(−8)− 15(2) + 6(8) = −5(33) + 52 + 26. Suppose the statement
is true for all n ≤ N . Then, f(N + 1) = 8f(N) − 15f(N − 1) + 6 · 2N+1 =
8(−5 · 3N + 5N−1 + 2N+3) − 15(−5 · 3N−1 + 5N−2 + 2N+2) + 6 · 2N+1 = −5 ·
3N+1 + 5N + 5 · 2N+3 + 3 · 23 = −5 · 3N+1 + 5N + 2N+4.
27. Prove Theorem 0.10. (That is, suppose that P (n) is a statement and
(i) for some n0 ∈ Z, P (n0) is true, and (ii) for each k ∈ Z, k ≥ n0, P (k) ⇒
P (k + 1). Then, prove that P (n) is true for all n ≥ n0).
Suppose that there is an N ≥ n0. Then, by contraposition, P (N − 1) is
false and so then is P (N − 2), etc. This eventually implies that P (n0) is false.
Contradiction.
*28. Prove the following modified version of the second principle of mathe-
matical induction: Let P (n) be a statement for each n ∈ Z. If
(a) P (n0), P (n0 + 1),..., P (m) is true, and
(b) for k ≥ m, if P (i) is true for n0 ≤ i ≤ k, then P (k+ 1) is true, then
P (n) is true for n ≥ n0, n ∈ Z.
Suppose that P (N) is false for some N . Then, there exists an i such that
n0 ≤ i ≤ N − 1 such that P (i) is false. Consequently, there exists ani(1) such
that n0 ≤ i(1) ≤ i − 1 such that P (i(1)) is false. Eventually, this implies that
there is an i˜ such that n0 ≤ i˜ ≤ m and P (˜i) is false. Contradiction.
29. Define f(n) as follows for n ∈ Z, n ≥ 0, f(0) = 7, f(1) = 4, and, for
n ≥ 2, f(n) = 6f(n− 2)− f(n− 1). Prove that f(n) = 5 · 2n + 2(−3)n for all
n ∈ Z, n ≥ 0.
For n = 2, f(2) = 6·7−4 = 38 = 5·22+2(−3)2. Suppose the statement is true
for all n < N . Then, f(N) = 6[5 · 2N−2 + 2(−3)N−2]− [5 · 2N−1 + 2(−3)N−1] =
5 · 2N + 2(−3)N .
0.4 Equivalent and Countable Sets
30. Prove Corollary 0.15. (That is, prove that any subset of a countable
subset is countable.)
Let X be a set with an uncountable subset K. Then, since K � N, there
cannot be a 1-1, onto function from K to N, so there consequently can’t be one
from X to N.
31. Find a 1-1 function f from N onto S where S is the set of all odd
integers.
CHAPTER 0. PREREQUISITES 10
Define f(n) =
{
n if n is odd
−(n− 1) if n is even . Then, the function is 1-1 by its
linear nature. It is also onto: if m ∈ S is negative, there is an even natural that
is its preimage. If m ∈ S is positive, then there is an odd natural that is its
preimage.
32. Let Pn be the set of all polynomials of degree n with integer coefficients.
Prove that Pn is countable.
Consider the function f : Nn+1 → Pn defined by f(a0, a1, ..., an) = anxn +
an−1xn−1 + · · ·+a1x+a0. We have established a 1-1, onto map from Nn+1 onto
Pn. Since we know that N is countable, by Theorem 0.16, Nn+1is countable.
Thus, Pn ∼ Nn+1 ∼ N.
33. Use Exercise 32 to show that the set of all polynomials with integer
coefficients is a countable set.
For any n, we have shown that Pn is countable. Thus,
∞⋃
k=0
Pk is countable
by Theorem 0.17.
34. Prove the following generalization of Theorem 0.17: If S is a count-
able set and {As}s∈S is an indexed family of countable sets, then ∪s∈SAs is a
countable set.
Let s1be the element of S that is in 1-1 correspondence with 1. Similarly,
let sn be the element in S that is in 1-1 correspondence with n. Then, set
An = Asn . Then, ∪s∈SAs = ∪∞n=1An which is a countable union of countable
sets which is countable.
35. For each p ∈ Pn, define B(p)= {x : p(x) = 0}. Prove that ∪p∈PnB(p)
is countable.
We see that B(p) is just the set of roots of the polynomial p(x) which is
a finite set for any p(x). We have already shown that Pn is countable, so
{B(p)}p∈Pn is a countable indexed family of finite sets, which by Exercise 34 is
countable.
36. An algebraic number is any number that is the root of a polynomial
equation p(x) = 0 where the coefficients of p are integers. Show that the set of
algebraic numbers is a countable set.
Certainly A ⊆ ∪p∈PnB(p). Any subset of a countable set is countable.
CHAPTER 0. PREREQUISITES 11
37. For a set A, let P (A) be the set of all subsets of A. Prove that A is not
equivalent to P (A).
Let A be finite such that |A| = n. Then, |P (A)| = 2n. (For any subset, each
element can either be in the subset or not, so the total number of ways to build
a subset is 2n.) Clearly no 1-1 correspondence can exist.
IfA is countably infinite, generate a subsetAk by ak,n =
{
1 if the nth element of A is in Ak
0 otherwise
.
Suppose that {Ak} is countable. Then we can enumerate the subsets via the
sequences:
A1 ↔ a1,1 a1,2 a1,3 · · ·
A2 ↔ a2,1 a2,2 a2,3 · · ·
...
...
...
...
...
...
Ak ↔ ak,1 ak,2 ak,3 · · ·
...
...
...
...
...
...
.
But then the sequence {an,n} is another unique sequence that doesn’t corre-
spond to any Ai. Thus, P (A) is uncountable. Thus, there can be no 1-1 map
from A into P (A).
Finally, if A is uncountable, then, any countable subset of A cannot be
mapped 1-1 with the set of subsets (by the above argument), so again there can
be no 1-1 map from A onto P (A).
38. Let a, b, c, and d be any real numbers such that a < b and c < d. Prove
that [a, b] is equivalent to [c, d].
Define f : [a, b] → [c, d] via f(x) = c + x−ab−a · (d − c). This function maps
the position of x in [a, b] to a position in [c, d] such that x−ab−a =
x−c
d−c (preserves
percentage of interval. This function is linear and thus 1-1 and onto.
0.5 Real Numbers
*39. If x < y, prove that x < x+y2 < y.
We see that x = x2 +
x
2 <
x
2 +
y
2 =
x+y
2 . Similarly, y =
y
2 +
y
2 >
x
2 +
y
2 =
x+y
2 .
*40. If x ≥ 0 and y ≥ 0, prove that √xy ≤ x+y2 .
Let x and y be two dimensions. Then, 2x + 2y gives the perimeter of the
rectangle of length x and width y. We see that 4
√
xy gives the perimeter of
a square with the same area. The perimeter of a square is always less than a
perimeter of a proper rectangle of the same area. Thus, 4
√
xy ≤ 2x + 2y ⇔√
xy ≤ x+y2 .
CHAPTER 0. PREREQUISITES 12
*41. If 0 < a < b, prove that 0 < a2 < b2 and 0 <
√
a <
√
b.
Let b = a + δ with δ > 0. Then, b2 = a2 + δ2 + 2aδ > a. Next, suppose
c2 = a and d2 = b = a + δ. Then, d2 − δ = a. Thus, √a = √d2 − δ, while√
b = d. Thus,
√
a <
√
b.
42. If x, y, a, and b are greater than zero and xy <
a
b , prove that
x
y <
x+a
y+b <
a
b .
First, x+ay+b =
x
y+b +
a
y+b <
x
y +
a
b . Now, since
x
y <
a
b , this implies
x
y <
x+a
y+b <
a
b .
43. Let A = {r : r is a rational number and r2 < 2}. Prove that A has no
largest member.
Let ab ∈ A. Suppose that 2 − a
2
b2 =
c
d . Then,
(
a
b +
e
f
)2
= a
2
b2 +
e2
f2 + 2
ae
bf .
This is less than two precisely when e
2
f2 + 2
ae
bf =
e
f
(
e
f + 2
a
b
)
< cd . Thus,
1 + 2ab <
c
d · fe ⇔ cd + 2acbd < fe .
*44. If x = supS, show that, for each � > 0, there is a ∈ S such that
x− � < a ≤ x.
Suppose that there is an �0 > 0 such that there is no a between x− �0 and
x. Then, this implies that x− �0 is an upper bound. Contradiction.
*45. If y = inf S, show that, for each � > 0, there is a ∈ S such that
y ≤ a < y + �.
Suppose that there is an �0 > 0 such that there is no a between y + �0 and
y. Then, y + �0 is a lower bound. Contradiction.
Chapter 1
Sequences
1.1 Sequences and Convergence
1. Show that [0, 1] is a neighborhood of 23– that is, there is � > 0 such that(
2
3 − �, 23 + �
) ⊂ [0, 1].
Choose � = 13 . Then
(
2
3 − 13 , 23 + 13
)
=
(
1
3 , 1
) ⊂ [0, 1].
*2. Let x and y be distinct real numbers. Prove there is a neighborhood P
of x and a neighborhood Q of y such that P ∩Q = ∅.
Choose � = |x−y|2 . Then, let P = (x− �, x+ �) and Q = (y− �, y+ �). Then,
P ∩Q = ∅.
*3. Suppose x is a real number and � > 0. Prove that (x − �, x + �) is a
neighborhood of each of its members; in other words, if y ∈ (x− �, x + �), then
there is δ > 0 such that (y − δ, y + δ) ⊂ (x− �, x+ �).
Let δ = min
(
|y−(x+�)|
2 ,
|y−(x−�)|
2
)
. Then, (y − δ, y + δ) ⊂ (x− �, x+ �).
4. Find upper and lower bounds for the sequence
{
3n+7
n
}∞
n=1
.
First, 3n+7n = 3 +
7
n . Thus, the lower bound is 3 and the upper bound is 10.
5. Give an example of a sequence that is bounded but not convergent.
Let an = (−1)n. Then, this sequence alternates between −1 and 1, but
never converges.
13
CHAPTER 1. SEQUENCES 14
6. Use the definition of convergence to prove that each of the following
sequences converges:
(a)
{
5 + 1n
}∞
n=1
(b)
{
2−2n
n
}∞
n=1
(c) {2−n}∞n=1
(d)
{
3n
2n+1
}∞
n=1
(a) Let � > 0 be given. Let N =
⌈
1
�
⌉
. Then,
∣∣5 + 1n − 5∣∣ = ∣∣ 1n ∣∣ < ∣∣ 1N ∣∣ < �.
