Intro to Analysis Solutions

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 gN Natural numbers (symbolized as J in the book)

LHS Left-hand sideRHS Right-hand sideincr. Increasingdecr. Decreasingcts. Continuousdiff. Differentiable

intgbl. Integrablelg(x) log2(x)dxe If if a < x ≤ a+ 1 where a is a positive integer, then, dxe = a+ 1.

an → A limn→∞

an = A.

lim an limn→∞

an

IVT Intermediate Value TheoremH l’Hospital’s Rule

MVT Mean Value TheoremIFT Inverse Funcion TheoremMi 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) =

∑ni=0Mi(xi − xi−1).

L(P, f) Similar to U(P, f), but replace Mi with mi

µ(P ) supk≤n|xk − xk+1| for partition P0 and refinements {Pn}. Called the mesh of P .

S(P, f) S(P, f) =∑ni=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±ε= b b− ε < a < b+ ε

1

Contents

0 Prerequisites 40.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.2 Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . 60.3 Mathematical Induction and Recursion . . . . . . . . . . . . . . . 70.4 Equivalent and Countable Sets . . . . . . . . . . . . . . . . . . . 90.5 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1 Sequences 131.1 Sequences and Convergence . . . . . . . . . . . . . . . . . . . . . 131.2 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3 Arithmetic Operations on Sequences . . . . . . . . . . . . . . . . 171.4 Subsequences and Monotone Sequences . . . . . . . . . . . . . . 19Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2 Limits of Functions 232.1 Definition of the Limit of a Function . . . . . . . . . . . . . . . . 232.2 Limits of Functions and Sequences . . . . . . . . . . . . . . . . . 242.3 Algebra of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4 Limits of Monotone Functions . . . . . . . . . . . . . . . . . . . . 27Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3 Continuity 303.1 Continuity of a Function at a Point . . . . . . . . . . . . . . . . . 303.2 Algebra of Continuous Functions . . . . . . . . . . . . . . . . . . 323.3 Uniform Continuity: Open, Closed, and Compact Sets . . . . . . 343.4 Properties of Continuous Functions . . . . . . . . . . . . . . . . . 38Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4 Differentiation 414.1 The Derivative of a Function . . . . . . . . . . . . . . . . . . . . 414.2 The Algebra of Derivatives . . . . . . . . . . . . . . . . . . . . . 444.3 Rolle’s Theorem and the Mean Value Theorem . . . . . . . . . . 454.4 L’Hospital’s Rule and the Inverse-Function Theorem . . . . . . . 48Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2

CONTENTS 3

5 The Riemann Integral 525.1 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . 525.2 Classes of Integrable Functions . . . . . . . . . . . . . . . . . . . 535.3 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.4 The Fundamental Theorem of Integral Calculus . . . . . . . . . . 565.5 Algebra of Integrable Functions . . . . . . . . . . . . . . . . . . . 565.6 Derivatives of Integrals . . . . . . . . . . . . . . . . . . . . . . . . 585.7 Mean-Value and Change-of-Variable Theorems . . . . . . . . . . 59Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

6 Infinite Series 636.1 Convergence of Infinite Series . . . . . . . . . . . . . . . . . . . . 636.2 Absolute Convergence and the Comparison Test . . . . . . . . . 656.3 Ratio and Root Tests . . . . . . . . . . . . . . . . . . . . . . . . 666.4 Conditional Convergence . . . . . . . . . . . . . . . . . . . . . . . 696.5 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, thenx ∈ (A ∪B) ∩ (A ∪ C) since A ⊆ (A ∪B) ∩ (A ∪ C). If x /∈ A, then x ∈ B andx ∈ 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 orx ∈ 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 andx /∈ C. Thus, x ∈ (A \B) ∩ (A \ C).

(⊇) Let x ∈ (A \ B) ∩ (A \ C). Then, (x ∈ A and x /∈ B) and (x ∈ A andx /∈ 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 istrue, 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 = Bwhich 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 toSection 0.1.

Russell’s paradox points out a flaw in naive set theory. If we examine theset A of all sets, then A ∈ A. But then, there must be another set containingA. This is the paradox. Since the theory in this section are based on naive settheory (the treatment of “set” as an undefined term), it shows that the theoryis not complete.

10. Describe each of the following sets as the empty set, as R, or in intervalnotation, as appropriate:

(a)∞⋂n=1

(− 1n ,

1n

)


(b)∞⋃n=1

(−n, n)

(c)∞⋂n=1

(− 1n , 1 + 1

n

)(d)

∞⋃n=1

(− 1n , 2 + 1

n

)(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 andx /∈ 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 thefollowing sets:

(a) R \∞⋂n=1

(− 1n ,

1n

)(b)

∞⋃n=1

(R \[

1n , 2 + 1

n

])

(a) R \∞⋂n=1

(− 1n ,

1n

)=⋃{(

−∞,− 1n

]∪[

1n ,∞

)}= R.

(b)∞⋃n=1

(R \[

1n , 2 + 1

n

]) = R \

⋂[1n , 2 + 1

n

]= 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? Isf 1-1? Is f onto? If f has an inverse, find the domain of the inverse and givea 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. Sincef is 1-1 and im f = {odd natural numbers}, f−1 : {odd natural numbers} → Nexists. To find it, set y = f−1(n) and write n = 2y − 1⇔ f−1(n) = n+1

2 .

14. What is the domain of f(x) = xx+2? What is im f? Is f injective? If

so, find the inverse.


We see thatdom f = R \ {−2}. Since xx+2 = x+2−2

x+2 = x+2x+2 −

2x+2 = 1− 2

x+2 ,we see that im f = (1,∞)∪ (−∞, 1). We see that f is injective, so we find thatf−1(x) = 2

x−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 neednot be a function.)

16. Give an example of f : A → B that has an inverse, and show theinverse.

Let f(n) = n, if 2 ≤ n ≤ 5, and f(1) = 6. Then f−1 : B \{7} → A is definedby 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 gis 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 and7 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 alla ∈ 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 isalso 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 followsfrom 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.


For n = 1, 1 = 12. Suppose the statement is true for all n ≤ N . Examine1 + 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 statementis 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. SupposeN is even. Then, 3N has 6 as a factor and 6 | 3N(N +1). Now, suppose that Nis 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 thesecond 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 alln ≤ 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 = 182N+1 + 1

42N+1 + 122N+1 = 7

82N+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 forthis exercise.

For n = 2, f(2) =√

3 + 2 ≤ 2.4. Suppose the statement is true for alln ≤ 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.


For f(3) = 8(−8)− 15(2) + 6(8) = −5(33) + 52 + 26. Suppose the statementis 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) isfalse 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 thatn0 ≤ i ≤ N − 1 such that P (i) is false. Consequently, there exists ani(1) suchthat n0 ≤ i(1) ≤ i − 1 such that P (i(1)) is false. Eventually, this implies thatthere 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, forn ≥ 2, f(n) = 6f(n− 2)− f(n− 1). Prove that f(n) = 5 · 2n + 2(−3)n for alln ∈ Z, n ≥ 0.

For n = 2, f(2) = 6·7−4 = 38 = 5·22+2(−3)2. Suppose the statement is truefor 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 countablesubset is countable.)

Let X be a set with an uncountable subset K. Then, since K � N, therecannot be a 1-1, onto function from K to N, so there consequently can’t be onefrom X to N.

31. Find a 1-1 function f from N onto S where S is the set of all oddintegers.


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 thatis its preimage. If m ∈ S is positive, then there is an odd natural that is itspreimage.

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 integercoefficients 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 acountable 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, setAn = Asn . Then, ∪s∈SAs = ∪∞n=1An which is a countable union of countablesets 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 isa 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 iscountable.

36. An algebraic number is any number that is the root of a polynomialequation p(x) = 0 where the coefficients of p are integers. Show that the set ofalgebraic numbers is a countable set.

Certainly A ⊆ ∪p∈PnB(p). Any subset of a countable set is countable.


37. For a set A, let P (A) be the set of all subsets of A. Prove that A is notequivalent to P (A).

Let A be finite such that |A| = n. Then, |P (A)| = 2n. (For any subset, eachelement can either be in the subset or not, so the total number of ways to builda 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 Ak0 otherwise

.

Suppose that {Ak} is countable. Then we can enumerate the subsets via thesequences:

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 mapfrom A into P (A).

Finally, if A is uncountable, then, any countable subset of A cannot bemapped 1-1 with the set of subsets (by the above argument), so again there canbe 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. Provethat [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 (preservespercentage 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 <x2 + y

2 = x+y2 . Similarly, y = y

2 + y2 >

x2 + y

2 = x+y2 .

*40. If x ≥ 0 and y ≥ 0, prove that√xy ≤ x+y

2 .

Let x and y be two dimensions. Then, 2x + 2y gives the perimeter of therectangle 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 aperimeter of a proper rectangle of the same area. Thus, 4

√xy ≤ 2x + 2y ⇔

√xy ≤ x+y

2 .


*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, supposec2 = 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 <

ab , prove that x

y <x+ay+b <

ab .

First, x+ay+b = x

y+b + ay+b <

xy + a

b . Now, since xy <

ab , this implies x

y <x+ay+b <

ab .

43. Let A = {r : r is a rational number and r2 < 2}. Prove that A has nolargest member.

Let ab ∈ A. Suppose that 2 − a2

b2 = cd . Then,

(ab + e

f

)2

= a2

b2 + e2

f2 + 2aebf .

This is less than two precisely when e2

f2 + 2aebf = ef

(ef + 2ab

)< c

d . Thus,

1 + 2ab <cd ·

fe ⇔

cd + 2acbd <

fe .

*44. If x = supS, show that, for each ε > 0, there is a ∈ S such thatx− ε < a ≤ x.

Suppose that there is an ε0 > 0 such that there is no a between x− ε0 andx. Then, this implies that x− ε0 is an upper bound. Contradiction.

*45. If y = inf S, show that, for each ε > 0, there is a ∈ S such thaty ≤ a < y + ε.

Suppose that there is an ε0 > 0 such that there is no a between y + ε0 andy. Then, y + ε0 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 ε > 0 such that(

23 − ε,

23 + ε

)⊂ [0, 1].

Choose ε = 13 . Then

(23 −

13 ,

23 + 1

3

)=(

13 , 1)⊂ [0, 1].

*2. Let x and y be distinct real numbers. Prove there is a neighborhood Pof x and a neighborhood Q of y such that P ∩Q = ∅.

Choose ε = |x−y|2 . Then, let P = (x− ε, x+ ε) and Q = (y− ε, y+ ε). Then,

P ∩Q = ∅.

*3. Suppose x is a real number and ε > 0. Prove that (x − ε, x + ε) is aneighborhood of each of its members; in other words, if y ∈ (x− ε, x + ε), thenthere is δ > 0 such that (y − δ, y + δ) ⊂ (x− ε, x+ ε).

Let δ = min(|y−(x+ε)|

2 , |y−(x−ε)|2

). Then, (y − δ, y + δ) ⊂ (x− ε, x+ ε).

4. Find upper and lower bounds for the sequence{

3n+7n

}∞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, butnever converges.

13

CHAPTER 1. SEQUENCES 14

6. Use the definition of convergence to prove that each of the followingsequences converges:

(a){

5 + 1n

}∞n=1

(b){

2−2nn

}∞n=1

(c) {2−n}∞n=1

(d){

3n2n+1

}∞n=1

(a) Let ε > 0 be given. Let N =⌈

1ε

⌉. Then,

∣∣5 + 1n − 5

∣∣ =∣∣ 1n

∣∣ < ∣∣ 1N

∣∣ < ε.

(b) Let ε > 0 be given. Let N =⌈

2ε

⌉. Then,

∣∣ 2−2nn + 2

∣∣ =∣∣ 2−2n+2n

n

∣∣ =∣∣ 2n

∣∣ < ∣∣ 2N

∣∣ < ε.

(c) Let ε > 0 be given. Let N =⌈lg(

1ε

)⌉. Then, |2−n| =

∣∣ 12n

∣∣ < ∣∣ 12N

∣∣ < ε.

(d) Let ε > 0 be given. Let N =⌈

34ε

⌉Then,

∣∣∣ 3n2n+1 −

32

∣∣∣ =∣∣∣ 6n−6n−3

4n+2

∣∣∣ =∣∣∣ −34n+2

∣∣∣ < ∣∣ 34n

∣∣ < ∣∣ 34N

∣∣ < ε.