(b) Let � > 0 be given. Let N =
⌈
2
�
⌉
. Then,
∣∣ 2−2n
n + 2
∣∣ = ∣∣ 2−2n+2nn ∣∣ =∣∣ 2
n
∣∣ < ∣∣ 2N ∣∣ < �.
(c) Let � > 0 be given. Let N =
⌈
lg
(
1
�
)⌉
. Then, |2−n| = ∣∣ 12n ∣∣ < ∣∣ 12N ∣∣ < �.
(d) Let � > 0 be given. Let N =
⌈
3
4�
⌉
Then,
∣∣∣ 3n2n+1 − 32 ∣∣∣ = ∣∣∣ 6n−6n−34n+2 ∣∣∣ =∣∣∣ −34n+2 ∣∣∣ < ∣∣ 34n ∣∣ < ∣∣ 34N ∣∣ < �.
*7. Show that {an}∞n=1 converges to A iff {an −A}∞n=1 converges to 0.
We see that |(an −A)− 0| = |an −A| < �.
8. Suppose {an}∞n=1 converges to A, and define a new sequence {bn}∞n=1 by
bn =
an+an+1
2 for all n. Prove that {bn}∞n=1 converges to A.
Let � > 0 be given. We see that
∣∣∣an+an+12 −A∣∣∣ = ∣∣∣an+an+1−2A2 ∣∣∣ = ∣∣∣an−A+an+1−A2 ∣∣∣ ≤
|an−A|
2 +
|an+1−A|
2
∃N
<
s.t.
�
2 +
�
2 = �. Thus, bn → A.
*9. Suppose {an}∞n=1, {bn}∞n=1, and {cn}∞n=1 such that {an}∞n=1 converges
to A, {bn}∞n=1 converges to A, and an ≤ cn ≤ bn for all n. Prove that {cn}∞n=1
converges to A.
Since an ≤ cn ≤ bn, we must have an−A ≤ cn−A ≤ bn−A. By convergence
and definition of absolute value, −� < an−1−A ≤ cn−A ≤ bn−A < �. Hence,
|cn −A| < �. Thus, cn → A. (We will call this result the Squeeze Theorem.)
*10. Prove that, if {an}∞n=1 converges to A, then {|an|}∞n=1 converges to
|A|. Is the converse true? Justify your conclusion.
We see that ||an| − |A||
TI
< ||an −A|| = |an −A| < �. The converse is not
true. For instance, |(−1)n| → 1, but {(−1)n} diverges.
*11. Let {an}∞n=1 be a sequence such that there exist numbers α and N such
that, for n ≥ N , an = α. Prove that {an}∞n=1 converges to α.
We see that |an − α| ≤ |aN − α| = 0 < � for all � > 0.
CHAPTER 1. SEQUENCES 15
12. Give an alternate proof of Theorem 1.1 along the following lines. Choose
� > 0. There is N1 such that for n ≥ N1, |an −A| < �2 , and there is N2 such
that for n ≥ N2, |an −B| < �2 . Use the triangle inequality to show that this
implies that |A−B| < �.
Let N = max(N1, N2). Then, � > |an −A| + |an −B| = |an −A| +
|B − an| > |an −A+B − an| = |B −A|. Thus, |A−B| < �.
13. Let x be any positive real number, and define a sequence {an}∞n=1 by
an =
[x] + [2x] + · · ·+ [nx]
n2
where [x] is the largest integer less than or equal to x. Prove that {an}∞n=1
converges to x/2.
Let � > 0 be given and set N = x2� . Then,
∣∣an − x2 ∣∣ ≤ ∣∣x+2x+···+nxn2 − x2 ∣∣ =∣∣∣x(1+···+n)n2 − x2 ∣∣∣ = ∣∣∣xn(n+1)2n2 − x2 ∣∣∣ = ∣∣∣xn22n2 + xn2n2 − x2 ∣∣∣ = ∣∣ x2n ∣∣ < ∣∣ x2N ∣∣ < �.
1.2 Cauchy Sequences
14. Prove that every Cauchy sequence is bounded. (Theorem 1.4)
Suppose that {an}is not bounded. Then, for any k, there is an nk such that
|ank | > k. Then, {ank}is an unbounded sequence. Then, for any N , there exist
ank and an`such that |ank − an` | > |ank |− |an` | = k− ` where k− ` > N . Thus,
{an} is not Cauchy.
15. Prove directly (do not use Theorem 1.8) that, if {an}∞n=1 and {bn}∞n=1
are Cauchy, so is {an + bn}∞n=1.
Since {an} and {bn} are Cauchy, then for all � > 0, there exist N1 and N2
such that |an − am| < �2 for all m,n > N1 and |bn − bm| < �2 for all m,n > N2.
ChooseN = max(N1, N2). Then, |an + bn − (am + bm)| = |an − am + bn − bm| <
|an − am|+ |bn − bm| < �2 + �2 = � for all m,n > N .
16. Prove directly (do not use Theorem 1.9) that, if {an}∞n=1 and {bn}∞n=1
are Cauchy, so is {anbn}∞n=1. You will want to use Theorem 1.4.
Since {bn} is Cauchy, then it is bounded (by Exercise 14). Thus, |bn| < M for
some M . Since {an} is Cauchy, then for all � > 0, there exist N |an − am| < �M
for all m,n > N and |bn − bm| < �2 for all m,n > N2. Let � > 0 be given. Then,|anbn − ambm| < |anM − amM | = |M(an − am)| < �.
CHAPTER 1. SEQUENCES 16
17. Prove that the sequence
{
2n+1
n
}∞
n=1
is Cauchy.
Let � > 0 be given. ChooseN = 1� . Then,
∣∣ 2n+1
n − 2m+1m
∣∣ = ∣∣ 2mn+m−2mn−nnm ∣∣ =∣∣m−n
nm
∣∣ = ∣∣ mmn − nmn ∣∣ < ∣∣ NN2 ∣∣ = ∣∣ 1N ∣∣ < �.
18. Give an example of a sequence with exactly two accumulation points.
Let an =
{
1/n , if n is even
1 + 1/n , if n is odd
. Then, an has accumulation points at
0 and 1.
19. Give an example of a set with a countably infinite set of accumulation
points.
The setQ has the property that every element is an accumulation point, since
for any ab ∈ Q, the sequence
{
a
b +
1
n
}
converges to ab . Since Q is countable, we
have found the desired set.
20. Give an example of a set that contains each of its accumulation points.
The set [0, 1] contains all of its accumulation points.
21. Determine the accumulation points of the set
{
2n + 1k : n and k are positive integers
}
.
The set {2n : n ∈ Z+}∪{∞} is the set of accumulation points since 2n+ 1k →
2n as k →∞ and 2n + 1k →∞ as n→∞.
22. Let S be a nonempty set of real numbers that is bounded from above
(below) and let x = supS (inf S). Prove that either x belongs to S or x is an
accumulation point of S.
It is clear that x ∈ S is a possibility. Suppose x /∈ S. Then, by Exercise
0.44, for any � > 0, there is an a ∈ S such that x− � < a < x. Thus, for all n,
there exists an an ∈ S such that x − 1n < an < x. Since x − 1n → x, we have
an → x. Thus, x is an accumulation point of S.
23. Let a0 and a1 be distinct real numbers. Define an =
an−1+an−2
2 for each
positive integer n ≥ 2. Show that {an}∞n=1 is a Cauchy sequence. You may want
to use induction to show that
an+1 − an =
(
−1
2
)n
(a1 − a0)
and then use the result from Example 0.9 of Chapter 0.
CHAPTER 1. SEQUENCES 17
The statement an+1 − an =
(− 12)n (a1 − a0) is obviously true for n =
1. Suppose that it is true for n < N . Then, aN+2 − aN+1 = aN+1+aN2 −[(− 12)N (a1 − a0) + aN] = aN+1+aN−2aN2 + (− 12)N+1 (a1 − a0) = aN+1−aN2 +(− 12)N+1 (a1−a0) = (− 12)N (a1−a0)+(− 12)N+1 (a1−a0) =[(− 12)N + (− 12)N+1] (a1−
a0) =
[(− 12)N + (− 12)N (− 12)] (a1 − a0) = [(− 12 + 1) (− 12)N] (a1 − a0) =(− 12)N+1 (a1 − a0). Thus, the statement is proven by induction.
Now, let � > 0 and n,m ∈ N be given. Choose N =
⌈
lg
(
|a1−a0|(n−m)
�
)⌉
.
Then,
∣∣(− 12)n (a1 − a0)∣∣ < �n−m for n ≥ N. Thus, we see that |an − am| =|an − an−1 + an−1 − an−2 + · · ·+ am+1 − am| < |an − an−1|+· · ·+|am+1 − am| <
�
n−m +
�
n−m + · · ·+ �n−m = �.
24. Suppose {an}∞n=1 converges to A and {an : n ∈ N} is an infinite set.
Show that A is an accumulation point of {an : n ∈ N}.
LetNk =
(
A− 1k , a+ 1k
)
. By convergence, there is anN such that |an −A| <
1
k for all n ≥ N . Thus, there is an element of {an : n ∈ N} in Nk for every k.
Now, every neighborhood of A has Nk as a subset for some k and since there are
an infinity of Nk’s, we have an infinity of members of {an} in any neighborhood.
1.3 Arithmetic Operations on Sequences
25. Suppose {an}∞n=1 and {bn}∞n=1 are sequences such that {an}∞n=1 and
{an + bn}∞n=1 converge. Prove that {bn}∞n=1 converges.
Suppose an → A and an + bn → C. Then, {bn} = {an + bn − an}. Thus, by
Theorem 1.8, bn → C −A.
26. Give an example in which {an}∞n=1 and {bn}∞n=1 do not converge but
{an + bn}∞n=1 converges.
Let an = (−1)n and bn = (−1)n+1. We know that an and bn don’t converge,
but an + bn → 0.
27. Suppose {an}∞n=1 and {bn}∞n=1 are sequences such that {an}∞n=1 con-
verges to A 6= 0 and {anbn}∞n=1. Prove that {bn}∞n=1 converges.
Suppose anbn → C. Then, {bn} =
{
anbn
an
}
, so by Theorem 1.9, bn → CA .
28. If {an}∞n=1 converges to a with an ≥ 0 for all n, show
{√
an
}∞
n=1
converges to
√
a.
CHAPTER 1. SEQUENCES 18
Let � > 0 be given. Then, there is an N such that |an − a| < �√a . Then,∣∣√an − a∣∣ = ∣∣∣ an−a√an+√a ∣∣∣ = |an − a| ∣∣∣ 1√an+√a ∣∣∣ < |an − a| ∣∣∣ 1√a ∣∣∣ < � for all n ≥ N .