*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| < ε.

8. Suppose {an}∞n=1 converges to A, and define a new sequence {bn}∞n=1 bybn = an+an+1

2 for all n. Prove that {bn}∞n=1 converges to A.

Let ε > 0 be given. We see that∣∣∣an+an+1

2 −A∣∣∣ =

∣∣∣an+an+1−2A2

∣∣∣ =∣∣∣an−A+an+1−A

2

∣∣∣ ≤|an−A|

2 + |an+1−A|2

∃N<s.t.

ε2 + ε

2 = ε. Thus, bn → A.

*9. Suppose {an}∞n=1, {bn}∞n=1, and {cn}∞n=1 such that {an}∞n=1 convergesto 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 convergenceand definition of absolute value, −ε < an−1−A ≤ cn−A ≤ bn−A < ε. Hence,|cn −A| < ε. 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| < ε. 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 suchthat, for n ≥ N , an = α. Prove that {an}∞n=1 converges to α.

We see that |an − α| ≤ |aN − α| = 0 < ε for all ε > 0.


12. Give an alternate proof of Theorem 1.1 along the following lines. Chooseε > 0. There is N1 such that for n ≥ N1, |an −A| < ε

2 , and there is N2 suchthat for n ≥ N2, |an −B| < ε

2 . Use the triangle inequality to show that thisimplies that |A−B| < ε.

Let N = max(N1, N2). Then, ε > |an −A| + |an −B| = |an −A| +|B − an| > |an −A+B − an| = |B −A|. Thus, |A−B| < ε.

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 ε > 0 be given and set N = x2ε . Then,

∣∣an − x2

∣∣ ≤ ∣∣x+2x+···+nxn2 − x

2

∣∣ =∣∣∣x(1+···+n)n2 − x

2

∣∣∣ =∣∣∣xn(n+1)

2n2 − x2

∣∣∣ =∣∣∣xn2

2n2 + xn2n2 − x

2

∣∣∣ =∣∣ x

2n

∣∣ < ∣∣ x2N ∣∣ < ε.

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 existank 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 ε > 0, there exist N1 and N2

such that |an − am| < ε2 for all m,n > N1 and |bn − bm| < ε

2 for all m,n > N2.ChooseN = max(N1, N2). Then, |an + bn − (am + bm)| = |an − am + bn − bm| <|an − am|+ |bn − bm| < ε

2 + ε2 = ε 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 forsome M . Since {an} is Cauchy, then for all ε > 0, there exist N |an − am| < ε

Mfor all m,n > N and |bn − bm| < ε

2 for all m,n > N2. Let ε > 0 be given. Then,|anbn − ambm| < |anM − amM | = |M(an − am)| < ε.


17. Prove that the sequence{

2n+1n

}∞n=1

is Cauchy.

Let ε > 0 be given. ChooseN = 1ε . Then,

∣∣ 2n+1n − 2m+1

m

∣∣ =∣∣ 2mn+m−2mn−n

nm

∣∣ =∣∣m−nnm

∣∣ =∣∣ mmn −

nmn

∣∣ < ∣∣ NN2

∣∣ =∣∣ 1N

∣∣ < ε.

18. Give an example of a sequence with exactly two accumulation points.

Let an =

{1/n , if n is even1 + 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 accumulationpoints.

The setQ has the property that every element is an accumulation point, sincefor any a

b ∈ Q, the sequence{ab + 1

n

}converges to a

b . Since Q is countable, wehave 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 anaccumulation point of S.

It is clear that x ∈ S is a possibility. Suppose x /∈ S. Then, by Exercise0.44, for any ε > 0, there is an a ∈ S such that x− ε < a < x. Thus, for all n,there exists an an ∈ S such that x − 1

n < 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 eachpositive integer n ≥ 2. Show that {an}∞n=1 is a Cauchy sequence. You may wantto 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.


The statement an+1 − an =(− 1

2

)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−2aN

2 +(− 1

2

)N+1(a1 − a0) = aN+1−aN

2 +(− 1

2

)N+1(a1−a0) =

(− 1

2

)N(a1−a0)+

(− 1

2

)N+1(a1−a0) =

[(− 1

2

)N+(− 1

2

)N+1]

(a1−

a0) =[(− 1

2

)N+(− 1

2

)N (− 12

)](a1 − a0) =

[(− 1

2 + 1) (− 1

2

)N](a1 − a0) =(

− 12

)N+1(a1 − a0). Thus, the statement is proven by induction.

Now, let ε > 0 and n,m ∈ N be given. Choose N =⌈lg(|a1−a0|(n−m)

ε

)⌉.

Then,∣∣(− 1

2

)n(a1 − a0)

∣∣ < εn−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| <ε

n−m + εn−m + · · ·+ ε

n−m = ε.

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− 1

k , a+ 1k

). By convergence, there is anN such that |an −A| <

1k 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 arean 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, byTheorem 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} ={anbnan

}, so by Theorem 1.9, bn → C

A .

28. If {an}∞n=1 converges to a with an ≥ 0 for all n, show{√

an}∞n=1

converges to√a.


Let ε > 0 be given. Then, there is an N such that |an − a| < ε√a. Then,∣∣√an − a∣∣ =

∣∣∣ an−a√an+√a

∣∣∣ = |an − a|∣∣∣ 1√

an+√a

∣∣∣ < |an − a| ∣∣∣ 1√a

∣∣∣ < ε for all n ≥ N .

29. Prove that (n+ kk

)(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 ε > 0 be

given. ChooseN =⌈

1−kε

⌉. Then,

∣∣∣ (n+k−1)···(n+1)k!(n+k)k−1 − 1

k!

∣∣∣ < ∣∣∣ (n+k)k−2(n+1)k!(n+k)k−1 − 1

k!

∣∣∣ =∣∣∣ n+1k!(n+k) −

1k!

∣∣∣ =∣∣∣n+1−n−kk!(n+k)

∣∣∣ =∣∣∣ 1−kk!(n+k)

∣∣∣ < ∣∣ 1−kn ∣∣ < ε for all n ≥ N .

30. Prove the following variation on Lemma 1.10. If {bn}∞n=1 converges toB 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 anexample in which {αn}∞n=1 converges, but {an}∞n=1 does not.

Let ε > 0 be given. There is an N1 such that |an −A| < ε2 . for all n > N .

Let M = |a1 −A|+ |a2 −A|+ · · ·+ |aN1 −A|. Then, there is an N2 such that∣∣Mn

∣∣ < ε2 for all n > N2. Let N = max(N1, N2). Then,

∣∣a1+a2+···+ann −A

∣∣ =∣∣a1+···+aN+···+an−nAn

∣∣ < |a1−A|+···+|aN−A|n + |aN+1−A|+···+|an−A|

n = Mn + |aN+1−A|+···+|an−A|

n <ε2 + ε

2 = ε. (A quicker way to show this would be to observe that by the definitionof convergence, |an| < ε

2 for all n > N for some N . Then, sinceLet an = (−1)n. Then, as we have seen before, {an} diverges, but a1+a2+···+an

n →0.

32. Find the limit of the sequences with general term as given:

(a) n2+4nn2−5

(b) cosnn

(c) sinn2√n


(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− 1

n−2n =√

4n2 − n−√

4n2 = 4n2−n−4n2√

4n2−n+√

4n2∼

−n4n . Thus, limit is − 1

4 .(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−2

2 =(− 1

2n

)3 + 2. Since − 1

2n → 0, the limitmust 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 isa subsequence of {an}∞n=1 that converges to x.

Since x is an accumulation point, every neighborhood about x contains aninfinity of {an}. Thus, let ank be a member of {an : n ∈ N} ∩

(x− 1

k , x+ 1k

).

Then, for any ε > 0, there is a K such that 1k < ε 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 of values or {an} has an infinite number ofvalues. For the former, there must be some value x for which there are infinitelymany k such that ank = x. Thus, ank → x. For the latter, the sequence is a


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 ε > 0 be given. Then, there are a1−M

ε intervals that sequence valuesmay 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 thesequence is decreasing, the last region must contain an infinity of values; thatis, an ∈ (M,M + ε) for all n > N for some N . Since ε was arbitrarily chosen,the proof is complete. The case for when {an} has only finitely many values iseasy.

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. Definea 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 > 1is convergent and find its limit.

To find the limit L, set L =√

6 + L⇔ L2 = 6 + L⇔ L2 − L− 6 = 0. Thesolutions are −2 and 3. The only solution that works is 3. Thus, an → 3. Weprove 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, thesequence 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 subsequentiallimits of that sequence. By Exercise 36, E is nonempty. Prove that E is boundedand contains both supE and inf E.

Since {xn} is bounded (by M), its limit points must be such that they arewithin ε distance of some sequence values. Thus, limit points must be withinthe same bounds as {xn} or within ε distance of the boundary for any ε. Thus,E is bounded (by, say M + 1). We must ensure that members of E do not forma sequence themselves that converges to a non-limit point. So, suppose thereis a sequence {en} of limit points. Then, for every ε, there is an N such that


|en − xn| < ε for all n > N . Thus, xn → en. Thus, all sequences of E convergein 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. Provethat if {xn}∞n=1 converges to x, then {xT (n)}∞n=1 also converges to x. Explainhow 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 allsubsequences converge, we must have xT (n) → x. Let T : N → N be any 1-1function 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 ordiverge? If the sequence converges, find the limit.

The sequence does not converge in general. For instance, if a = 1 andb = −1, then the sequence becomes {[1n+(−1)n]1/n}. Taking even indexes, thelimit is 1, and taking odd indexes, the limit is 0. Thus, not all subsequencesconverge 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∑kn=1

1√k2+n

<∑∞n=1

1k = 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 rationalnumbers 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 bewithin 10k distance for any integer k and n > N).

*46. Show that if x is any real number, there is a sequence of irrationalnumbers converging to x.

If x is already irrational, define xn ≡ x. Clearly, xn → x. If x is rational,define xn = x+ π

n . Then, xn ∈ R \Q for all n and xn → x.


47. Suppose that {an}∞n=1 converges to A and that B is an accumulationpoint of {an : n ∈ N}. Prove that A = B.

If B is an accumulation point, then(B − 1

n , B + 1n

)contains a member of

{an} for all n. Thus, one can construct a subsequence of these members thatconverges 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 realnumbers. We say that an is O(bn) (read as “big oh” of bn) if there is an integerN 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 canyou say if L = 0? Illustrate with examples.

Since an/bn → L, we guess that there is an N such that an ≤ (L + 1) · bnfor all n ≥ N . We now prove this assertion. First, for any ε > 0, there isan N for which an/bn ∈ (L − ε, L + ε) for all n ≥ N . Thus, for the same N ,an ∈ (bn[L− ε], bn[L+ ε]). Thus, an ≤ bn(L + 1). Thus, an is O(bn) if N ischosen to correspond to ε = 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, n2

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 ε > 0 be given. Choose δ = ε. Then∣∣∣x2−4x+2 + 4

∣∣∣ = |x− 2 + 4| = |x+ 2| <δ = ε. Thus, lim

x→−2f(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 ε > 0 be given. Choose δ = ε2 . Then

∣∣∣ 2x2+3x−2x+2 + 5

∣∣∣ =∣∣∣ (x+2)(2x−1)

x+2 + 5∣∣∣ =

|2x− 1 + 5| = |2x+ 4| = 2 |x+ 2| < 2δ = ε.

3. Give an example of a function f : (0, 1) → R that has a limit at everypoint except 1

2 . Use the definition of limit of a function to justify the example.

Let f(x) =

{0, 0 < x < 1

21, 1

2 ≤ x < 1. Then, the limit is clearly 0 for every input

x ∈(0, 1

2

)and 1 for every input x ∈

(12 , 1). Let ε0 = 1. Let δ > 0 be given.

Then, in the neighborhood(

12 − 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 intervalsuch that f(x) = 1. Now, the only possibilities for the limit would be either 0or 1. However, the maximum difference between function values on the intervaland either limit is 1 which is equal to ε0. Thus, the limit does not exist.

4. Give an example of a function f : R→ R that is bounded and has a limitat 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 similarproof.

*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 ε > 0 be given. Then, there are δ1 and δ2 such that |f(x)− L1| < ε2 and

|f(x)− L2| < ε2 . Then, |L1 − L2| = |L1 − f(x) + f(x)− L2| < |L1 − f(x)| +

|f(x)− L2| < ε2 + ε

2 = ε. 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 frominputs x ∈ (0, δ). Thus,

∣∣cos 1x − L

∣∣ can be made greater than some ε. Thus, nolimit 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 limx→0

f(x) = 0. Let ε > 0 be given. Choose δ = ε. Then,∣∣x cos(

1x

)∣∣ < |x| < δ = ε.

8. Define f : (0, 1) → R by f(x) = x3−x2+x−1x−1 . Prove that f has a limit at

1.

We see limx→1

x3−x2+x−1x−1 = lim

x→1

x2(x−1)+(x−1)x−1 = lim

x→1

(x2+1)(x−1)x−1 = lim

x→1x2 + 1 =

2. Let ε > 0 be given. Choose δ =√ε. Then,

∣∣∣x3−x2+x−1x−1 − 2

∣∣∣ =∣∣x2 + 1− 2

∣∣ =∣∣x2 − 1∣∣ = |x+ 1| |x− 1| < δ2 = ε.

9. Define f : (−1, 1) → R by f(x) = x+1x2−1 . Does f have a limit at 1?

Justify.

We see limx→1

x+1x2−1 = lim

x→1

x+1(x+1)(x−1) = lim

x→1

1x−1 . From what we know from

calculus, we know that the limit does not exist. To be more precise, inputs in(1, 1 + ε) yield outputs which become arbitrarily large as ε → 0 and inputs in(1− ε, 1) yield outputs which become arbitrarily negative as ε→ 0.

2.2 Limits of Functions and Sequences

10. Consider f : (0, 2) → R defined by f(x) = xx. Assume that f has alimit at 0 and find that limit.


Examine xn = log(1 + 1

n

). Clearly, xn → 0. Since we assumed that f had

a limit, it must be the case that f(xn)→ limx→0

f(x). Then, f(xn) = log(1 + 1

n

)2which clearly has milt 0.

*11. Suppose f , g, and h : D → R where x0 is an accumulation point ofD, f(x) ≤ g(x) ≤ h(x) for all x ∈ D, and f and h have limits at x0 withlimx→x0

f(x) = limx→x0h(x). Prove that g has a limit at x0 and

limx→x0

f(x) = limx→x0

g(x) = limx→x0

h(x).

Since f and h have limits, limn→∞

f(xn) = limn→∞

g(xn) = limn→∞

h(xn) for any

xn → x0 by the Squeeze Theorem. Thus, limx→x0

f(x) = limx→x0

g(x) = limx→x0

h(x) by

Theorem 2.1.

*12. Suppose f : D → R has a limit at x0. Prove that |f | : D → R has alimit at x0 and that limx→x0

|f(x)| = |limx→x0f(x)|.

Let ε > 0 be given. Then by definition of limit, there is a δ > 0 such thatfor all x ∈ (x0 − δ, x0 + δ), |f(x)− x0| < ε. There are three cases to consider.If x0 > 0, then |f(x)| ≡ f(x) on the interval of interest and so the limit wouldbe that of f . If x0 < 0, then, |f(x)| ≡ −f(x) on the interval of interest and|−f(x) + x0| = |−1| |f(x)− x0| < ε for any x ∈ (x0 − δ, x0 + δ). Finally, ifx0 = 0, then ||f(x)|| = |f(x)| < ε 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 fhas a limit and justify your conclusions.

We see that x− [x] leaves us with the decimal part of x. Thus, the functionis continuous for (k, k + 1) for every integer k. It is not have a limit at theendpoints as there will be a jump of 1 unit.

14. Define f : R→ R as follows:f(x) = 8x if x is a rational numberf(x) = 2x2 + 8 if x is an irrational number.

Use subsequences to guess at which points f has a limit, then use ε’s and δ’s tojustify your conclusions.

We can only hope to find a limit at an accumulation point and since there canalways be found a rational number between two irrationals (and vice versa), wesee that the only possibility is at x = 2. We suppose that the limit at this point is16. Now, let ε > 0 be given. Choose δ1 = ε

2Then |8x− 16| = 8 |x− 2| < 8δ1 = ε.

Then, let δ2 = ε2 . Then,

∣∣2x2 + 8− 16∣∣ =

∣∣2x2 − 8∣∣ = 2

∣∣x2 − 4∣∣ < 2 |x− 4|2 =


2 · ε2 = ε. Choose δ = min(δ1, δ2) to see that the function has a limit of 16 atx = 2.

15. Let f : D → R with x0 as an accumulation point of D. Prove that f hasa limit at x0 if for each ε > 0, there is a neighborhood Q of x0 such that, forany x, y ∈ Q ∩D, x 6= x0, y 6= x0, we have |f(x)− f(y)| < ε.

Let ε > 0 be given. Since f has a limit L at x = x0, there is a δ such that forany x ∈ (x0 − δ, x0 + δ), |f(x)− L| < ε

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| = ε

2 + ε2 = ε.

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(x2+6x+1)x(x−6) . Thus, the limit is lim

x→0

xx ·

limx→0

(x2+6x+1)

limx→

(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 wasdetermined that the function had no limits for any x ∈ Z. We now observe thatx− [x+ 1] gives us the negative value of the decimal part of 1−x. We still havelimits in the intervals (k, k + 1), but approaching the left endpoint of an oddnumbered interval from the left, the limit is 1, while the limit from the right is1. Thus, the limit is defined at odd left-endpoints. Similarly, all the endpointshave limits. Thus, f has a limit at every point in R.

18. Define g : (0, 1) → R by g(x) =√

1+x−1x . Prove that g has a limit at 0

and find it.

We see that, for x 6= 0, g(x) =√

1+x−1x ·

√1+x+1x = 1+x−1

x = xx . The limit of

this function is clearly 1. (This is a proof because of the multiplication of limitsis the limit of multiplications.)

19. Define f : (0, 1)→ R by f(x) =√

9−x−3x . Prove that f has a limit at 0

and find it.


We see that for x 6= 0, f(x) =√

9−x−3x ·

√9−x+3x = 9−x−9

x = −xx . The limitis 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 iff(x) ≤ g(x) for all x ∈ D, then lim

x→x0

f(x) ≤ limx→x0

g(x).]

Suppose limx→x0

f(x) > limx→x0

g(x). Then, we must have limx→x0

f(x)− limx→x0

g(x) =

limx→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= 0for 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. [Lemma1.10 states: “If {bn}∞n=1 converges to B and B 6= 0, then there is a positive realnumber 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 realnumber M and a δ > 0 such that, if x ∈ (x0 − δ, x0 + δ), then |g(x)| ≥M .

Proof. Suppose that limx→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, itis 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, limx→0

(f + g)(x) = 0. Also,

if f is the function defined in Exercise 14, then f has no limit any where exceptat x = 2. If g ≡ 1

f , then limx→0

(fg)(x) = 1. For the same f , choosing g ≡ f gives

limx→1

fg = 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 atx0 ∈ (α, β) iff U(x0) = L(x0), and in this case lim

x→x0

f(x) = f(x0) = U(x0) =

L(x0).”]


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 limx→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 − 1

n

)→ 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 ε > 0, there exists a t1 ∈ (x0, β] suchthat U(x0) ≤ f(t1) < U(x0) + ε. Similarly, there exists a t2 ∈ [α, x0) suchthat L(x0) ≥ f(t2) > L(x0) − ε. Since we have U(x0) = L(x0) = M , wehave M + ε ≥ f(t1) ≥ M ≥ f(t2) > M − ε. Let δ = |t1 − t2|. Then forany x ∈ (x0 − δ, x0 + δ), we have |f(x)− f(x0)| < |f(t1)− f(t2)| < ε. Thus,limx→x0

f(x) = f(x0).

24. Let f : [a, b] → R be monotone. Prove that f has a limit both at a andb.

Without loss of generality, suppose that f is increasing. Then, the only waythat f might not have a limit at a is if U(a) > a (L(a) = a since there are nofunction values defined for inputs less than a). Suppose that U(a) = f(a+δ0) =f(a) + ε0. 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 largestfunction 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 ≡ fon (x0 − δ, x0) and g ≡ f(x?) on (x0, x0 + δ). Either way, since f has a limit atx0 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 atall 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 allx, y ∈ R. If f has a limit at zero, prove that f has a limit at every point andeither lim

x→0f(x) = 1 or f(x) = 0 for all x ∈ R.


If f has a limit at 0, then for any an → 0, f(an) → limx→0

f(x). Then, since

any input y can be written as 0 + y, we see that f (y + an) = f(y)f(an) →limx→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 limx→x0

f(x + y) =

limx→x0

[f(x)f(y)] = limx→x0

f(x) · limx→x0

f(y). In particular, limx→x0

f(x+ 0) = limx→0

f(x) ·limx→x0

f(x). Thus, limx→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 ε > 0 such that D ∩ [x0 − ε, x0 + ε] = E ∩ [x0 − ε, x0 + ε]. Iff(x) = g(x) for all x ∈ D ∩ E ∩ [x0 − ε, x0 + ε], prove that f has a limit at x0

iff g has a limit at x0.

Let ε > 0 be given. Then, limx→x0

f(x) = L exists ⇔ |f(x)− L| < ε2 . At the

same time, |f(x)− g(x)| < ε2 . If g(x) has a limit at x0, it must be equal to L.

To see this, suppose |g(x)− L| ≥ ε0 for some x ∈ D∩E∩[x0−δ, x0+δ]. We thenhave |g(x)− f(x)| = |g(x)− L+ L− f(x)| > |g(x)− L| − |f(x)− L| > ε0 − ε.Contradiction. The argument is similar if we assume that g(x) has a limit atx0.

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 at2.

Let ε > 0 be given. Choose δ =√ε. We see that

∣∣3x2 − 2x+ 1− 9∣∣ =∣∣3x2 − 2x− 8

∣∣ =∣∣x+ 4

3

∣∣ |x− 2| <∣∣x+ 4

3

∣∣2 < δ2 = ε.

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 ε > 0 be given. Choose δ = ε2 . Then,

∣∣∣ 2x2−18x+3 + 3

∣∣∣ =∣∣∣ 2(x2−9)

x+3 + 3∣∣∣ =

|2(x− 3) + 3| = |2x− 3| < 2 |x− 3| < 2δ = ε.

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{en+1n

}={e · e 1

n

}. By Theorem 3.1,

e1n → 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 εN such that|f(xn)− f(x0)| ≥ εN 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+1x . 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 numberr. Determine f(

√2) and justify your conclusion.

Since f is continuous, any sequence of numbers converging to√

2 convergesto 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 rationalnumber 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 sequenceof 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 ε > 0 be given and choose δ = ε. We then seethat

∣∣x sin 1x

∣∣ = |x|∣∣sin 1

x

∣∣ < |x| < δ = ε for any x ∈ (−δ, δ).

*10. Suppose f : E → R is continuous at x0 and x0 ∈ F ⊂ E. Defineg : 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 off at x0.

Let ε > 0 be given. Then, since f is continuous at x0, there is a δ > 0such that for all x ∈ (x0 − δ, x0 + δ) we have |f(x)− f(x0)| < ε. Then, chooseδ′ = |sup[(x0 − δ, x0 + δ) ∩ F ]− x0|. Then, (x0 − δ′, x0 + δ′) ⊆ (x0 − δ, x0 + δ),so we must have |g(x)− g(x0)| < ε.

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


E = R and F = {2}. Then, g(x) is trivially continuous at 2, but f(x) is notcontinuous 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 isirrational. Prove from the definition that f is continuous at 2 and discontinuousat 1.

Let ε > 0 be given. Choose δ1 = ε8 . Then, |8x− 16| = 8 |x− 2| ≤ 8δ1 =

ε. Choose δ2 = min(2, ε8 ). Then,∣∣2x2 + 8− 16

∣∣ =∣∣2x2 − 8

∣∣ = 2∣∣x2 − 4

∣∣ =2 |x+ 2| |x− 2| < 8 |x− 2| < 8δ = ε. Thus, choosing δ = min(δ1, δ2) gives|f(x)− 16| < ε.

Let ε0 = 2. Let δ > 0 be given. Then, there is always an irrational numberx ∈ (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 = ε0.