29. Prove that 
(
n+ k
k
)
(n+ k)k

∞
n=1
converges to 1k! , where (
n+ k
k
)
=
(n+ k)!
n!k!
.
We see that
{
(n+k)!
n!k!(n+k)k
}
=
{
(n+k−1)(n+k−2)···(n+1)
k!(n+k)k−1
}
. Now, let � > 0 be
given. ChooseN =
⌈
1−k
�
⌉
. Then,
∣∣∣ (n+k−1)···(n+1)k!(n+k)k−1 − 1k! ∣∣∣ < ∣∣∣ (n+k)k−2(n+1)k!(n+k)k−1 − 1k! ∣∣∣ =∣∣∣ n+1k!(n+k) − 1k! ∣∣∣ = ∣∣∣n+1−n−kk!(n+k) ∣∣∣ = ∣∣∣ 1−kk!(n+k) ∣∣∣ < ∣∣ 1−kn ∣∣ < � for all n ≥ N .
30. Prove the following variation on Lemma 1.10. If {bn}∞n=1 converges to
B 6= 0 and bn 6= 0 for all n, then there is M > 0 such that |bn| ≥M for all n.
Choose M = |bn| /2. Then, the statement holds.
31. Consider a sequence {an}∞n=1 and, for each n, define
αn =
a1 + a2 + · · ·+ an
n
.
Prove that if {an}∞n=1 converges to A, then {αn}∞n=1 converges to A. Give an
example in which {αn}∞n=1 converges, but {an}∞n=1 does not.
Let � > 0 be given. There is an N1 such that |an −A| < �2 . for all n > N .
Let M = |a1 −A|+ |a2 −A|+ · · ·+ |aN1 −A|. Then, there is an N2 such that∣∣M
n
∣∣ < �2 for all n > N2. Let N = max(N1, N2). Then, ∣∣a1+a2+···+ann −A∣∣ =∣∣a1+···+aN+···+an−nA
n
∣∣ < |a1−A|+···+|aN−A|n + |aN+1−A|+···+|an−A|n = Mn + |aN+1−A|+···+|an−A|n <
�
2 +
�
2 = �. (A quicker way to show this would be to observe that by the definition
of convergence, |an| < �2 for all n > N for some N . Then, since
Let an = (−1)n. Then, as we have seen before, {an} diverges, but a1+a2+···+ann →
0.
32. Find the limit of the sequences with general term as given:
(a) n
2+4n
n2−5
(b) cosnn
(c) sinn
2√
n
CHAPTER 1. SEQUENCES 19
(d) nn2−3
(e)
(√
4− 1n − 2
)
n
(f) (−1)n
√
n
n+7
(a) 1
(b) 0
(c) 0
(d) 0
(e)
(√
4− 1n − 2
)
n = n
√
4− 1n−2n =
√
4n2 − n−
√
4n2 = 4n
2−n−4n2√
4n2−n+
√
4n2
∼
−n
4n . Thus, limit is − 14 .
(f) 0
33. Find the limit of the sequence in Exercise 23 when a0 = 0 and a1 = 3.
You might want to look at Example 0.10.
Example 0.10 says that an−1+an−22 =
(− 12n ) 3 + 2. Since − 12n → 0, the limit
must be 2.
1.4 Subsequences and Monotone Sequences
34. Find a convergent subsequence of the sequence{
(−1)n
(
1− 1
n
)}∞
n=1
Let nk = 2n. Then, the subsequence is
{
1− 12n
}
which converges to 0.
35. Suppose x is an accumulation point of {an : n ∈ N}. Show that there is
a subsequence of {an}∞n=1 that converges to x.
Since x is an accumulation point, every neighborhood about x contains an
infinity of {an}. Thus, let ank be a member of {an : n ∈ N} ∩
(
x− 1k , x+ 1k
)
.
Then, for any � > 0, there is a K such that 1k < � for all k > K. Thus, ank → x.
36. Let {an}∞n=1 be a bounded sequence of real numbers. Prove that {an}∞n=1
has a convergent subsequence.
Either{an} has a finite number ofvalues or {an} has an infinite number of
values. For the former, there must be some value x for which there are infinitely
many k such that ank = x. Thus, ank → x. For the latter, the sequence is a
CHAPTER 1. SEQUENCES 20
bounded infinite set of real numbers, so by the Bolzano-Weierstrass Theorem,
an has a convergent subsequence.
*37. Prove that if {an}∞n=1 is decreasing and bounded, then {an}∞n=1 con-
verges.
Assume that {an} attains an infinite number of values. Suppose that inf an =
M . Let � > 0 be given. Then, there are a1−M� intervals that sequence values
may fall. Since this is a finite number and there are an infinite number of values,
at least one region must contain an infinite number of function values. Since the
sequence is decreasing, the last region must contain an infinity of values; that
is, an ∈ (M,M + �) for all n > N for some N . Since � was arbitrarily chosen,
the proof is complete. The case for when {an} has only finitely many values is
easy.
38. Prove that if c > 1, then { n√c}∞n=1 converges to 1.
It is clear that n
√
c > n+1
√
c. Thus, { n√c} is a monotone decreasing sequence.
Also, n
√
c > 1, so by Theorem 1.16, n
√
c→ 1.
*39. Suppose {xn}∞n=1 converges to x0 and {yn}∞n=1 converge to x0. Define
a sequence {zn}∞n=1 as follows: z2n = xn and z2n−1 = yn. Prove that {zn}∞n=1
converges to x0.
Both subsequences of {zn} converge to x0. Thus, by Theorem 1.14, zn → x0.
40. Show that the sequence defined by a1 = 6 and an =
√
6 + an−1 for n > 1
is convergent and find its limit.
To find the limit L, set L =
√
6 + L⇔ L2 = 6 + L⇔ L2 − L− 6 = 0. The
solutions are −2 and 3. The only solution that works is 3. Thus, an → 3. We
prove that {an} is decreasing. Since √6 + an−1 <
√
6 +
√
an−1 and we know
that
√
an−1 is decreasing, we see that the whole sequence is decreasing. Also,
square roots must be greater than 0, so the sequence is bounded. Thus, the
sequence is bounded below and decreasing and is thus convergent.
41. Let {xn}∞n=1 be a bounded sequence and let E be the set of subsequential
limits of that sequence. By Exercise 36, E is nonempty. Prove that E is bounded
and contains both supE and inf E.
Since {xn} is bounded (by M), its limit points must be such that they are
within � distance of some sequence values. Thus, limit points must be within
the same bounds as {xn} or within � distance of the boundary for any �. Thus,
E is bounded (by, say M + 1). We must ensure that members of E do not form
a sequence themselves that converges to a non-limit point. So, suppose there
is a sequence {en} of limit points. Then, for every �, there is an N such that
CHAPTER 1. SEQUENCES 21
|en − xn| < � for all n > N . Thus, xn → en. Thus, all sequences of E converge
in E (since they are estimated by subsequences of {xn}. Thus, supE, inf E ∈ E.
42. Let {xn}∞n=1 be any sequence and T : N→ N be any 1-1 function. Prove
that if {xn}∞n=1 converges to x, then {xT (n)}∞n=1 also converges to x. Explain
how this relates to subsequences. Define what one might call a “rearrangement”
of a sequence. What does the result imply about rearrangements of sequences?
We see that {xT (n)} = {xn1 , xn2 , ...} and is a subsequence of {xn}. Since all
subsequences converge, we must have xT (n) → x. Let T : N → N be any 1-1
function and let {xn} be a sequence. Then, {xT (n)} is called a rearrangement.
The result implies that if {xn} converges, then so does {xT (n)} for any T .
43. Assume 0 ≤ a ≤ b. Does the sequence {(an + bn)1/n}∞n=1 converge or
diverge? If the sequence converges, find the limit.
The sequence does not converge in general. For instance, if a = 1 and
b = −1, then the sequence becomes {[1n+(−1)n]1/n}. Taking even indexes, the
limit is 1, and taking odd indexes, the limit is 0. Thus, not all subsequences
converge to the same limit point, so the sequence is not convergent.
44. Does the sequence {
k∑
n=1
(
1√
k2 + n
)}∞
k=1
diverge or converge? If the sequence converges, find the limit.
We see that
∑k
n=1
1√
k2+n
<
∑∞
n=1
1
k =
k
k = 1. Also,
∑∞
n=1
1√
k2+n
>∑∞
n=1
1√
k2+k
= k√
k2+k
→ 1. Thus, the sequence must converge to 1 by the
Squeeze Theorem (established in Exercise 9).
*45. Show that if x is any real number, there is a sequence of rational
numbers converging to x.
Let a0.a1a2 · · · be the decimal expansion for x. Then, define xn = a0.a1a2 · · · an.
Then, xn ∈ Q for all n and xn → x (for there exists an N for which xn can be
within 10k distance for any integer k and n > N).
*46. Show that if x is any real number, there is a sequence of irrational
numbers converging to x.
If x is already irrational, define xn ≡ x. Clearly, xn → x. If x is rational,
define xn = x+
pi
n . Then, xn ∈ R \Q for all n and xn → x.
CHAPTER 1. SEQUENCES 22
47. Suppose that {an}∞n=1 converges to A and that B is an accumulation
point of {an : n ∈ N}. Prove that A = B.
If B is an accumulation point, then
(
B − 1n , B + 1n
)
contains a member of
{an} for all n. Thus, one can construct a subsequence of these members that
converges to B, and since {an} is convergent, we must have A = B.
Miscellaneous
48. Suppose that {an}∞n=1 and {bn}∞n=1 are two sequences of positive real
numbers. We say that an is O(bn) (read as “big oh” of bn) if there is an integer
N and a real number M such that for n ≥ N , an ≤ M · bn. Prove that if
{an/bn}∞n=1 converges to L 6= 0, then an is O(bn) and bn is O(an). What can
you say if L = 0? Illustrate with examples.
Since an/bn → L, we guess that there is an N such that an ≤ (L + 1) · bn
for all n ≥ N . We now prove this assertion. First, for any � > 0, there is
an N for which an/bn ∈ (L − �, L + �) for all n ≥ N . Thus, for the same N ,
an ∈ (bn[L− �], bn[L+ �]). Thus, an ≤ bn(L + 1). Thus, an is O(bn) if N is
chosen to correspond to � = 1. Similarly, bn is O(an).
If L = 0, then either an → 0 and {bn} is bounded or bn → ∞ and {an} is
bounded. For instance, n
2
n3 → 0 and 1/n2+1/n = 0. (This result is called the Limit
Comparison Test.)
Chapter 2
Limits of Functions
2.1 Definition of the Limit of a Function
1. Define f : (−2, 0) → R by f(x) = x2−4x+2 . Prove that f has a limit at −2
and find it.
Let � > 0 be given. Choose δ = �. Then
∣∣∣x2−4x+2 + 4∣∣∣ = |x− 2 + 4| = |x+ 2| <
δ = �. Thus, lim
x→−2
f(x) = −4.
2. Define f : (−2, 0) → R by f(x) = 2x2+3x−2x+2 . Prove that f has a limit at−2 and find it.
Let � > 0 be given. Choose δ = �2 . Then
∣∣∣ 2x2+3x−2x+2 + 5∣∣∣ = ∣∣∣ (x+2)(2x−1)x+2 + 5∣∣∣ =
|2x− 1 + 5| = |2x+ 4| = 2 |x+ 2| < 2δ = �.
3. Give an example of a function f : (0, 1) → R that has a limit at every
point except 12 . Use the definition of limit of a function to justify the example.
Let f(x) =
{
0, 0 < x < 12
1, 12 ≤ x < 1
. Then, the limit is clearly 0 for every input
x ∈ (0, 12) and 1 for every input x ∈ ( 12 , 1). Let �0 = 1. Let δ > 0 be given.
Then, in the neighborhood
(
1
2 − δ, 12 + δ
)
, there is an input to the left-half of the
interval so that f(x) = 0 and there is an input in the right-half of the interval
such that f(x) = 1. Now, the only possibilities for the limit would be either 0
or 1. However, the maximum difference between function values on the interval
and either limit is 1 which is equal to �0. Thus, the limit does not exist.
4. Give an example of a function f : R→ R that is bounded and has a limit
at every point except −2. Use the definition to justify the example.
23
CHAPTER 2. LIMITS OF FUNCTIONS 24
Change the parameters of the solution given to Exercise 3 to obtain a similar
proof.
*5. Suppose f : D → R with x0 an accumulation point of D. Assume L1
and L2 are limits of f at x0. Prove L1 = L2.
Let � > 0 be given. Then, there are δ1 and δ2 such that |f(x)− L1| < �2 and|f(x)− L2| < �2 . Then, |L1 − L2| = |L1 − f(x) + f(x)− L2| < |L1− f(x)| +|f(x)− L2| < �2 + �2 = �. Thus, L1 = L2.
6. Define f : (0, 1)→ R by f(x) = cos 1x . Does f have a limit at 0? Justify.
For any δ > 0, any function value between 0 and 1 can be obtained from
inputs x ∈ (0, δ). Thus, ∣∣cos 1x − L∣∣ can be made greater than some �. Thus, no
limit exists at x = 0.
7. Define f : (0, 1)→ R by f(x) = x cos ( 1x). Does f have a limit at x = 0?
Justify.
We guess that lim
x→0
f(x) = 0. Let � > 0 be given. Choose δ = �. Then,∣∣x cos ( 1x)∣∣ < |x| < δ = �.
8. Define f : (0, 1) → R by f(x) = x3−x2+x−1x−1 . Prove that f has a limit at
1.
We see lim
x→1
x3−x2+x−1
x−1 = limx→1
x2(x−1)+(x−1)
x−1 = limx→1
(x2+1)(x−1)
x−1 = limx→1
x2 + 1 =
2. Let � > 0 be given. Choose δ =
√
�. Then,
∣∣∣x3−x2+x−1x−1 − 2∣∣∣ = ∣∣x2 + 1− 2∣∣ =∣∣x2 − 1∣∣ = |x+ 1| |x− 1| < δ2 = �.
9. Define f : (−1, 1) → R by f(x) = x+1x2−1 . Does f have a limit at 1?
Justify.
We see lim
x→1
x+1
x2−1 = limx→1
x+1
(x+1)(x−1) = limx→1
1
x−1 . From what we know from
calculus, we know that the limit does not exist. To be more precise, inputs in
(1, 1 + �) yield outputs which become arbitrarily large as � → 0 and inputs in
(1− �, 1) yield outputs which become arbitrarily negative as �→ 0.
2.2 Limits of Functions and Sequences
10. Consider f : (0, 2) → R defined by f(x) = xx. Assume that f has a
limit at 0 and find that limit.
CHAPTER 2. LIMITS OF FUNCTIONS 25
Examine xn = log
(
1 + 1n
)
. Clearly, xn → 0. Since we assumed that f had
a limit, it must be the case that f(xn)→ lim
x→0
f(x). Then, f(xn) = log
(
1 + 1n
)2
which clearly has milt 0.
*11. Suppose f , g, and h : D → R where x0 is an accumulation point of
D, f(x) ≤ g(x) ≤ h(x) for all x ∈ D, and f and h have limits at x0 with
limx→x0 f(x) = limx→x0 h(x). Prove that g has a limit at x0 and
lim
x→x0
f(x) = lim
x→x0
g(x) = lim
x→x0
h(x).
Since f and h have limits, lim
n→∞f(xn) = limn→∞g(xn) = limn→∞h(xn) for any
xn → x0 by the Squeeze Theorem. Thus, lim
x→x0
f(x) = lim
x→x0
g(x) = lim
x→x0
h(x) by
Theorem 2.1.
*12. Suppose f : D → R has a limit at x0. Prove that |f | : D → R has a
limit at x0 and that limx→x0 |f(x)| = |limx→x0 f(x)|.
Let � > 0 be given. Then by definition of limit, there is a δ > 0 such that
for all x ∈ (x0 − δ, x0 + δ), |f(x)− x0| < �. There are three cases to consider.
If x0 > 0, then |f(x)| ≡ f(x) on the interval of interest and so the limit would
be that of f . If x0 < 0, then, |f(x)| ≡ −f(x) on the interval of interest and
|−f(x) + x0| = |−1| |f(x)− x0| < � for any x ∈ (x0 − δ, x0 + δ). Finally, if
x0 = 0, then ||f(x)|| = |f(x)| < � for any x ∈ (x0− δ, x0 + δ), so the limit exists.
13. Define f : R → R by f(x) = x − [x]. Determine the points at which f
has a limit and justify your conclusions.
We see that x− [x] leaves us with the decimal part of x. Thus, the function
is continuous for (k, k + 1) for every integer k. It is not have a limit at the
endpoints as there will be a jump of 1 unit.
14. Define f : R→ R as follows:
f(x) = 8x if x is a rational number
f(x) = 2x2 + 8 if x is an irrational number.
Use subsequences to guess at which points f has a limit, then use �’s and δ’s to
justify your conclusions.
We can only hope to find a limit at an accumulation point and since there can
always be found a rational number between two irrationals (and vice versa), we
see that the only possibility is at x = 2. We suppose that the limit at this point is
16. Now, let � > 0 be given. Choose δ1 =
�
2Then |8x− 16| = 8 |x− 2| < 8δ1 = �.
Then, let δ2 =
�
2 . Then,
∣∣2x2 + 8− 16∣∣ = ∣∣2x2 − 8∣∣ = 2 ∣∣x2 − 4∣∣ < 2 |x− 4|2 =
CHAPTER 2. LIMITS OF FUNCTIONS 26
2 · �2 = �. Choose δ = min(δ1, δ2) to see that the function has a limit of 16 at
x = 2.
15. Let f : D → R with x0 as an accumulation point of D. Prove that f has
a limit at x0 if for each � > 0, there is a neighborhood Q of x0 such that, for
any x, y ∈ Q ∩D, x 6= x0, y 6= x0, we have |f(x)− f(y)| < �.
Let � > 0 be given. Since f has a limit L at x = x0, there is a δ such that for
any x ∈ (x0 − δ, x0 + δ), |f(x)− L| < �2 . Let Q = (x0 − δ, x0 + δ). We see that|f(x)− f(y)| = |f(x)− L+ L− f(y)| < |f(x)− L|+ |f(y)− L| = �2 + �2 = �.
2.3 Algebra of Limits
16. Define f : (0, 1) → R by f(x) = x3+6x2+xx2−6x . Prove that f has a limit at
0 and find that limit.
We see that f(x) = x(x
2+6x+1)
x(x−6) . Thus, the limit is limx→0
x
x ·
lim
x→0
(x2+6x+1)
lim
x→(x−6)
which
we certainly see is equal to − 16 .
17. Define f : R→ R as follows:
f(x) = x− [x] if [x] is even.
f(x) = x− [x+ 1] if [x] is odd.
Determine those points where f has a limit and justify your conclusions.
As discussed in Exercise 13, x − [x] gives us the decimal part of x. It was
determined that the function had no limits for any x ∈ Z. We now observe that
x− [x+ 1] gives us the negative value of the decimal part of 1−x. We still have
limits in the intervals (k, k + 1), but approaching the left endpoint of an odd
numbered interval from the left, the limit is 1, while the limit from the right is
1. Thus, the limit is defined at odd left-endpoints. Similarly, all the endpoints
have limits. Thus, f has a limit at every point in R.
18. Define g : (0, 1) → R by g(x) =
√
1+x−1
x . Prove that g has a limit at 0
and find it.
We see that, for x 6= 0, g(x) =
√
1+x−1
x ·
√
1+x+1
x =
1+x−1
x =
x
x . The limit of
this function is clearly 1. (This is a proof because of the multiplication of limits
is the limit of multiplications.)
19. Define f : (0, 1)→ R by f(x) =
√
9−x−3
x . Prove that f has a limit at 0
and find it.
CHAPTER 2. LIMITS OF FUNCTIONS 27
We see that for x 6= 0, f(x) =
√
9−x−3
x ·
√
9−x+3
x =
9−x−9
x = −xx . The limit
is clearly 1.
20. Prove Theorem 2.5. [That is, suppose f : D → R and g : D → R, x0
is an accumulation point of D, and f and g have limits at x0. Prove that if
f(x) ≤ g(x) for all x ∈ D, then lim
x→x0
f(x) ≤ lim
x→x0
g(x).]
Suppose lim
x→x0
f(x) > lim
x→x0
g(x). Then, we must have lim
x→x0
f(x)− lim
x→x0
g(x) =
lim
x→x0
(f(x) − g(x)). We see that the left-hand side must be positive, but the
right-hand side is at most 0. Contradiction.
21. Suppose g : D → R with x0 an accumulation point of D and g(x) 6= 0
for all x ∈ D. Further assume that g has a limit at x0 and limx→x0 g(x) 6= 0.
State and prove a lemma similar to Lemma 1.10 for such a function. [Lemma
1.10 states: “If {bn}∞n=1 converges to B and B 6= 0, then there is a positive real
number M and a positive integer N such that, if n ≥ N , then |bn| ≥M .”]