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 thus definedwill 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 ε > 0 be given. Let M = maxx∈(x0−1,x0+1)

p(x)q(x) . Choose

δ = εM . Then, |f(x)| < |M(x− x0)| < δM = ε. 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 ε0 = 1. Let δ > 0 be

given and m = minx∈(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 1

x−x0gets arbitrarily close

to zero, there will always be an x ∈ (x− δ, x+ δ) such that 1x−x0

> y+1m . Thus,

|f(x)− y| >∣∣∣m(y+1)

m − y∣∣∣ = 1 = ε0. 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 anda 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 ε > 0, there is a δε > 0,such that if x ∈ (x0− δε, x0 + δε), we have −ε < f(x)−f(x0) < ε⇔ f(x0)− ε <


f(x) < f(x0) + ε. Thus, let M = maxε>0{f(x0) ± ε} and Q = (x0 − δε, x0 + δε),

where ε is the value chosen in the definition of M . Then, by construction, wehave 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 ε > 0 be given. Since f(x) is continuous at x0, there exists a δ > 0such that |f(x)− f(x0)| < ε for any x ∈ (x0 − δ, x0 + δ). Now we see that||f(x)| − f(x0)| < |f(x)− f(x0)| < ε for any x ∈ (x0 − δ, x0 + δ).

15. Suppose f, g : D → R are both continuous on D. Define h : D → R byh(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). Supposethat there is an ε0 > 0 such that for every δ > 0, we have an x ∈ (x0− δ, x0 + δ)such that |h(x)− f(x0)| ≥ ε0. Since f(x) is continuous at x0, there is a δ0such that |f(x)− f(x0)| < ε0 for all x ∈ (x0 − δ0, x0 + δ0). Take δn = 1

k+n

where 1k < δ0. Construct {xn} by selecting xn ∈ (x0 − δn, x0 + δn) such that

|h(xn)− f(x0)| ≥ ε0. 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 plugin x = 2 to see that there is no equality). I believe that the author intendedto write xx = ex ln x. In this case, we know that the function k(x) = x iscontinuous, 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 iscontinuous at x0, then

√f is continuous at x0.

Let ε > 0 be given. Then, there is a δ > 0 such that for any x ∈ (x0−δ, x0+δ),

we have |f(x)− f(x0)| < ε. Now,∣∣∣√f(x)−

√f(x0)

∣∣∣ < ∣∣∣√f(x)−√f(x0)

∣∣∣ ∣∣∣√f(x) +√f(x0)

∣∣∣ =

|f(x)− f(x0)| < ε 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.


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 thelimit. 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 isuniformly continuous. What can be said about fg? Justify.

Let ε > 0 be given. Then, there is some δ > 0 such that |f(x)− f(y)| <ε2 and |g(x)− g(y)| < ε

2 if |x− y| < δ. Now, |f(x) + g(x)− f(y)− g(y)| ≤|f(x)− f(y)|+ |g(x)− g(y)| < ε

2 + ε2 = ε.

Now, fg is not necessarily uniformly continuous. For example, let f(x) =g(x) = x. Let ε0 = 1. Assume without loss of generality that y > x. Then, forany 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 · 1

4δ · 2δ = 1 = ε0. Thus, f(x)g(x) is notuniformly continuous.

20. Let f : A → B and g : B → C be uniformly continuous. What can besaid about g ◦ f : A→ C? Justify.

The function g ◦ f is not necessarily uniformly continuous if A 6= A. Forinstance g : (1, 2)→

(1, 1

2

)defined by g(x) = 1

x is uniformly continuous on (1, 2).Also, f(x) = x−1 is uniformly continuous everywhere. However, g(f(x)) = 1

x−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 ofExample 3.2.

Let ε > 0 be given. Choose δ = 0.162 ε. 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 = ε.

22. Define f : (2, 7) → R by f(x) = x3 − x + 1. Show that f is uniformlycontinuous on (2, 7) without using Theorem 3.8– that is, use the methods ofExample 3.3.

Let ε > 0 be given. Choose δ = ε148 . 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 = ε.


23. A function f : R → R is periodic iff there is a real number h 6= 0 suchthat f(x + h) = f(x) for all x ∈ R. Prove that if f : R → R is periodic andcontinuous, then f is uniformly continuous.

Define fn : R→ R by fn(x) = f(x− nh) if x ∈ [(n− 1)h, nh] and fn(x) = 0otherwise. Then, fn is continuous on [(n−1)h, nh] which is compact, making ituniformly 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 functionthat is continuous on A but not uniformly continuous. Give an example of aset that is not compact, but every function continuous on that set is uniformlycontinuous.

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) = 1

x−x0will

not be uniformly continuous. (Since it is an accumulation point, inputs can getarbitrarily close to x0 making 1

x−x0arbitrarily 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 onN. (Since δ < 1 implies that x = y and |f(x)− f(x)| = 0 < ε.)

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+1if 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 isa sequence {xn}∞n=1 of points of E converging to x0, it is true that x0 ∈ E. Inother words, prove E is closed if it contains all limits of sequences of membersof E.

Suppose that for every xn → x0, we have x0 ∈ E. Then, by the propertiesof convergence and the fact that xn ∈ E for all xn, for any ε > 0, we have axk ∈ {xn} ⊆ E such that xk ∈ (x0−ε, x0 +ε)∩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 setof the form {x : a ≤ x ≤ b} is closed.


Let x0 ∈ {x : a < x < b} = (a, b). Then, choose ε0 = min(x0 − a, b − x0).Then by construction, we have (x0 − ε0, x0 + ε0) ⊂ (a, b). Thus, (a, b) is open.

Let x0 ∈ {x : a ≤ x ≤ b} = [a, b]. Let ε > 0 be given. Then, x0 ∈(x0 − ε, x0 + ε) ∩ [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. Provethat D = D ∪ D′ is closed and if F is any closed set that contains D, thenD ⊂ 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 itsaccumulation points. Since D ⊂ F , we know that F must contain all of theaccumulation 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 ε0 = 50, we see that (x0− ε0, x0 + ε0)∩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}. Letx0 ∈ {x : f(x) 6= r0}. Choose x ∈ f←(r0) such that |x− x0| is minimal. Chooseε0 = inf{|x− y| : x, y ∈ f←(r0)}. Then, by the continuity of f , there is a δ0 > 0such that |f(x)− r0| < ε0. 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. LetT = {x : f(x) = g(x)}. Prove that T is closed.

Let x0 ∈ T and let ε > 0 be given. Then, x0 ∈ (x0 − ε, x0 + ε). Thus, T isclosed. (This problem is quite trivial.)

32. If D ⊂ R, then x ∈ D is said to be an interior point of D iff there is aneighborhood Q of x such that Q ⊂ D. Define D◦ to be the set of interior pointsof D. Prove that D◦ is open and that if S is any open set contained in D, thenS ⊂ D◦ ⊂ D. D◦ is called the interior of D.

Since for any point x0 ∈ D◦, there is a ε > 0 such that (x0− ε, x0 + ε) ⊂ D◦,we know that D◦ is open. Let S ⊂ D be open. Then, for any x0 ∈ S, there mustbe ε > 0 such that (x0− ε, x0 + ε) ⊂ S. Since S ⊂ D, we have (x0− ε, x0 + ε) ⊂S ⊂ D◦. Thus, S ⊂ D◦ ⊂ D.


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− 1

n , 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 thatsupE and inf E both belong to E.

Since E is compact, it is closed and bounded. Now, supE and inf E areboth 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, ifx0 ∈ ∪Ei, then x0 ∈ Ei for some i. Since each Ei is closed, x0 is a point ofclosure. 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 isbounded.

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 ε0 = min (f(x0), 1− f(x0)), there is a δ for which x ∈ (x0 − δ, x0 + δ) ∩Dimplies that |f(x)− f(x0)| < ε. Since this is true of all x in this interval, defineδn = δ/n. We can then construct {xn} such that xn ∈ (x0 − δn, x0 + δn) for alln. 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 eachε > 0, there is M > 0 such that if |x| > M , then |f(x)| < ε. Show that f isuniformly continuous.

Let ε > 0 be given. Then, since |f(x)| < ε for all x ∈ (−∞,−M) ∪ (M,∞),|f(x)− f(y)| < |f(x)| < ε for all x, y ∈ (−∞,−M) ∪ (M,∞). Thus, f is


uniformly continuous on (−∞,−M) ∪ (M,∞). Then, by Theorem 3.8, f isuniformly continuous on [−M,M ] (since it is compact). Thus, f is uniformlycontinuous.

40. Give an example of a function f : R→ R that is continuous and boundedbut 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 closerand closer to each other. Therefore, for any delta we choose, we can find xvalues 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’sTheorem, [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 thatthere 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 trivialproblem.)

44. Suppose f : [a, b]→ [a, b] is continuous. Prove that there is at least onefixed 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 forsome 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.)


First, f is monotone. To see this, suppose not. Then, there is some pointx0 where f(x) < f(x0) on one side and f(x) > f(x0) on the other for valuesx ∈ (x0 − δ, x0 + δ) for some δ > 0. Without loss of generality, let us assumethat it is increasing to the left of x0 and increasing to the right of x0. Bythe intermediate value property, f attains all values in [f(x0 − δ), f(x0)] and[f(x0), f(x0 + δ)]. Let ε = min (|f(x0)− f(x0 − δ)| , |f(x0)− f(x0 + δ)|). Bythe intermediate value property, every function value in (f(x0)− ε, f(x0)) mustbe 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] andlet ε > 0 be given. Since ε > 0 is only of interest when it is small, suppose(f(x0)− ε, f(x0) + ε) ⊂ (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)−ε and f(x2) =f(x0)+ε. Take δ = min (|x0 − x1| , |x0 − x2|). Then, for every x ∈ (x0−δ, x0+δ)we have |f(x0)− f(x)| < ε. Thus, f is continuous at x0.

46. Prove that there is no continuous function f : R→ R such that, foreach c ∈ R, the equation f(x) = c has exactly two solutions.

Since f is continuous, it has the intermediate value property. From this, wecan narrow down the form that f can have. We see that f must be increasing,then decreasing or vice versa. If f changes its increasing/decreasing status anymore, some function value will have 3 elements in its preimage. Also, if f ismonotone, 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 intervalsince 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) andto the right, f(x0) > f(x) (again, we do not have equality because of the abovediscussion). 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 areM > 0 and a > 0 such that if x ∈ [−a, a], then |f(x)| ≤ M . Prove that f isuniformly continuous. In particular, prove that there is a real number m suchthat f(x) = mx for all x ∈ R.

From Project 2.1, we know that an additive function that fits the abovecriteria 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


ε > 0, there is a δ > 0 such that |f(x)| < ε for any x ∈ (−δ, δ). Now, since f isadditive, we have f(x) = f(0 +x) = f(0) + f(x). Thus, f(x) = 0. Finally, sincef is continuous at each point, it is continuous at x = 0. Thus, there is δ > 0such that |f(x)− 0| = |f(x)| < ε. 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 = mso 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 ε > 0 and x0 ∈[a, b] be given. By the continuity of f , there is a δ > 0 for which |f(x)− f(x0)| <ε. Now, if g ≡ f on (x0−δ, x0+δ), then g is clearly continuous. If g(x) 6= f(x) forsome 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)| < ε for all x ∈ (x0 − δ′, x0 + δ′).

49. Suppose that g : D → R is continuous at x0 and that x0 is also anaccumulation point of D. Define D0 = {x : g(x) = 0}. If g(x) 6= 0, prove thatx0 is an accumulation point of D0.

Choose x to be such that g(x) = 0 and |x− x0| is minimum. Then, bythe 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, anyneighborhood 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 thatthere is ε > 0 such that D∩ [x0− ε, x0 + ε] = E∩ [x0− ε, x0 + ε] and f(x) = g(x)for all x ∈ D ∩ E ∩ [x0 − ε, x0 + ε]. Prove that f is continuous at x0 iff g iscontinuous at x0.