Lemma. If g(x) has a limit at x0 and x0 6= 0, then there is a positive real
number M and a δ > 0 such that, if x ∈ (x0 − δ, x0 + δ), then |g(x)| ≥M .
Proof. Suppose that lim
x→x0
g(x) = L. Then, there is a δ > 0 such that for
x ∈ (x0 − δ, x0 + δ), we have g(x) ∈ (x0 − δ, x0 + δ). Thus, setting M = x0 − δ,
we see that |g(x)| ≥M .
22. Show by example that, even though f and g fail to have limits at x0, it
is possible for f + g to have a limit at x0. Give similar examples for fg and
f
g .
We have seen that the function f(x) = 1x does not have a limit at 0. Thus,
g(x) = −f(x) will also not have a limit. Nonetheless, lim
x→0
(f + g)(x) = 0. Also,
if f is the function defined in Exercise 14, then f has no limit any where except
at x = 2. If g ≡ 1f , then limx→0(fg)(x) = 1. For the same f , choosing g ≡ f gives
lim
x→1
f
g = 1.
2.4 Limits of Monotone Functions
23. State and prove a lemma similar to Lemma 2.7 for decreasing functions.
[Lemma 2.7 states: “Let f : [α, β] → R be increasing. Let U(x) = inf{f(y) :
x < y} and L(x) = sup{f(y) : y < x} for x ∈ (α, β). Then f has a limit at
x0 ∈ (α, β) iff U(x0) = L(x0), and in this case lim
x→x0
f(x) = f(x0) = U(x0) =
L(x0).”]
CHAPTER 2. LIMITS OF FUNCTIONS28
Lemma. Let f : [α, β] → R be decreasing. Let L(x) = sup{f(y) : y < x}
and U(x) = inf{f(y) : y > x} for x ∈ (α, β). Then f has a limit at x0 ∈ (α, β)
iff U(x0) = L(x0), and in this case lim
x→x0
f(x) = f(x0) = U(x0) = L(x0).
Proof. (⇒) Suppose that lim
x→x0
f(x) = L. Since f is decreasing, if x ≤ y,
we have f(x) ≤ f(y). Thus, for every δ > 0, f(x0) ≥ L(x0) ≥ f(x0 + δ).
Similarly, f(x0) ≤ U(x0) ≤ f(x0 + δ) . We also note that since the limit exists,
f
(
x0 +
1
n
)
, f
(
x0 − 1n
)→ L = f(x0). Thus, L(x0) = L and U(x0) = L.
(⇐) Suppose that U(x0) = L(x0). Then, inf{f(y) : y > x0} = sup{f(y) :
y < x0}. This implies that for every � > 0, there exists a t1 ∈ (x0, β] such
that U(x0) ≤ f(t1) < U(x0) + �. Similarly, there exists a t2 ∈ [α, x0) such
that L(x0) ≥ f(t2) > L(x0) − �. Since we have U(x0) = L(x0) = M , we
have M + � ≥ f(t1) ≥ M ≥ f(t2) > M − �. Let δ = |t1 − t2|. Then for
any x ∈ (x0 − δ, x0 + δ), we have |f(x)− f(x0)| < |f(t1)− f(t2)| < �. Thus,
lim
x→x0
f(x) = f(x0).
24. Let f : [a, b] → R be monotone. Prove that f has a limit both at a and
b.
Without loss of generality, suppose that f is increasing. Then, the only way
that f might not have a limit at a is if U(a) > a (L(a) = a since there are no
function values defined for inputs less than a). Suppose that U(a) = f(a+δ0) =
f(a) + �0. But then, f(a) < f
(
a+δ0
2
)
< U(a) by the fact that f is increasing.
Contradiction. Similar for the limit at b. (Use L(b) and argue similarly.)
25. Suppose f : [a, b]→ R and define g : [a, b]→ R as follows:
g(x) = sup{f(t) : a ≤ t ≤ x}
Prove that g has a limit at x0 if f has a limit at x0 and limt→x0 f(t) = f(x0).
We note that g(x) is increasing. Let x? be such that f(x?) is the largest
function value on [a, x0]. There are two cases to consider: (1) f(x0) = f(x
?) or
(2) f(x0) < f(x
?). For (1), either g ≡ f on (x0 − δ, x+ δ) for some δ or g ≡ f
on (x0 − δ, x0) and g ≡ f(x?) on (x0, x0 + δ). Either way, since f has a limit at
x0 and constant functions have limits at any point, g(x) has a limit at x0. For
(2), g ≡ f(x?) on (x0 − δ, x0 + δ) for some δ. Constant functions have limits at
all points, so g(x) has a limit at x0.
Miscellaneous
26. Assume that f : R → R is such that f(x + y) = f(x)g(x) for all
x, y ∈ R. If f has a limit at zero, prove that f has a limit at every point and
either lim
x→0
f(x) = 1 or f(x) = 0 for all x ∈ R.
CHAPTER 2. LIMITS OF FUNCTIONS 29
If f has a limit at 0, then for any an → 0, f(an) → lim
x→0
f(x). Then, since
any input y can be written as 0 + y, we see that f (y + an) = f(y)f(an) →
lim
x→0
f(x) · f(y), showing that a limit exists for all inputs y.
If we have a limit for any input, then we should have lim
x→x0
f(x + y) =
lim
x→x0
[f(x)f(y)] = lim
x→x0
f(x) · lim
x→x0
f(y). In particular, lim
x→x0
f(x+ 0) = lim
x→0
f(x) ·
lim
x→x0
f(x). Thus, lim
x→0
f(x) = 1 or f ≡ 0.
27. Suppose f : D → R, g : E → R, x0 is an accumulation point of D ∩ E,
and there is � > 0 such that D ∩ [x0 − �, x0 + �] = E ∩ [x0 − �, x0 + �]. If
f(x) = g(x) for all x ∈ D ∩ E ∩ [x0 − �, x0 + �], prove that f has a limit at x0
iff g has a limit at x0.
Let � > 0 be given. Then, lim
x→x0
f(x) = L exists ⇔ |f(x)− L| < �2 . At the
same time, |f(x)− g(x)| < �2 . If g(x) has a limit at x0, it must be equal to L.
To see this, suppose |g(x)− L| ≥ �0 for some x ∈ D∩E∩[x0−δ, x0+δ]. We then
have |g(x)− f(x)| = |g(x)− L+ L− f(x)| > |g(x)− L| − |f(x)− L| > �0 − �.
Contradiction. The argument is similar if we assume that g(x) has a limit at
x0.
Chapter 3
Continuity
3.1 Continuity of a Function at a Point
1. Define f : R→ R by f(x) = 3x2 − 2x+ 1. Show that f is continuous at
2.
Let � > 0 be given. Choose δ =
√
�. We see that
∣∣3x2 − 2x+ 1− 9∣∣ =∣∣3x2 − 2x− 8∣∣ = ∣∣x+ 43 ∣∣ |x− 2| < ∣∣x+ 43 ∣∣2 < δ2 = �.
2. Define f : [−4, 0] → R by f(x) = 2x2−18x+3 for x 6= −3 and f(−3) = 12.
Show that f is continuous at −3.
Let � > 0 be given. Choose δ = �2 . Then,
∣∣∣ 2x2−18x+3 + 3∣∣∣ = ∣∣∣ 2(x2−9)x+3 + 3∣∣∣ =
|2(x− 3) + 3| = |2x− 3| < 2 |x− 3| < 2δ = �.
3. Use Theorem 3.1 to prove that
{
n
√
en+1
}∞
n=1
is convergent and find the
limit. You may assume that the function f(x) = ex is continuous on R.
First, our sequence is equivalent to
{
e
n+1
n
}
=
{
e · e 1n
}
. By Theorem 3.1,
e
1
n → e0 = 1, so the limit of the sequence is e.
4. If x0 ∈ E, x0 is not an accumulation point of E, and f : E → R,
prove that, for every sequence {xn}∞n=1 converging to x0 with xn ∈ E for all n,
{f(xn)}∞n=1 converges to f(x0).
Suppose that f(xn) 9 f(x0). Then, for any N , there is an �N such that
|f(xn)− f(x0)| ≥ �N for some n > N . Thus, there always exists an xn within
30
CHAPTER 3. CONTINUITY 31
δN distance of x0 such that |xn − x0| ≥ δN . Thus, xn 9 x0.
5. Define f : (0, 1) → R by f(x) = 1√
x
−
√
x+1
x . Can one define f(0) to
make f continuous at 0?
We see that f(x) = 1−x−1x = −xx
x 6=0
= 1. Thus, extending f so that f(0) = 1
will allow the function value to be equal to the limit, ensuring continuity.
6. Prove that f(x) =
√
x is continuous for all x ≥ 0.
Let xn → x0. By Exercise 28 of Chapter 1, √xn → √x0. Thus, f(x) is
continuous for all x ≥ 0.
7. Suppose f : R→ R is continuous and f(r) = r2 for each rational number
r. Determine f(
√
2) and justify your conclusion.
Since f is continuous, any sequence of numbers converging to
√
2 converges
to f(
√
2). By Exercise 45 from Chapter 1,
√
2 has a sequence of rationals that
converges to it. Thus, f(
√
2) = 2.
8. Suppose f : (a, b) → R is continuous and f(r) = 0 for each rational
number r ∈ (a, b). Prove that f(x) = 0 for all x ∈ (a, b).
By the same argument as above, since any irrational number has a sequence
of rationals converging to it, f(j) = 0 for all irrational numbers in (a, b).
9. Define f : (0, 1) → R by f(x) = x sin 1x . Can one define f(0) to make f
continuous at 0? Explain.
Yes. Define f(0) = 0. Let � > 0 be given and choose δ = �. We then see
that
∣∣x sin 1x ∣∣ = |x| ∣∣sin 1x ∣∣ < |x| < δ = � for any x ∈ (−δ, δ).
*10. Suppose f : E → R is continuous at x0 and x0 ∈ F ⊂ E. Define
g : F → R by g(x) = f(x) for all x ∈ F . Prove that g(x) is continuous at x0.
Show by example that the continuity of g at x0 need not imply the continuity of
f at x0.
Let � > 0 be given. Then, since f is continuous at x0, there is a δ > 0
such that for all x ∈ (x0 − δ, x0 + δ) we have |f(x)− f(x0)| < �. Then, choose
δ′ = |sup[(x0 − δ, x0 + δ) ∩ F ]− x0|. Then, (x0 − δ′, x0 + δ′) ⊆ (x0 − δ, x0 + δ),
so we must have |g(x)− g(x0)| < �.