If f is continuous at x0, then for all ε′ > 0, there is a δ > 0 such that forall x ∈ (x0 − δ, x0 + δ), we have |f(x)− f(x0)| < ε′. If δ < ε, then we haveg ≡ f on (x0 − δ, x0 + δ) since D ∩ [x0 − δ, x0 + δ] = E ∩ [x0 − δ, x0 + δ]. Ifδ ≥ ε, then for all x ∈ (x0 − ε, x0 + ε), we also have |f(x)− f(x0)| < ε′. (Since(x0 − ε, x0 + ε) ⊆ (x0 − δ, x0 + δ).) Thus, f(x) and f(x0) can be replaced byg(x) and g(x0) respectively. The argument is similar if we assume that g(x) iscontinuous 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) = x2.For x0 6= 0, find the equation of the line tangent to f at that point by finding aline that intersects the curve in exactly one point. Do not use the derivative tofind this line.

We need x20 = mx0 +b⇔ x2

0−mx0−b = 0. Since we want only one solution,

we must have the discriminant (−m)2 − 4(−b) = m2 + 4b = 0. Thus, b = −m2

4

and the equation is x20 −mx0 + m2

4 = 0. This implies that m = 2x0. Thus, theequation of the tangent line is y − x2

0 = 2x0(x− x0).

2. Prove that the definition of the derivative and the alternate definition ofthe derivative are equivalent.

Set x + h = x0. Then, since as x → x0, h → 0, we have limx→x0

f(x)−f(x0)x−x0

=

limh→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?

limh→0

√x+h−

√x

h

(√x+h+

√x√

x+h+√x

)= lim

h→0

x+h−xh(√x+h+

√x)

= limh→0

1√x+h+

√x

= 12√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

limh→0

(x+h)2−x2

h = limh→0

x2+h2+2xh−x2

h = limh→0

h(h+2x)h = lim

h→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 aderivative at one point. You may use the rules for differentiating products, sumsand quotients of elementary functions that you learned in calculus.

We see that f ′(x) = 3x2 sin 1x + cos 1

x ·(− 1x2

)· x3(?). Now, lim

h→0

x3 sin(x+h)h =

limh→0

x3 sin x sinh−cos x coshh = lim

h→0x3 sinx · sinh

h − 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 thelimit 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

limh→0

f(x+ h)− f(x− h)

2h

exists and equals f ′(x). Give an example where this limit exists, but the functionis not differentiable.

If limh→0

f(x+h)−f(x)h exists, then by the same token, lim

h→0

f(x)−f(x−h)h exists.

Thus, limh→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 limh→0

f(x+h)−f(x−h)2h must also exist. (It’s just the last limit multiplied

by 1/2.) Then, since limh→0

f(x+h)−f(x)h = lim

h→0

f(x)−f(x−h)h = f ′(x), we see that

limh→0

f(x+h)+f(x−h)2h = 2f ′(x)

2 = f ′(x).

Let f(x) = |x|. Then, limh→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 ε > 0 such that |x− y| < ε and y ∈ (a, b) imply that|f(x)− f(y)| < M |x− y|. Give an example of a function that fails to satisfya Lipschitz condition at a point of continuity. If f is differentiable at x, provethat 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.


We see that at x = 0, the graph has a vertical tangent line, so as x→ 0, theaverage rate of change from 0 to x will go to ∞. Thus, f is not Lipschitz atx = 0.

Now, if f is differentiable at x, we have limy→x

f(x)−f(y)x−y = M . Thus, for

any ε > 0, there is a δ > 0 such that for any y ∈ (x − δ, x + δ), we have∣∣∣ f(x)−f(y)x−y −M

∣∣∣ < ε. Then,∣∣∣ f(x)−f(y)

x−y

∣∣∣−M ≤ ∣∣∣ f(x)−f(y)x−y −M

∣∣∣, so∣∣∣ f(x)−f(y)

x−y

∣∣∣ <ε+M and |f(x)− f(y)| < |x− y| (ε+M). Thus f is Lipschitz at x.

8. A function f : (a, b) → R is said to be uniformly differentiable iff f isdifferentiable on (a, b) and for each ε > 0, there is δ > 0 such that 0 < |x− y| <δ and x, y ∈ (a, b) imply that∣∣∣∣f(x)− f(y)

x− y− f ′(x)

∣∣∣∣ < ε.

Prove that if f is uniformly differentiable on (a, b), then f ′ is continuous on(a, b).

Since f is uniformly continuous, for every ε > 0, there is a δ for which if |x−y| < δ we have −ε/2 < f(x)−f(y)

x−y − f ′(x) < ε/2. Thus, for the same δ, f(x)−f(y)x−y −

ε/2 < f ′(x) < f(x)−f(y)x−y + ε/2. By the same token, f(x)−f(y)

x−y − ε/2 < f ′(y) <f(x)−f(y)

x−y + ε/2. Thus, |f ′(x)− f ′(y)| <∣∣∣ f(x)−f(y)

x−y − ε/2−(f(x)−f(y)

x−y + ε/2)∣∣∣ = ε.

Thus, f ′(x) is continuous on (a, b).

9. Suppose f : (a, b) → R is continuous on (a, b) and differentiable atx0 ∈ (a, b). Define

g(x) =f(x)− f(x0)

x− x0for x ∈ (a, b) \ {x0}, g(x0) = f ′(x0).

Prove that g is continuous on (a, b).


We see that g(x) represents the average rate of change from the variable pointx 6= x0 to the set point x0. Since f itself is continuous, g is also continuous here.At x = x0, g is the derivative, which is defined to be the limit of the averagerate of change as x → x0. Thus, limit = function value and g is continuous atx0.

10. Suppose f, g, and h are defined on (a, b) and a < x0 < b. Assume fand h are differentiable at x0, f(x0) = h(x0), and f(x) ≤ g(x) ≤ h(x) for allx ∈ (a, b). Prove that g is differentiable at x0 and f ′(x0) = g′(x0) = h′(x0).

Define xn = x0 − 1/n. Then, we see that f(xn)−f(x0)xn−x0

→ f ′(x0) = h′(x0) ←g(xn)−g(x0)xn−x0

as n → ∞. Thus, since f(xn)−f(x0)xn−x0

≤ g(xn)−g(x0)xn−x0

≤ h(xn)−h(x0)xn−x0

for all n, we see that the middle sequence must converge to f ′(x0) = h′(x0).Define xn = x0 + 1/n, repeat the same argument (except that the inequalities

will be reversed with g(xn)−g(x0)xn−x0

in the middle). By proving the convergenceof this sequence, there is a δ = 1/N (where N is chosen to prove that the

sequences converge) such that∣∣∣ g(x)−g(x0)

x−x0

∣∣∣ < ε. Thus, g′(x0) exists and is equal

to f ′(x0) = h′(x0).

4.2 The Algebra of Derivatives

11. Prove f : (0, 1) → R defined by f(x) =√

2x2 − 3x+ 6 is differentiableon (0, 1) and compute the derivative.

We know that√x is differentiable on (0,∞). We know that 2x2 − 3x + 6

is differentiable on R. Thus,√

2x2 − 3x+ 6 is differentiable for all x such that2x2 − 3x + 6 > 0. This is always the case, so the function is differentiable andthe derivative is 1/2(2x2 − 3x+ 6)−1/2(4x− 3).

12. Suppose f : [a, b] → [c, d], g : [c, d] → [p, q], and h : [p, q] → R, withf differentiable at x0 ∈ [a, b], g differentiable at f(x0), and h differentiable atg(f(x0)). Prove that h ◦ (g ◦ f) is differentiable at x0 and find the derivative.

The chain rule implies that [h(g(f(x0)))]′ = h′(g(f(x0)) · g′(f(x0)) · f ′(x0).Done.

13. Suppose f : [a, b] → [c, d] and g : [c, d] → R are differentiable on [a, b]and [c, d], respectively. Suppose

f ′ : [a, b]→ R and g′ : [c, d]→ R

are also differentiable on [a, b] and [c, d], respectively. Show that (g◦f)′ : [a, b]→R is differentiable and find the derivative.


By the chain rule, [g(f(x))]′ = g′(f(x)) · f ′(x). Since this is defined for allpossible inputs, this must indeed be the derivative.

14. Suppose f : R → R is differentiable and define g(x) = x2f(x3).Showthat g is differentiable and compute g′.

Chain rule + product rule: g′(x) = 2x · f(x3) + x2 · f ′(x3) · 3x2.

15. Define f(x) =

√x+

√x+√x for x ≥ 0. Determine where f is

differentiable and compute the derivative.

Chain rule. f ′(x) = 1/2(x+√x+√x)−1/2 ·(1+1/2(x+

√x)−1/2 ·(1+1/2x−1/2)).

Now, x cannot be 0 since there will be a factor in the denominator.

4.3 Rolle’s Theorem and the Mean Value The-orem

16. Define f : [0, 2] → R by f(x) =√

2x− x2. Show that f satisfies theconditions of Rolle’s theorem and find c such that f ′(c) = 0.

The function is clearly continuous on the interval, and the chain rule willshow that the function is differentiable on (0, 2). Also, f(0) = f(2) = 0.

Next, f ′(x) = 1/2(2x− x2)(2− 2x). This equals 0 at x = 1/2.

17. Define f : R → R by f(x) = 1/(1 + x2). Prove that f has a maximumvalue and find the point at which the maximum occurs.

f ′(x) = − 2x(1+x2)2 , so f has its only extremum at x = 0. Since the slope

of the tangent line is positive to the left of x = 0 and negative to the right, itfollows that f attains its maximum value at x = 0.

18. Prove that the equation x3 − 3x + b = 0 has at most one root in theinterval [−1, 1].

The function f(x) = x3 − 3x + b has derivative f ′(x) = 3x2 − 3. Thus, theextrema occur at ±1. This means that the slope of the tangent line switchesfrom negative to positive only once on this interval. Thus, only one root (at themost) exists.

19. Show that cosx = x3 + x2 + 4x has exactly one root in[0, π2

].


Since cosx is decreasing on the interval and x3 + x2 + 4x is increasing onthe interval, their graphs can intersect but once.

20. Suppose f : [0, 2] → R is differentiable, f(0) = 0, f(1) = 2, andf(2) = 2. Prove that

1. there is c1such that f ′(c1) = 0,

2. there is c2 such that f ′(c2) = 2, and

3. there is c3 such that f ′(c3) = 32 .

(1) Since f(1) = f(2) = 2, Rolle’s Theorem implies that there is c1 such thatf ′(c1) = 0.

(2) By MVT, there is c2 such that f ′(c) = f(1)−f(0)1 = 2−0

1 = 2.(3) Since f ′ is continuous, IVT implies that there is c3 such that f ′(c3) = 3/2.

21. Let f : [0, 1] → R and g : [0, 1] → R be differentiable with f(0) = g(0)and f ′(x) > g′(x) for all x ∈ [0, 1]. Prove that f(x) > g(x) for all x ∈ (0, 1].

Define D(x) = f(x) − g(x) for all x ∈ [0, 1]. We already have D(0) = 0.Suppose that g(x0) ≥ f(x0) for some x0. Then, there is c for which f(c) = g(c)by IVT. Thus, D(c) = 0. Then, by Rolle’s Theorem, there is ξ ∈ (0, c) for whichD′(ξ) = f ′(ξ)− g′(ξ) = 0. This implies that f ′(ξ) = g′(ξ). Contradiction.

22. Use the Mean-Value Theorem to prove that

nyn−1(x− y) ≤ xn − yn ≤ nxn−1(x− y)

if n ≥ 1 and 0 ≤ y ≤ x.

Define f(z) = zn. By MVT, there is c ∈ (x, y) for which f ′(c) = f(x)−f(y)x−y ⇔

ncn−1(x − y) = xn − yn. The only critical point of f is at z = 0. Thus, thisequality holds for all x, y. Since 0 ≤ y ≤ x, this means that the lower boundfor xn − yn is nyn−1(x − y), while the upper bound is nxn−1(x − y). Hence,nyn−1(x− y) ≤ xn − yn ≤ nxn−1(x− y).

23. Use the Mean-Value Theorem to prove that

√1 + h < 1 + 1/2h for h > 0.

Since LHS, RHS > 1, we square both sides and obtain 1+h?< 1+1/4h+h

h>0⇔0

X< 1/4h. I don’t know how or why MVT would be used to solve this.


24. Generalize Exercise 23 as follows: If 0 < p < 1 and h > 0, then showthat

(1 + h)p < 1 + ph.

You may assume the usual rules about differentiating powers.

By MVT, there is always a c ∈ (0, h) such that f ′(c) = f(h)−f(0)h . Thus,

if f ′(c) < 0, we have f(h) < f(0). Define f(x) = (1 + px) − (1 + x)p. Wesee f(0) = 0 and f ′(x) = p − p(1 + x)p−1 = p[1 − (1 + x)p−1]. For c > 0(the only possible ones), we clearly have f ′(c) > 0, so f(h) > f(0). Thus,(1 + ph)− (1 + h)p > 0⇔ (1 + h)p < 1 + px.

25. Suppose f : (a, b) → R is differentiable and |f ′(x)| ≤ M for all x ∈(a, b). Prove that f is uniformly continuous on (a, b). Give an example of afunction f : (0, 1)→ R that is differentiable and uniformly continuous on (0, 1),but such that f ′ is unbounded.

By MVT, there is c ∈ (x, y) for which f ′(c) = f(x)−f(y)x−y for all x, y ∈ (a, b).

Let ε > 0 be given and choose δ = ε/M Now, |f(x) − f(y)| ≤ f ′(c)|x − y| ≤M |x− y| < Mδ = ε. Thus, f is uniformly continuous.

The function f(x) =√

1− x2 is differentiable on (0, 1). It is also uniformlycontinuous on (0, 1) (we have seen why). But, we have also shown (Q7) thatthe tangent lines to the graph tend to a vertical line as x→ 1− and x→ 0−, sof ′(x)→∞ as x→ 0+, 1− and is unbounded.

26. Suppose f is differentiable on (a, b) except possibly at x0 ∈ (a, b), and iscontinuous on [a, b]; assume limx→x0 f

′(x) exists. Prove that f is differentiableat x0 and f ′ is continuous at x0.

Let δ be given. Then, by MVT, there is cδ such that f ′(cδ) = f(x)−f(x0)x−x0

for

x ∈ (x0 − δn, x0 + δn). Then, limδ→0

f ′(cδ) = limδ→0

f(x)−f(x0)x−x0

= limx→x0

f(x)−f(x0)x−x0

=

f ′(x). Thus f is differentiable at x0 and since f ′(x0) = limx→x0

f ′(x), f ′ is also

continuous.

27. Define f(x) = x + 2x2 sin 1x for x 6= 0 and f(0) = 0. Prove that f is

differentiable everywhere. Show that there exists a number a such that f ′(a) > 0,but there does not exist a neighborhood of a in which f is increasing.

First, f is continuous since limx→0

f(x) = 0 = f(0). Thus, by Q26, f ′(0) exists.

Thus, f is differentiable everywhere.

By MVT, there is c ∈ (0, a) such that f ′(c) = f(a)−f(0)a−0 = f(a)

a . Thus.f ′(c) < 0 precisely when f(a) < 0. There are an infinity of such points. Now,

near 0, f ′(x) = 1 + 4x sin 1x −

2x2 cos 1x

x2 = 1 + 4x sin 1x − 2 cos 1

x . We thus havehorizontal tangent lines at 2/kπ for each k.


28. Prove that the function f(x) = 2x3 +3x2−36x+5 is 1-1 on the interval[−1, 1]. Is f increasing or decreasing.

f ′(x) = 6x2 + 6x−36 = (x+ 3)(x−2). This function has no roots in [−1, 1],so f is 1-1.

f ′(0) = −6 < 0, so f is decreasing.

29. Show that the function f(x) = x3 − 3x2 + 17 is not 1-1 on the interval[−1, 1].

f ′(x) = 3x2 − 6x = 3x(x − 2) which has a root at x = 0 ∈ [−1, 1] and so fis not 1-1.

30. Give an example of a function f : R→ R that is differentiable and 1-1,but f ′(x) = 0 for some x ∈ R.

Let f(x) = x3. Then, f ′(x) = 3x2 which has a root at x = 0. As we know,x3 is a 1-1 function.

31. If f : [a, b] → R is differentiable at c, a < c < b and f ′(c) > 0, provethat there is x, c < x < b, such that f(x) > f(c).

If f ′(c) > 0, then f is increasing on some interval about c and so there is xsuch that f(x) > f(c).

4.4 L’Hospital’s Rule and the Inverse-FunctionTheorem

32. Assume the rules for differentiating the elementary functions, andL’Hospital’s Rule and find the following limits:

a. limx→1

ln xx−1

b. limx→0

xex−1

c. limx→0

sin xx

(a) limx→1

ln xx−1

H= limx→1

1/x1 = 1.

(b) limx→0

xex−1

H= limx→0

1ex = 1.

(c) limx→0

sin xx

H= limx→0

cos x1 = 1.


33. Use L’Hospital’s Rule to find the limit:

limx→0

x2 sinx

sinx− x cosx.

limx→0

x2 sin xsin x−x cos x

H= limx→0

2x sin x+x2 cos xx sin x

H= limx→0

4x cos x+(2−x2) sin xx cos x+sin x

H= limx→0

(6−x2) cos x−6x sin x2 cos x−x sin x =

3.

34. Prove the following variant of Theorem 4.14. Suppose f : [a, b] → R is1-1. If f is differentiable at c, f ′(c) 6= 0, and f−1 is continuous at d = f(c),then f−1 is differentiable at d and

(f−1)′(d) =1

f ′(c).

Suppose that the tangent line to f at c is y − d = f ′(c)(x− c). The inverseof this tangent line is given by the equation x − d = f ′(c)(y − c) ⇔ y − c =

1f ′(c) (x− d). Thus, the slope of the tangent line passing through (d, c) is 1

f ′(c) .

Done.

35. Find the equation for the line tangent to the graph of f−1at the point(3, 1) if

f(x) = x3 + 2x2 − x+ 1.

f ′(x) = 3x2 + 4x− 1. f ′(1) = 3 + 4− 1 = 6. Thus, (f−1(3))−1 = 1/6.

36. Use the Inverse-Function Theorem to derive the formula for the deriva-tive of the inverse of sinx on the interval

[−π2 ,

π2

]. You may assume the usual

facts about the function f(x) = sinx.

By IFT, (sin−1)′(sinx) = 1(sin x)′ = 1

cos(x) = sec(x).

37. Suppose both f and f−1 are twice-differentiable functions. Derive aformula for (f−1)′′.

By IFT, (f−1)′(f(x)) = 1f ′(x) . Thus, (f−1)′′(f(x)) =

(1

f ′(x)

)= − 1

f ′(x)2 ·

f ′′(x) = − f ′′(x)(f ′(x))2 .

Miscellaneous


38. Suppose f : (a, b) → Ris differentiable at x0 ∈ (a, b) with {αn}∞n=1 and{βn}∞n=1 two sequences in (a, b) \ {x0}converging to x0 such that the sequence{

βn − x0

βn − αn

}∞n=1

is bounded. Prove that {f(βn)− f(αn)

βn − αn

}∞n=1

converges to f ′(x0).

For each n, we have cn between αn and βn such that f ′(c) = f(βn)−f(αn)βn−αn .

Since αn, βn → x0 and since f ′ is continuous, f(cn)→ f(x0).

39. Suppose f : R→ R is such that f(x+ y) = f(x)f(y), f is differentiableat zero, and f is not identically zero. Prove that f is differentiable everywhereand that f ′(x) = f(x)f ′(0). Assuming the properties of the exponential function,prove that f(x) = ecxwhere c = f ′(0).

We know that limh→0

f(0)−f(h)h = f ′(0). Then, lim

h→0

f(x)−f(x+h)h = lim

h→0

f(0)f(x)−f(x)f(h)h =

limh→0

f(x)(f(0)−f(h))h = f(x)f ′(0) for any x. Thus, f is differentiable everywhere.

Next, if f(x) = ef′(0)x, then f ′(x) = f ′(0)ef

′(0)x = f ′(0)f(x). Thus, f(x) =ef′(0)x + K, where K ∈ R. But, f(x) = f(x + 0) = f(0)f(x) for all x, so

f(0) = 1. Thus, 1 = ef′(0)(0) +K = e0 +K. Thus, K = 0. Thus, f(x) = ef

′(0)x.

40. Suppose f : [a, b]→ R is differentiable and f ′′ exists at t ∈ (a, b). Provethat

f ′′(t) = limh→0

f(t+ h)− 2f(t) + f(t− h)

h2.

Give an example where this limit exists, but f ′ is not differentiable at t.

First, limh→0

f(t+h)−2f(t)+f(t−h)h2

H= lim

h→0

f ′(t+h)−f ′(t−h)2h

H= lim

h→0

f ′′(t+h)+f ′′(t−h)2 =

2f ′′(t)2 = f ′′(t).

Let f(x) =

{1/2x2, if x ≥ 0−1/2x2, if x < 0

. Then at x = 0, we see that limh→0−

−1/2h2+1/2h2

h2 =

0. Also, limh→0+

1/2h2−1/2h2

h2 = 0. Thus, the limit at 0 is 0. But f ′(x) = |x| which

we know is not differentiable at x = 0.

41. Suppose f : D → R, g : E → R, x0 ∈ D ∩ E, x0 an accumulation pointof D ∩ E. Suppose further that there is ε > 0 such that

D ∩ [x0 − ε, x0 + ε] = E ∩ [x0 − ε, x0 + ε]


with f(x) = g(x) for all x ∈ D∩E∩ [x0−ε, x0 +ε]. Prove that f is differentiableat x0 iff g is differentiable at x0.

f ′(x0) = limh→0

f(x0)−f(x0+h)h

|h|<ε= lim

h→0

g(x0)−g(x0+h)h = g′(x0).

Chapter 5

The Riemann Integral

5.1 The Riemann Integral

1. Use Theorem 5.2 to prove directly that the function f(x) = x3 is integrableon [0, 1].

Let P = {0, 1/n, 2/n, ..., 1}. Then, U(P, f) =∑nk=0

(k+1n

)3 · 1n and L(P, f) =∑n

k=0

(kn

)3 · 1n . Thus, U(P, f) − L(P, f) =

∑nk=0

(3k2+3k+1

n4

)= (n+1)3

n4 → 0 as

n→∞. Thus, f(x) is integrable.

2. Use Theorem 5.2. to prove directly that f(x) = x is integrable on [0, 1].Find the integral of f by finding a number A such that L(P, f) ≤ A ≤ U(P, f)for all partitions of [0, 1].

Let P = {0, 1/n, 2/n, ..., 1}. Then, U(P, f) =∑nk=0

k+1n ·

1n and L(P, f) =∑n

k=0kn ·

1n . Thus, U(P, f) − L(P, f) =

∑nk=0

1n2 = n+1

n2 → 0 as n → ∞. Thus,f(x) is integrable.

Now, U(P, f) = (n+1)(n+2)2n2 → 1

2 ←n+12n = L(P, f), so

´ 1

0f(x) dx = 1

2 .

3. Define f(x) = x if x is rational and f(x) = 0 if x is irrational. Compute´ 1

0f(x) dx and

´ 1

0f(x) dx. Is f integrable on [0, 1]? You may wish to look at the

results of Exercise 2.

From Q2,´ 1

0f(x) dx = 1

2 . But,´ 1

0f(x) dx = lim

||P ||→0

∑mi(xi+1 − xi) = 0.

(mi = 0 since there is always an irrational number between two rational ones.To see the truth of this statement, consider two terminating or repeating decimalexpansions and see that one can easily find a non-terminating, non-repeatingdecimal expansion between them.) Thus, f(x) is not integrable on [0, 1].

52

CHAPTER 5. THE RIEMANN INTEGRAL 53

4. A set A ⊂ [0, 1] is dense in [0, 1] iff every open interval that intersects[0, 1] contains a point of A. Suppose f : [0, 1] → R is integrable and f(x) = 0

for all x ∈ A with A dense in [0, 1]. Show that´ 1

0f(x) dx = 0.

If f is integrable, then´

=´

. If´≥ 0, then

´= 0, because mi = 0 for any

possible partition. (There will always be a point of A in-between any possible

partition points of [0, 1] since A is dense.) Thus,´ 1

0f(x) dx = 0. Similarly, if´

≤ 0, then´

= 0 because Mi = 0 for similar reasons.

5. Define f : [0, 2] → R by f(x) = 1 for 0 ≤ x ≤ 1 and f(x) = 2 for1 < x ≤ 2. Show that f ∈ R(x) on [0, 2] and compute the integral.