If g(x) is continuous at x0 then f(x) is not necessarily continuous at x0.
Suppose for instance we have the function f(x) =
{
x if x ∈ R \ {2}
1 if x = 2
. Let
CHAPTER 3. CONTINUITY 32
E = R and F = {2}. Then, g(x) is trivially continuous at 2, but f(x) is not
continuous at 2 since its left- and right-hand limits are not equal.
11. Define f : R→ R by f(x) = 8x if x is rational and f(x) = 2x2 +8 if x is
irrational. Prove from the definition that f is continuous at 2 and discontinuous
at 1.
Let � > 0 be given. Choose δ1 =
�
8 . Then, |8x− 16| = 8 |x− 2| ≤ 8δ1 =
�. Choose δ2 = min(2,
�
8 ). Then,
∣∣2x2 + 8− 16∣∣ = ∣∣2x2 − 8∣∣ = 2 ∣∣x2 − 4∣∣ =
2 |x+ 2| |x− 2| < 8 |x− 2| < 8δ = �. Thus, choosing δ = min(δ1, δ2) gives
|f(x)− 16| < �.
Let �0 = 2. Let δ > 0 be given. Then, there is always an irrational number
x ∈ (1 − δ, 1 + δ) so that |f(x)− 8| = ∣∣2x2 + 8− 8∣∣ = ∣∣2x2∣∣ = 2 |x|2. Since
1 ∈ (x− δ, x+ δ) for all δ > 0, we see that there is always an x ∈ (1− δ, 1 + δ)
such that |f(x)− 8| = 2 |x|2 = 2 = �0.
3.2 Algebra of Continuous Functions
*12. Let p and q be polynomials and x0 be a zero of q of multiplicity m.
Prove that p/q can be assigned a value at x0 such that the function thusdefined
will be continuous iff x0 is a zero of p of multiplicity greater than or equal to m.
(⇐) If x0 is a zero of p of multiplicity n = m+ `, then set p˜(x) = p(x)(x−x0)m+`
and q˜(x) = q(x)(x−x0)m . We then have
p(x)
q(x) =
(x−x0)m+`p˜(x)
(x−x0)mq˜(x) =
(x−x0)`p˜(x)
q˜(x) . Define
p(x0)/q(x0) = 0. Let � > 0 be given. Let M = max
x∈(x0−1,x0+1)
p˜(x)
q˜(x) . Choose
δ = �M . Then, |f(x)| < |M(x− x0)| < δM = �. Thus, f(x) is continuous at x0.
(⇒, by way of contrapositive) Suppose that x0 is a zero of p of multiplicity
n = m − `. Then, as above, f(x) = p˜(x)
(x−x0)`q˜(x) . Let �0 = 1. Let δ > 0 be
given and m = min
x∈(x0−δ,x0+δ)
p˜(x)
q˜(x) . Then suppose f(x0) = y. We then have
|f(x)− y| =
∣∣∣ p˜(x)(x−x0)q˜(x) − y∣∣∣ > ∣∣∣ m(x−x0) − y∣∣∣. Since 1x−x0 gets arbitrarily close
to zero, there will always be an x ∈ (x− δ, x+ δ) such that 1x−x0 >
y+1
m . Thus,
|f(x)− y| >
∣∣∣m(y+1)m − y∣∣∣ = 1 = �0. Thus, f(x0) cannot be defined so as to be
continuous.
13. Let f : D → R be continuous at x0 ∈ D. Prove that there is M > 0 and
a neighborhood Q of x0 such that |f(x)| ≤M for all x ∈ Q ∩D.
Since f is continuous at x0, we have that for every � > 0, there is a δ� > 0,
such that if x ∈ (x0− δ�, x0 + δ�), we have −� < f(x)−f(x0) < �⇔ f(x0)− � <
CHAPTER 3. CONTINUITY 33
f(x) < f(x0) + �. Thus, let M = max
�>0
{f(x0) ± �} and Q = (x0 − δ�, x0 + δ�),
where � is the value chosen in the definition of M . Then, by construction, we
have that |f(x)| ≤M for every x ∈ Q ∩D.
14. If f : D → R is continuous at x0 ∈ D, prove that the function |f | : D →
R such that |f | (x) = |f(x)| is continuous at x0.
Let � > 0 be given. Since f(x) is continuous at x0, there exists a δ > 0
such that |f(x)− f(x0)| < � for any x ∈ (x0 − δ, x0 + δ). Now we see that
||f(x)| − f(x0)| < |f(x)− f(x0)| < � for any x ∈ (x0 − δ, x0 + δ).
15. Suppose f, g : D → R are both continuous on D. Define h : D → R by
h(x) = max{f(x), g(x)}. Show that h is continuous on D.
Let x0 ∈ D and without loss of generality, suppose h(x0) = f(x0). Suppose
that there is an �0 > 0 such that for every δ > 0, we have an x ∈ (x0− δ, x0 + δ)
such that |h(x)− f(x0)| ≥ �0. Since f(x) is continuous at x0, there is a δ0
such that |f(x)− f(x0)| < �0 for all x ∈ (x0 − δ0, x0 + δ0). Take δn = 1k+n
where 1k < δ0. Construct {xn} by selecting xn ∈ (x0 − δn, x0 + δn) such that|h(xn)− f(x0)| ≥ �0. By the construction of the sequence, we have h(xn) =
g(xn) for all n. Then, xn → x0, and since h(xn) = g(xn) and g(x) is continuous,
we have h(xn)→ h(x0) = g(x0) 6= f(x0). Contradiction.
16. Assume the continuity of f(x) = ex and g(x) = ln(x). Define h(x) = xx
by xx = xx ln x. Show that h is continuous for x > 0.
I’m pretty sure that this question contains a typo as xx 6= xx ln x (just plug
in x = 2 to see that there is no equality). I believe that the author intended
to write xx = ex ln x. In this case, we know that the function k(x) = x is
continuous, so h(x) = f(x) ◦ (k(x) · g(x)), and so is continuous.
17. Suppose that f : D → R with f(x) ≥ 0 for all x ∈ D. Show that, if f is
continuous at x0, then
√
f is continuous at x0.
Let � > 0 be given. Then, there is a δ > 0 such that for any x ∈ (x0−δ, x0+δ),
we have |f(x)− f(x0)| < �. Now,
∣∣∣√f(x)−√f(x0)∣∣∣ < ∣∣∣√f(x)−√f(x0)∣∣∣ ∣∣∣√f(x) +√f(x0)∣∣∣ =
|f(x)− f(x0)| < � for all x ∈ (x0 − δ, x0 + δ).
18. Define f : R→ R as follows:
f(x) = x− [x] if [x] is even.
f(x) = x− [x+ 1] if [x] is odd.
Determine where f is continuous. Justify.
CHAPTER 3. CONTINUITY 34
We have already seen in Exercise 2.17 that this function has limits every-
where, and the function values at those endpoints in question are equal to the
limit. Thus, the function is continuous everywhere.
3.3 Uniform Continuity: Open, Closed, and Com-
pact Sets
19. Let f, g : D → R be uniformly continuous. Prove that f + g : D → R is
uniformly continuous. What can be said about fg? Justify.
Let � > 0 be given. Then, there is some δ > 0 such that |f(x)− f(y)| <
�
2 and |g(x)− g(y)| < �2 if |x− y| < δ. Now, |f(x) + g(x)− f(y)− g(y)| ≤|f(x)− f(y)|+ |g(x)− g(y)| < �2 + �2 = �.
Now, fg is not necessarily uniformly continuous. For example, let f(x) =
g(x) = x. Let �0 = 1. Assume without loss of generality that y > x. Then, for
any δ > 0,
∣∣x2 − y2∣∣ = |x+ y| |x− y| > |x+ y| |y + δ − y + δ| = |x+ y| · 2δ >
2y · 2δ. If y = 14δ , we get
∣∣x2 − y2∣∣ > 2 · 14δ · 2δ = 1 = �0. Thus, f(x)g(x) is not
uniformly continuous.
20. Let f : A → B and g : B → C be uniformly continuous. What can be
said about g ◦ f : A→ C? Justify.
The function g ◦ f is not necessarily uniformly continuous if A 6= A¯. For
instance g : (1, 2)→ (1, 12) defined by g(x) = 1x is uniformly continuous on (1, 2).
Also, f(x) = x−1 is uniformly continuous everywhere. However, g(f(x)) = 1x−1
is not uniformly continuous on (1, 2) (for the same reason 1x isn’t uniformly
continuous on (0, 1)).
21. Define f : [3.4, 5] → R by f(x) = 2x−3 . Show that f is uniformly
continuous on [3.4, 5] without using Theorem 3.8– that is, use the methods of
Example 3.2.
Let � > 0 be given. Choose δ = 0.162 �. Then,
∣∣∣ 2x−3 − 2y−3 ∣∣∣ = 2 ∣∣∣ y−3−x+3(x−3)(y−3) ∣∣∣ =
2
∣∣∣ y−x(x−3)(y−3) ∣∣∣ ≤ 20.16 |x− y| ≤ 20.16δ = �.
22. Define f : (2, 7) → R by f(x) = x3 − x + 1. Show that f is uniformly
continuous on (2, 7) without using Theorem 3.8– that is, use the methods of
Example 3.3.
Let � > 0 be given. Choose δ = �148 . Then,
∣∣x3 − x+ 1− y3 + y − 1∣∣ =∣∣x3 − y3 + y − x∣∣ = ∣∣(x− y)(x2 + xy + y2) + x− y∣∣ = ∣∣(x− y)(x2 + xy + y2 + 1)∣∣ <
148 |x− y| < 148δ = �.
CHAPTER 3. CONTINUITY 35
23. A function f : R → R is periodic iff there is a real number h 6= 0 such
that f(x + h) = f(x) for all x ∈ R. Prove that if f : R → R is periodic and
continuous, then f is uniformly continuous.
Define fn : R→ R by fn(x) = f(x− nh) if x ∈ [(n− 1)h, nh] and fn(x) = 0
otherwise. Then, fn is continuous on [(n−1)h, nh] which is compact, making it
uniformly continuous. Then, f(x) =
∑
fn(x) is a sum of uniformly continuous
functions which is uniformly continuous by Exercise 19.
24. Suppose A is bounded and not compact. Prove that there is a function
that is continuous on A but not uniformly continuous. Give an example of a
set that is not compact, but every function continuous on that set is uniformly
continuous.