Let P = {0, 2/n, 4/n, ..., 1, ..., 2} where n = 2k. Then, U(P, f) − L(P, f) =∑nk=0(Mk − mk) · 2/n. By how we have insisted that our partition include a

point at x = 1, we know that mk = Mk for all k. Thus, this sum is 0. Hence,f ∈ R(x).

The integral is 3.

5.2 Classes of Integrable Functions

6. If f : [a, b]→ Ris decreasing, prove f ∈ R(x) on [a, b].

Let P = {x0 = a, x1 = a + b−an , x2 = a + 2(b−a)

n , ..., xn = b} be a partition.Then, Mk = f(xk) and mi = f(xk+1), since f is decreasing. Thus, U(P, f) −L(P, f)= lim

n→∞

n∑k=0

[f(xk) − f(xk+1)] · b−an = limn→∞

b−an ·[f (a) − f(x1)+f(x1) −

f(x2)+· · · − f(xn−1) + f(xn−1)− f(b)]. As we see, all the middle terms cancel,showing U(P, f) − L(P, f) = lim

n→∞b−an · [f(a) − f(b)] = 0. Thus, f ∈ R(x) on

[a, b].

*7. Suppose g : [a, b] → R is continuous except at x0 ∈ (a, b) and bounded.Prove that g(x) ∈ R(x) on [a, b]. See Exercises 24 and 25 for generalizations ofthis result.

First, even though g is not continuous at x0, g is bounded at x0. Thus,Mi = sup

xi≤x≤xi+1

g(x) and mi = infxi≤x≤xi+1

g(x) will still be defined for all i (since

there will be a right- and left-hand limit for g at x0). Thus, g ∈ R(x).

8. Find the integral of f(x) = x on [1, 3] using the techniques of Example5.5.


Consider the function g(x) = x2

2 and let P be any partition of [1, 3]. Then,∑f(ti)(xi − xi−1)

MVT=

∑[g(xi)− g(xi−1)] = g(3)− g(1) = 9/2− 1/2 = 4.

9. Assume f : [a, b]→ R is continuous and f(x) ≥ 0 for all x ∈ [a, b]. Prove

that if´ baf dx = 0, then f(x) = 0 for all x ∈ [a, b].

Let P0 be a partition of [a, b] and let {Pn} be a sequence of refinements.

Since´ baf(x) dx = 0, we have lim

n→∞

n∑i=0

supx∈(xi−1,xi)

f(x)(xi−xi−1) = 0. Now, since

f(x) ≥ 0 and (xi − xi−1) > 0 for all i, the only way that this sum can be 0 is iff(x) = 0 on each interval. But, since this f(x) is the supremum of the interval,this implies that f ≡ 0.

5.3 Riemann Sums

10. Prove Theorem 5.7. [Theorem 5.7 states “Suppose f : [a, b] → R isbounded. Then, f(x) ∈ R(x) on [a, b] iff, for each sequence {Pn}∞n=1 of markedpartitions with {µ(Pn)}∞n=1 converging to zero, the sequence {S(Pn, f)}∞n=1 isconvergent. If the condition is satisfied, then each of the sequences {S(Pn, f)}∞n=1will

converge to´ baf dx.”]

(⇒) Let P be a partition for [a, b] for which U = L. Then, since µ(Pn)→ 0,we know that there is N such that PN is a refinement of P (since partition pointsof Pn are required to get closer together, it follows that the intervals of Pn willeventually have to fit inside the established intervals of the fixed partition P )and so U(Pk, f) = L(Pk, f) for k ≥ N . Since L(Pk, f) ≤ S(Pk, f) ≤ U(Pk, f)

and L(Pk, f), U(Pk, f) →´ baf(x) dx, it follows that S(Pk, f) →

´ baf(x) dx by

the squeeze theorem.(⇐) Suppose that S(Pn, f) → S. Then, there is N such that |S(Pn, f) −

S)| < ε for all n > N . Thus, there exists P = PN such that |S(P, f) − S| <ε. Now, by definition of S and the fact that f is bounded, we can chooseti ∈ (xi−1, xi) such that S = U or S = L. (What we mean to say is thatthe convergence of S implies the convergence of U and L.) Thus, we get that

U(P, f) = L(P, f) as a natural consequence. Hence, S(P, f) =´ baf(x) dx.

11. Show that, for a > 1 and b > 1, the function f(x) = 1x is integrable

on [1, a] and on [b, ab]. Use the results of Section 5.3 to show that´ a

11x dx =´ ab

b1x dx.

The function f(x) is continuous and bounded on the intervals, so´ a

1dxx and´ ab

bdxx exist.


Choose {Pn} of partitions of [1, a] such that µ(Pn) → 0. We know thatS(Pn, f) →

´ a1dxx . Now, choose Qn = b · Pn. (By this we mean to set Qn

to be the partition obtained by multiplying every point of Pn by b.) Clearly,{Qn} is a sequence of partitions of [b, ab] and µ(Qn) → 0. Then, S(Pn, f) =∑ni=0

1ti

(xi−xi−1) while S(Qn, f) =∑ni=0

1bti·b(xi−xi−1) =

∑ni=0

1ti

(xi−xi−1).

Thus, S(Qn, f) = S(Pn, f) and since S(Pn, f)→´ a

1dxx and S(Qn, f)→

´ abb

dxx ,

we have that´ a

1dxx =

´ abb

dxx .

12. Suppose f ∈ R(x) on [0, 1]. Define

an =1

n

n∑k=1

f

(k

n

)

for all n. Prove {an}∞n=1 converges to´ 1

0f dt.

This is the integral of f over [0, 1] using the sequence of partitions Pn ={0, 1/n, 2/n, ..., 1} and ti = i/n:

∑ni=0 ti(xi−xi−1) =

∑nk=0 f

(kn

)·1/n = 1/n

∑nk=0 f

(kn

).

13. Suppose f : [a, b] → R is bounded and for each ε > 0 there is a parti-tion P such that for any refinements Q1and Q2of P , regardless of how marked,|S(Q1, f)− S(Q2, f)| < ε. Prove that f is integrable on [a, b].

If |S(Q1, f)−S(Q2, f)| < ε for any refinements of P , then there is a sequence{Qn} of refinements of P whose mesh is 0 for which |S(Qn, f)− S(Qm, f)| < ε

for all n,m. Thus {Qn} is Cauchy and so S(Qn, f) →´ baf(x) dx by Theorem

5.7.

14. Suppose f is integrable on [−b, b] and f is an odd function– that is,

f(−t) = −f(t) for all t ∈ [−b, b]. Prove that´ b−b f dx = 0. If f is even– that is,

f(−t) = f(t) for all t ∈ [−b, b]– prove that´ b−b f dx = 2

´ b0f dx.

Suppose f is odd. Let P = {x0 = 0, x1, ..., xn = b} be a partition of

[0, b]. Let´ b

0f(x)dx = lim

µ(P )→0

∑ni=0 f(x?i )(xi+1 − xi). Then,

´ 0

−b f(x) dx =

limµ(P )→0

∑ni=0 f(−x?i )(−xi−(−xi+1))

odd= limµ(P )→0

∑ni=0−f(x?i )(xi+1−xi) =−

´ b0f(x) dx =

−I. Hence,´ b−b f(x) = −I + I = 0.

Suppose f is even. Then,´ 0

−b f(x) dx = limµ(P )→0

∑ni=0 f(−x?i )(−xi−(−xi+1))

even=

limµ(P )

∑ni=0 f(x?i )(xi+1−xi)=

´ b0f(x) dx. Thus,

´ b−b f(x) dx = 2I = 2

´ b0f(x) dx.

15. Suppose that f : R → R is periodic and integrable on every closedinterval. If p is the period of f , prove that for any a ∈ R,

´ p0f dx =

´ a+p

af dx.


Write a = kp+ r where k ∈ Z and 0 ≤ r < p. Now,´ a+p

af(x)dx =

´ p0f(x+

a)dx=´ p

0f(x+kp+r)dx

per.=´ p

0f(x+r)dx=

´ p+rr

f(x)dx=´ prf(x)dx+

´ p+rp

f(x)dx=´ prf(x)dx +

´ r0f(x + p)dx

per.=´ prf(x)dx +

´ r0f(x)dx=

´ p0f(x)dx. (If the first

equality hasn’t been established yet, it is easily proven using the definition ofintegral and periodicity.)

5.4 The Fundamental Theorem of Integral Cal-culus

16. Use the Fundamental Theorem of Integral Calculus to compute thefollowing:

a.´ 3

0(x2 − x) dx

b.´ 4

−2(1− x3 − x2) dx

c.´ π/2

0x sinx2 dx

(a) F (x) = x3

3 −x2

2 ; F (3)− F (0) = 92 .

(b) F (x) = x− x4

4 −x3

3 ; F (4)− f(−2) = −78.

(c) F (x) = − cos(x2)2 ; F (π/2)− F (0) = 1

2 −cos(π2/4)

2 .

17. Define f : [0, 2]→ R by f(x) = 2x−x2 for 0 ≤ x ≤ 1 and f(x) = (x−2)2

for 1 < x ≤ 2. Prove that f is integrable on [0, 2] and find the integral of f over[0, 2]. Do not use Theorem 5.10, but rather find the integral by methods similarto those used in the proof of Theorem 5.8. [Theorem 5.8 is FTC and Theorem

5.10 is´ ba

=´ ca

+´ bc

.]

Let P = {x0 = 0, x1, ..., 1, ..., xn = 2} be a partition of [0, 2]. Then,´ 2

0f(x) dx = lim

µ(P )→0

∑ni=0 f(x?i )(xi+1−xi)= lim

µ(P )→0

(∑ki=0 f(x?i )(xi+1 − xi) +

∑ni=k f(x?i )(xi+1 − xi)

)where xk = 1.

Define the function F (x) =

{x2, 0 ≤ x ≤ 1(x−2)3

3 , 1 < x ≤ 2. We see that f(x) ≡

F ′(x). By MVT, we can select x?i such that f(x?i )(xi+1−xi) = F (xi+1)−F (xi).

Thus, limµ(P )→0

(∑ki=0 f(x?i )(xi+1 − xi) +

∑ni=k f(x?i )(xi+1 − xi)

)= limµ(P )→0

∑ki=0[F (xi+1)−

F (xi)] + limµ(P )→0

∑ni=k[F (xi+1)− F (xi)]

tel.=ser.

F (1−)− F (0) + F (2)− F (1+) = 2/3.

5.5 Algebra of Integrable Functions


*18. Suppose f and g are differentiable on [a, b] and f ′ and g′ are integrable

on [a, b]. Prove that f ′g and g′f are integrable on [a, b] and that´ baf ′g dx =

f(b)g(b)− f(a)g(a)−´ bag′f dx.

By the product rule, ddx (f(x)g(x)) = f ′(x)g(x) + g′(x)f(x). Integrating

both sides, f(x)g(x)|ba =´ baf ′(x)g(x) dx+

´ bag′(x)f(x) dx.

19. Suppose f ∈ R(x) on [a, b] and 1f is bounded on [a, b]. Prove that

1f ∈ R(x) on [a, b].

Since f ∈ R(x) and 1x ∈ R(x), 1

f(x) = 1x ◦ f(x) ∈ R(x).

20. Suppose f ∈ R(x) on [a, b] and f(x) ≥ 0 for all x ∈ [a, b]. Prove that√f ∈ R(x) on [a, b].

Since f ∈ R(x) and√x ∈ R(x) for x ≥ 0, we have

√f =√x ◦ f ∈ R(x).

21. Use Exercise 18 to calculate´ π/4

0x cosx dx. You may assume what you

need about the derivatives of the trigonometric functions.

By Q18,´ π/4

0x cosx dx = x2

2 cosx|π/40 −´ π/4

0sinx dx = π

√2

8 +√

22 − 1.

22. Suppose f is continuous on [0, 1]. Define gn = f(xn) for n = 1, 2, ....

Prove that{´ 1

0gn(x) dx

}∞n=1

converges to f(0).

For each n, we have min |f(x) ≤´ 1

0f(xn) dx ≤ max |f(xn)|. Thus, lim min |f(x)| ≤

lim´ 1

0f(xn) dx ≤ lim max |f(xn)| and since f is continuous, min |f (lim xn) | ≤´ 1

0f(xn) dx ≤ max |f (lim xn) |. Thus, f(0) ≤

´ 1

0f(x) dx ≤ f(0) so

´ 1

0f(x) dx =

f(0).