If A is bounded and not compact, then it is not closed (Heine-Borel The-
orem). Thus, suppose that x0 ∈ A¯ \ A. Then, the function f(x) = 1x−x0 will
not be uniformly continuous. (Since it is an accumulation point, inputs can get
arbitrarily close to x0 making
1
x−x0 arbitrarily large. Then, the function fails
to be uniformly continuous for the same reason that 1x fails to be uniformly
continuous.) It is nonetheless continuous (we’ve seen why before).
A set that is not compact would be N. Every function that is continuous on
N (which is to say any function defined on N) is also uniformly continuous on
N. (Since δ < 1 implies that x = y and |f(x)− f(x)| = 0 < �.)
25. Give an example of sets A and B and a continuous function f : A∪B →
R such that f is uniformly continuous on A and uniformly continuous on B,
but not uniformly continuous on A ∪B.
Let A = (−∞, 0] and B = (0,∞). Define f(x) = x if x ∈ A and f(x) = x+1
if x ∈ B. Then, f is not continuous at 0, let alone uniformly continuous.
Nevertheless, f is uniformly continuous on both A and B.
*26. Let E ⊂ R. Prove that E is closed if, for every x0 such that there is
a sequence {xn}∞n=1 of points of E converging to x0, it is true that x0 ∈ E. In
other words, prove E is closed if it contains all limits of sequences of members
of E.
Suppose that for every xn → x0, we have x0 ∈ E. Then, by the properties
of convergence and the fact that xn ∈ E for all xn, for any � > 0, we have a
xk ∈ {xn} ⊆ E such that xk ∈ (x0−�, x0 +�)∩E. Thus, x0 is a point of closure.
Thus, E is closed.*27. Prove that every set of the form {x : a < x < b} is open and every set
of the form {x : a ≤ x ≤ b} is closed.
CHAPTER 3. CONTINUITY 36
Let x0 ∈ {x : a < x < b} = (a, b). Then, choose �0 = min(x0 − a, b − x0).
Then by construction, we have (x0 − �0, x0 + �0) ⊂ (a, b). Thus, (a, b) is open.
Let x0 ∈ {x : a ≤ x ≤ b} = [a, b]. Let � > 0 be given. Then, x0 ∈
(x0 − �, x0 + �) ∩ [a, b], so x0 is a point of closure. Thus, [a, b] is closed.
28. Let D ⊂ R, and let D′ be the set of accumulation points of D. Prove
that D¯ = D ∪ D′ is closed and if F is any closed set that contains D, then
D¯ ⊂ F . D¯ is called the closure of D.
Since D¯ contains all the accumulation points of D (since D¯ = D ∪ D′),
D¯ is closed by Exercise 27. Now, since F is closed, F also contains all of its
accumulation points. Since D ⊂ F , we know that F must contain all of the
accumulation points of D. Thus, D¯ ⊂ F .
29. If D ⊂ R, prove that D¯ is bounded.
Suppose that D¯ is unbounded. Let M = supD and x0 ∈ D¯ \ D with
|x0 −M | > 100. Then, choosing �0 = 50, we see that (x0− �0, x0 + �0)∩D = ∅.
This contradicts that D¯ is closed. Thus, D¯ is bounded.
30. Suppose f : R→ R is continuous and let r0 ∈ R. Prove that {x : f(x) 6=
r0} is an open set.
Define the preimage of y under f to be f←(y) = {x ∈ R : f(x) = y}. Let
x0 ∈ {x : f(x) 6= r0}. Choose x¯ ∈ f←(r0) such that |x¯− x0| is minimal. Choose
�0 = inf{|x− y| : x, y ∈ f←(r0)}. Then, by the continuity of f , there is a δ0 > 0
such that |f(x)− r0| < �0. Let δ = min(|x− x¯| , δ0). Then, by construction,
(x0 − δ, x0 + δ) ⊆ {x : f(x) 6= r0}.
31. Suppose f : [a, b] → R and g : [a, b] → R are both continuous. Let
T = {x : f(x) = g(x)}. Prove that T is closed.
Let x0 ∈ T and let � > 0 be given. Then, x0 ∈ (x0 − �, x0 + �). Thus, T is
closed. (This problem is quite trivial.)
32. If D ⊂ R, then x ∈ D is said to be an interior point of D iff there is a
neighborhood Q of x such that Q ⊂ D. Define D◦ to be the set of interior points
of D. Prove that D◦ is open and that if S is any open set contained in D, then
S ⊂ D◦ ⊂ D. D◦ is called the interior of D.
Since for any point x0 ∈ D◦, there is a � > 0 such that (x0− �, x0 + �) ⊂ D◦,
we know that D◦ is open. Let S ⊂ D be open. Then, for any x0 ∈ S, there must
be � > 0 such that (x0− �, x0 + �) ⊂ S. Since S ⊂ D, we have (x0− �, x0 + �) ⊂
S ⊂ D◦. Thus, S ⊂ D◦ ⊂ D.
CHAPTER 3. CONTINUITY 37
33. Find an open cover of {x : x > 0} with no finite subcover.
Let Gn = (0, n]. Then,
⋃∞
n=1Gn is a cover of (0,∞) that has no finite
subcover.
34. Find an open cover of (1, 2) with no finite subcover.
Let Gn =
[
1− 1n , 2− 1n
]
. Then,
⋃∞
n=1Gn = (1, 2) that has no finite sub-
cover.
*35. Let E be compact and nonempty. Prove that E is bounded and that
supE and inf E both belong to E.
Since E is compact, it is closed and bounded. Now, supE and inf E are
both accumulation points of E, so since E must be bounded, inf E, supE ∈ E.
36. If E1, ..., En are compact sets, prove that E = ∪ni=1Ei is compact.
Since each Ei is bounded, we clearly have that ∪Ei is bounded. Next, if
x0 ∈ ∪Ei, then x0 ∈ Ei for some i. Since each Ei is closed, x0 is a point of
closure. Thus, ∪Ei is closed.
37. Let f : [a, b]→ R have a limit at each x ∈ [a, b]. Prove that f is bounded.
Suppose that for any n, there is an xn ∈ [a, b] such that f(xn) > n. Then,
f(xn) → ∞, but xn 9 ∞ since [a, b] contains all of its accumulation points.
This contradicts the fact that f has a limit for every x ∈ [a, b]. Thus, f is
bounded.
38. Suppose f : D → R is continuous with D compact. Prove that {x : 0 ≤
f(x) ≤ 1} is compact.
Since D is compact, it is closed and bounded. Thus, by Exercise 37, {x : 0 ≤
f(x) ≤ 1} is bounded. Let x0 ∈ {x : 0 ≤ f(x) ≤ 1}. Now, since f is continuous,
for �0 = min (f(x0), 1− f(x0)), there is a δ for which x ∈ (x0 − δ, x0 + δ) ∩D
implies that |f(x)− f(x0)| < �. Since this is true of all x in this interval, define
δn = δ/n. We can then construct {xn} such that xn ∈ (x0 − δn, x0 + δn) for all
n. Thus, the sequence xn → x0 (since δn → 0) is a sequence of members of D.
Thus, x0 is a point of closure. Thus, {x : 0 ≤ f(x) ≤ 1} is closed.
39. Suppose f : R→ R is continuous and has the property that for each
� > 0, there is M > 0 such that if |x| > M , then |f(x)| < �. Show that f is
uniformly continuous.
Let � > 0 be given. Then, since |f(x)| < � for all x ∈ (−∞,−M) ∪ (M,∞),
|f(x)− f(y)| < |f(x)| < � for all x, y ∈ (−∞,−M) ∪ (M,∞). Thus, f is
CHAPTER 3. CONTINUITY 38
uniformly continuous on (−∞,−M) ∪ (M,∞). Then, by Theorem 3.8, f is
uniformly continuous on [−M,M ] (since it is compact). Thus, f is uniformly
continuous.
40. Give an example of a function f : R→ R that is continuous and bounded
but not uniformly continuous.
Let f(x) = sin(x2). Then, f is clearly bounded and continuous. However,
the further away from the origin we go, we see that the maxima become closer
and closer to each other. Therefore, for any delta we choose, we can find x
values far enough away from the origin so that
∣∣sin(x2)− sin(y2)∣∣ = 1.
3.4 Properties of Continuous Functions
41. Find an interval of length 1 that contains a root of xex = 1.
We see that 0 · e0 − 1 = −1 and 1 · e1 − 1 = e − 1 > 0. Thus, by Bolzano’s
Theorem, [0,1] contains a root of xex = 1.
42. Find an interval of length 1 that contains a root of the equation x3 −
6x2 + 2.826 = 0.
We see that 03 − 6 · 02 + 2.826 = 2.826 while 13 − 6 · 12 + 2.826 < 0. Thus,
by Bolzano’s Theorem, [0, 1] contains a root of x3 − 6x+ 2.826 = 0.
43. Suppose f : [a, b] → R is continuous and f(b) ≤ y ≤ f(a). Prove that
there is c ∈ [a, b] such that f(c) = y.
If y ∈ (f(b), f(a)), then by the Intermediate Value Theorem, there is c ∈
(a, b) such that f(c) = y. Otherwise, f(a) = y or f(b) = y. (This is a trivial
problem.)
44. Suppose f : [a, b]→ [a, b] is continuous. Prove that there is at least one
fixed point in [a, b]– that is x such that f(x) = x.
Define F (x) = f(x) − x. We see that F (x) is also continuous. Now,
sup f(x) ≤ b, so F (x) ≤ 0 for some x = c1. Also, inf f(x) ≥ a, so F (x) ≥ 0 for
some c2. Then, by Bolzano’s Theorem, there is c ∈ [c1, c2] such that F (c) = 0.
Thus, f(x) has a fixed point.
45. If f : [a, b]→ R is 1-1 and has the intermediate value property– that is,
if y is between f(u) and f(v), there is x between u and v such that f(x) = y–
show that f is continuous. (Hint: First show that f is monotone.)
CHAPTER 3. CONTINUITY 39
First, f is monotone. To see this, suppose not. Then, there is some point
x0 where f(x) < f(x0) on one side and f(x) > f(x0) on the other for values
x ∈ (x0 − δ, x0 + δ) for some δ > 0. Without loss of generality, let us assume
that it is increasing to the left of x0 and increasing to the right of x0. By
the intermediate value property, f attains all values in [f(x0 − δ), f(x0)] and
[f(x0), f(x0 + δ)]. Let � = min (|f(x0)− f(x0 − δ)| , |f(x0)− f(x0 + δ)|). By
the intermediate value property, every function value in (f(x0)− �, f(x0)) must
be attained in both intervals, showing that at least one (many more than one,
actually) function value must have two elements in its preimage. Contradiction.
Without loss of generality, suppose that f is increasing. Let x0 ∈ [a, b] and
let � > 0 be given. Since � > 0 is only of interest when it is small, suppose
(f(x0)− �, f(x0) + �) ⊂ (f(a), f(b)). Then, by the intermediate value property,
this implies that there are x1, x2 ∈ [a, b] such that f(x1) = f(x0)−� and f(x2) =
f(x0)+�. Take δ = min (|x0 − x1| , |x0 − x2|). Then, for every x ∈ (x0−δ, x0+δ)
we have |f(x0)− f(x)| < �. Thus, f is continuous at x0.
46. Prove that there is no continuous function f : R→ R such that, for
each c ∈ R, the equation f(x) = c has exactly two solutions.
Since f is continuous, it has the intermediate value property. From this, we
can narrow down the form that f can have. We see thatf must be increasing,
then decreasing or vice versa. If f changes its increasing/decreasing status any
more, some function value will have 3 elements in its preimage. Also, if f is
monotone, then f is 1-1 (and so won’t fit the criteria outlined in the exercise).
Without loss of generality, assume f is increasing on some interval (−∞, x0)
then decreasing on (x0,∞). (Note that f cannot be constant at any interval
since this would result in f(x) = const. for infinitely many x.)
We examine f(x0). Now, we know that to the left of x0, f(x) < f(x0) and
to the right, f(x0) > f(x) (again, we do not have equality because of the above
discussion). Thus, f(x0) can have only one element in its preimage.
Miscellaneous
47. Let f : R→ R be additive. (See Project 2.1 at the end of Chapter 2.)
That is, f(x+ y) = f(x) + f(y) for all x, y ∈ R. In addition, assume there are
M > 0 and a > 0 such that if x ∈ [−a, a], then |f(x)| ≤ M . Prove that f is
uniformly continuous. In particular, prove that there is a real number m such
that f(x) = mx for all x ∈ R.
From Project 2.1, we know that an additive function that fits the above
criteria has a limit at each point and lim
x→x0
f(x) = f(x0). Now, notice that
|f(x)− f(y)| = |f(x− y)|. So, we actually only need to show that for any
CHAPTER 3. CONTINUITY 40
� > 0, there is a δ > 0 such that |f(x)| < � for any x ∈ (−δ, δ). Now, since f is
additive, we have f(x) = f(0 +x) = f(0) + f(x). Thus, f(x) = 0. Finally, since
f is continuous at each point, it is continuous at x = 0. Thus, there is δ > 0
such that |f(x)− 0| = |f(x)| < �. Thus, f is uniformly continuous.
By the axioms for the real numbers, if f(x) = y there is an m such that y =
mx. Now, suppose for x1 and x2, there are m1 and m2 for which f(x1) = m1x1
and f(x2) = m2x2. Then, since f is additive, we must have f(x1+x2) = f(x1)+
f(x2) = m1x1 +m2x2. By the distributive law, this implies that m1 = m2 = m
so that f(x1 + x2) = m(x+ y) = mx1 +mx2.
48. Let f : [a, b]→ R be continuous, and define g : [a, b]→ R by
g(t) = sup{f(x) : a ≤ x ≤ t}
Prove that g is continuous.
We see that g is either identical to f or constant. Now, let � > 0 and x0 ∈
[a, b] be given. By the continuity of f , there is a δ > 0 for which |f(x)− f(x0)| <
�. Now, if g ≡ f on (x0−δ, x0+δ), then g is clearly continuous. If g(x) 6= f(x) for
some x ∈ (x0− δ, x0 + δ), then g(x) = const. and we can find x¯ = inf{x ∈ (x0−
δ, x0 + δ) : g(x) 6= f(x)}. Choose δ′ = min(δ, |x¯− x0|). Then, by construction,
|g(x0)− g(x)| < � for all x ∈ (x0 − δ′, x0 + δ′).
49. Suppose that g : D → R is continuous at x0 and that x0 is also an
accumulation point of D. Define D0 = {x : g(x) = 0}. If g(x) 6= 0, prove that
x0 is an accumulation point of D0.
Choose x¯ to be such that g(x¯) = 0 and |x¯− x0| is minimum. Then, by
the Intermediate Value Theorem, g attains all values between g(x) and g(x0)
for any |x− x0| < |x¯− x0| (of which there are infinitely many). Thus, any
neighborhood of x0 contains an infinity of points of D0. Thus, x0 ∈ D′0.
50. Suppose f : D → R , g : E → R, and x0 ∈ D ∩E. Suppose further that
there is � > 0 such that D∩ [x0− �, x0 + �] = E∩ [x0− �, x0 + �] and f(x) = g(x)
for all x ∈ D ∩ E ∩ [x0 − �, x0 + �]. Prove that f is continuous at x0 iff g is
continuous at x0.
If f is continuous at x0, then for all �
′ > 0, there is a δ > 0 such that for
all x ∈ (x0 − δ, x0 + δ), we have |f(x)− f(x0)| < �′. If δ < �, then we have
g ≡ f on (x0 − δ, x0 + δ) since D ∩ [x0 − δ, x0 + δ] = E ∩ [x0 − δ, x0 + δ]. If
δ ≥ �, then for all x ∈ (x0 − �, x0 + �), we also have |f(x)− f(x0)| < �′. (Since
(x0 − �, x0 + �) ⊆ (x0 − δ, x0 + δ).) Thus, f(x) and f(x0) can be replaced by
g(x) and g(x0) respectively. The argument is similar if we assume that g(x) is
continuous at x0.
Chapter 4
Differentiation
4.1 The Derivative of a Function
1. Let (x0, y0) be an arbitrary point on the graph of the function f(x) = x
2.
For x0 6= 0, find the equation of the line tangent to f at that point by finding a
line that intersects the curve in exactly one point. Do not use the derivative to
find this line.
We need x20 = mx0 +b⇔ x20−mx0−b = 0. Since we want only one solution,
we must have the discriminant (−m)2 − 4(−b) = m2 + 4b = 0. Thus, b = −m24
and the equation is x20 −mx0 + m
2
4 = 0. This implies that m = 2x0. Thus, the
equation of the tangent line is y − x20 = 2x0(x− x0).
2. Prove that the definition of the derivative and the alternate definition of
the derivative are equivalent.
Set x + h = x0. Then, since as x → x0, h → 0, we have lim
x→x0
f(x)−f(x0)
x−x0 =
lim
h→0
f(x+h)−f(x)
h .
3. Use the definition to find the derivative of f(x) =
√
x, for x > 0. Is f
differentiable at zero?
lim
h→0
√
x+h−√x
h
(√
x+h+
√
x√
x+h+
√
x
)
= lim
h→0
x+h−x
h(
√
x+h+
√
x)
= lim
h→0
1√
x+h+
√
x
= 1
2
√
x
. The
limit does not exist at x = 0, so f is not differentiable there.
4. Use the definition to find the derivative of g(x) = x2.
41
CHAPTER 4. DIFFERENTIATION 42
lim
h→0
(x+h)2−x2
h = limh→0
x2+h2+2xh−x2
h = limh→0
h(h+2x)
h = limh→0
(h+ 2x) = 2x.
5. Define f(x) = x3 sin 1x for x 6= 0 and h(0) = 0. Show that h is dif-
ferentiable everywhere and that h′ is continuous everywhere but fails to have a
derivative at one point. You may use the rules for differentiating products, sums
and quotients of elementary functions that you learned in calculus.
We see that f ′(x) = 3x2 sin 1x + cos
1
x ·
(− 1x2 ) · x3(?). Now, limh→0x3 sin(x+h)h =
lim
h→0
x3 sin x sinh−cos x cosh
h = limh→0
x3 sinx · sinhh − cosx · coshh = 0 at x = 0. Thus,
f ′(0) = 0 and is defined everywhere. Now, from (?), f ′is continuous at every-
where except possibly at x = 0. Using elementary calculus, we see that the
limit of f ′ as x→ 0 is −∞. Thus, f ′ is not continuous at x = 0.
6. Suppose f : (a, b)→ R is differentiable at x ∈ (a, b). Prove that
lim
h→0
f(x+ h)− f(x− h)
2h
exists and equals f ′(x). Give an example where this limit exists, but the function
is not differentiable.
If lim
h→0
f(x+h)−f(x)
h exists, then by the same token, limh→0
f(x)−f(x−h)
h exists.
Thus, lim
h→0
[
f(x+h)−f(x)
h +
f(x)−f(x−h)
h
]
= lim
h→0
f(x+h)+f(x−h)
h exists by the limit
laws, so lim
h→0
f(x+h)−f(x−h)
2h must also exist. (It’s just the last limit multiplied
by 1/2.) Then, since lim
h→0
f(x+h)−f(x)
h = limh→0
f(x)−f(x−h)
h = f
′(x), we see that
lim
h→0
f(x+h)+f(x−h)
2h =
2f ′(x)
2 = f
′(x).
Let f(x) = |x|. Then, lim
h→0
f(x+h)+f(x−h)
2h =
2h
2h = 1 near x = 0. Thus, the
limit exists, but from what we know of calculus, f is not differentiable at x = 0.
7. A function f : (a, b) → R satisfies a Lipschitz condition at x ∈ (a, b)
iff there is M > 0 and � > 0 such that |x− y| < � and y ∈ (a, b) imply that
|f(x)− f(y)| < M |x− y|. Give an example of a function that fails to satisfy
a Lipschitz condition at a point of continuity. If f is differentiable at x, prove
that f satisfies a Lipschitz condition at x.
We first make the observation that |f(x)− f(y)| ≤ |x− y|M ⇔ |f(x)−f(y)||x−y| ≤
M (if x 6= y), so essentially the criterion is that the average rate of change be-
tween x and y is bounded. Define f(x) =
{ √
1− (x− 1)2 if x ∈ [0, 2)√
1− (x+ 1)2 if x ∈ (−2, 0] .
The following is the graph of the function.
CHAPTER 4. DIFFERENTIATION 43
We see that at x = 0, the graph has a vertical tangent line, so as x→ 0, the
average rate of change from 0 to x will go to ∞. Thus, f is not Lipschitz at
x = 0.
Now, if f is differentiable at x, we have lim
y→x
f(x)−f(y)
x−y = M . Thus, for
any � > 0, there is a δ > 0 such that for any y ∈ (x − δ, x + δ), we have∣∣∣ f(x)−f(y)x−y −M ∣∣∣ < �. Then, ∣∣∣ f(x)−f(y)x−y ∣∣∣−M ≤ ∣∣∣ f(x)−f(y)x−y −M ∣∣∣, so ∣∣∣ f(x)−f(y)x−y ∣∣∣ <
�+M and |f(x)−