23. Define γn = 1+ 12 + 1

3 + · · ·+ 1n−´ n

11t dt. Prove that {γn}∞n=1 converges.

Notice that 1 + 1/2 + 1/3 + · · ·+ 1/n−´ n

1dtt →

∑[1k −´ k+1

kdt/t]

as n→∞.

By visualizing∑

1k as the sum of areas of rectangles of base 1 and height k, we

find that∑[

1k −´ k+1

kdt/t]

is just the difference between the aforementioned

rectangles and the curve 1/t. Next, 1k −´ k+1

kdtt

decr.< 1

k −1k+1 = 1

k(k+1) . Thus,∑[1k −´ k+1

kdtt

]<∑

1k(k+1)

calc< ∞. Thus, the sequence converges by the

comparison test (proven in ch. 1).

*24. Suppose g : [a, b]→ R is bounded and continuous except at x1, ..., xn ∈[a, b]. Prove that g ∈ R(x) on [a, b].


We have continuity on the intervals [a, x1), (x1, x2), ..., (xn, b], so clearly´ x1

ag(x) dx+

´ x2

x1g(x) dx+· · ·+

´ bxng(x) dx (?) exists. Now, let P = {a, x1, x2, ..., xn, b}

be a partition of [a, b]. We can easily find refinements Pn of P such thatµ(Pn) → 0 as n → ∞. But these refinements must be unions of refinements of

{a, x1}, {x1, x2}, ..., {xn, b} by how P was defined. Thus,´ bag(x) dx exists and

equals (?).

*25. Suppose {an}∞n=1 is a sequence of members of [a, b] converging to x0

in [a, b]. If f is bounded on [a, b] and continuous on [a, b] except at x0 and thepoints of the sequence {an}∞n=1, prove that f is integrable on [a, b].

We repeat a very similar argument to Q24. Let Pk = {an}kn=1 ∪ {a, a +(b−a)/n, ..., b}. Clearly, µ(Pk) → 0 as k → ∞. Since we have continuity at allpoints between partition points, the integral exists.

26. Prove that 13√

2≤´ 1

0x2

√1+x2

dx ≤ 13 .

On the interval (0, 1), we find that x2√

2≤ x2√

1+x2≤ x2. Integrating each

function over (0, 1), we establish the inequality 13√

2≤´ 1

0x2

√1+x2

dx ≤ 13 .

27. Suppose f and g are integrable on [a, b]. Define h(x) = max{f(x), g(x)}.Prove that h is integrable on [a, b].

´ bah(x) dx =

´Gg(x) dx +

´Ff(x) dx where F = {x|f(x) ≥ g(x)} and G =

{x|f(x) < g(x)}(= F c). Since the only places at which h is discontinuous isat the points where f(x) = g(x), and f(x−) 6= g(x−), we can invoke Q25 tosubstantiate our claim since this can only happen countably many times.

5.6 Derivatives of Integrals

28. Suppose f : [0, 1] → R is continuous and´ x

0f(t) dt =

´ 1

xf(t) dt for all

x ∈ [0, 1]. Prove that f(x) = 0 for all x ∈ [0, 1].

First, ddx

´ x0f(t) dt = f(x). Also, d

dx

´ 1

xf(t) dt = − d

dx

´ x1f(t) dt = −f(x).

Thus, f(x) = −f(x) which implies that f ≡ 0.

29. Suppose f and g are continuous on [a, b] and´ baf(x) dx =

´ bag(x) dx.

Prove that there is c ∈ [a, b] such that f(c) = g(c).

Let F (x) =´ x

0f(t) dt and G(x) =

´ x0g(t) dt. By Cauchy’s MVT, there is

a c ∈ (a, b) such that (F (b) − F (a))G′(c) = (G(b) − G(a))F ′(c). Since F (b) −F (a) =

´ baf(x) dx =

´ bag(x) dx = G(b) − G(a) and G′(x) = g(x) and F ′(x) =

f(x), the result follows.


30. Find f ′ where f is defined on [0, 1] as indicated:a. f(x) =

´ x0

√t2 + 1 dt

b. f(x) =´ 1

xcos 1

t+1 dt

c. f(x) =´ 2x

x2 sin t2 dt

d. f(x) =´√xx

11+t3 dt

(a) f ′(x) =√x2 + 1

(b) −f(x) =´ x

1cos dt

t+1 ⇒−f′(x) = cos 1

x+1 ⇒f′(x) = − cos 1

x+1 .

(c) f(x) =´ 0

x2 sin t2 dt+´ 2x

0sin t2 dt⇒f ′(x) = − d

dx

´ x2

0sin t2 dt+ d

dx

´ 2x

0sin t2 dt⇒f ′(x)

chain=

rule

− sin((x2)2) · (2x) + sin(2x2) · 2.

(d) f(x) = −´ x

0dt

1+t3 +´√x

0dt

1+t3 ⇒f′(x) = − 1

1+x3 · 1 + 11+(√x)3· 1/2x−1/2.

31. Let f : R→ R be continuous and δ > 0. Define g(t) =´ t+δt−δ f(x) dx for

all t ∈ R. Prove that g is differentiable and compute g′.

We see that g(t) =´ t+δt−δ f(x) dx =

´ 0

t−δ f(x) dx +´ t+δ

0f(x) dx= −

´ tδf(x−

δ) dx+´ t−δ f(x+ δ) dx. The last equality is a valid integral since f(t+ const.)

is continuous, and thus integrable. The resulting function is thus differentiable(as all integrals of continuous functions are). Thus, g′(x) = f(t+ δ)− f(t− δ).

*32. Suppose f : [a, b] → R is continuous and g : [c, d] → [a, b] is differen-

tiable. Define F (x) =´ g(x)

af(t) dt for x ∈ [c, d]. Prove that F is differentiable

and compute F ′.

Since we have continuity of f , we can use FTC and since g is differentiable,we can use the chain rule. Thus, F ′(x) = f(g(x))g′(x).

5.7 Mean-Value and Change-of-Variable Theo-rems

33. Suppose f : R → R is continuous and has period p so that f(x + p) =

f(x) for all x ∈ R. Show that´ x+p

xf(t) dt is independent of x in that, for all

x, y,´ x+p

xf(t) dt =

´ y+p

yf(t) dt. Show, then, that

´ p0

[f(x+a)−f(x)]dx = 0 forany real number a. Conclude that for any real number a, there is x such thatf(x+ a) = f(x).

By Q15,´ p

0=´ a+p

afor any a ∈ R. Thus, for all x, y,

´ x+p

x=´ y+p

y=´ p

0.

Then,´ p

0[f(x+ a)− f(x)]dx =

´ p0f(x+ a) dx−

´ p0f(x) dx=

´ a+p

af(x) dx−


´ a+p

af(x) = 0. The first term of the last equality comes from the change-of-

variable theorem and the second comes from Q15.

*34. Prove the following variation on Theorem 5.18. Assume that φ :[a, b] → R is differentiable, 1-1, and increasing with φ(a) = c and φ(b) = d. Iff : [c, d]→ R is integrable on [c, d] and (f ◦φ)φ′ is also integrable on [a, b], then´ dcf(x) dx =

´ baf(φ(t))φ′(t) dt.

It is easily seen that f(φ(b))φ′(b)−f(φ(a))φ′(a) =´ baf(φ(t))φ′(t) dt=

´ b0f(φ(t))φ′(t) dt−´ a

0f(φ(t))φ′(t) dt=

´ dcf(x) dx. The last equality comes from FTC and Q32.

35. Use Theorem 5.18 to evaluate the integrals [Theorem 5.18 is the change-of-variable theorem]:

a.´ 3

03√

1 + x2x dx

b.´ 4

1(√x+2)3√x

dx

c.´√3

1

√x2−9x dx

d.´ 1

0x2

√1−x2

dx

(a)´ 3

0(1 + x2)1/3 · x dx = 1/2

´ 10

1u1/3du = 1/2

[34u

4/3]10

1=

3(

10·101/3−1

)8 .

(b)´ 4

1(√x+2)3√x

dx = 2´ 4

3u3du = 1/2u4|43 = 175

2 .

(c) Clearly, something is wrong. Both 1 and√

3 have squares less than orequal to 9, so this result is complex. It isn’t even a nice complex integral either.The answer is about ei(1.57) · 1.47.

(d) Recognizing ddx (sin−1(x)) = 1√

1−x2, let u = sin−1(x), then, x2 = sin2(u)

and the integral becomes´ π/2

0sin2(u) du = π/4.

36. Use Theorem 5.18 to establish the result in Exercise 11.

1b

´ a1dxx =´ a

1dxbx =

´ abb

duu .

37. Use Theorem 5.18 to establish the result in Example 5.6. [Example 5.6

establishes the identity´ 1

adxx2+1 =

´ 1/a

1dxx2+1 .]

´ 1

adxx2+1 =

´ 1/a

1a·du

(au)2+1 = a ·´ 1/a

1du

(au)2+1 = a ·´ 1/a

1dx

a[(x)2+1] =´ 1/a

1dxx2+1 .

Miscellaneous

38. Assume f : [a, b] → R is continuous, f(x) ≥ 0 for all x ∈ [a, b], and

M = sup{f(x) : x ∈ [a, b]}. Show that

{[´ ba

[f(x)]ndx]1/n}∞

n=1

converges to


M .

Let Iδ = {x ∈ [a, b] : |f(x)| ≥ M − δ}. (The set exists by MVT.) Now,

M(b− a)1/n ≥(´ b

a[f(x)]ndx

)1/n

≥(´

Iδ[f(x)]ndx

)1/n

≥ (M − δ)`(Iδ)1/n where

`(Iδ) is the sum of lengths of intervals composing Iδ. (The first inequality is theupper bound of the integral over (a, b), and the third is a similar lower boundfor the integral over Iδ.) Since any set real number raised to 1/n tends to 1 as

n → ∞, we see that M ≥ lim(´ b

a[f(x)]ndx

)1/n

≥ (M − δ). Taking δ → 0, we

see that lim(´ b

a[f(x)]ndx

)1/n

= M .

39. For each positive integer n, define an = 1n

[(1n

)2+(

2n

)2+ · · ·+

(nn

)2].

Find the limit of the sequence {an}∞n=1.

an = 1n

∑nk=1

(kn

)2= (n+1)(2n+1)

6n2 → 13 . (The second equality was proven

somewhere in Chapter 0. I don’t know what this has to do with integration.)

40. For each positive integer n, define an = 1n

[sin π

n + sin 2πn + · · ·+ sin nπ

n

].

Find the limit of the sequence {an}∞n=1.

an =∑nk=1

1/n · sin(kπn

)which is easily seen to be a left-handed Riemann

sum. Thus, lim an =´ 1

0sin(πx) dx = 2

π .

41. If f and g are integrable on [a, b], show that[´ bafg dx

]2≤´ baf2 dx

´ bag2 dx.

´ ba

(f(x) + tg(x))2dx =

´ ba

(f(x))2dx+2´ baf(x)g(x)dx+ t2

´ ba

(g(x))2dx. Set

t =

´ ba

(f(x))2dx´ baf(x)g(x)dx

(unless fg ≡ 0, in which case the inequality is apparent).

Then, by performing straightforward, careful algebra using this value of t, we

find that´ ba

[f(x)+tg(x)]dx =(´ b

a(f(x))2dx

)´ ba (f(x))2dx´ ba

(g(x))2dx(´ baf(x)g(x)dx

)2 − 1

.

Since LHS≥0 and´ ba

(f(x))2dx ≥ 0, it follows that(´ b

af(x)g(x)dx

)2

≤´ ba

(f(x))2dx´ ba

(g(x))2dx.

42. Use Exercise 41 to prove the Minkowski inequality– that is, if f and

g are integrable on [a, b], then[´ ba

[f(x) + g(x)]2dx]1/2

≤[´ ba

[f(x)]2dx]1/2

+[´ ba

[g(x)]2dx]1/2

.

´ ba

(f(x)+g(x))2dx =´ ba

(f(x))2dx+2´ baf(x)g(x)dx+

´ ba

(g(x))2dx≤´ ba

(f(x))2dx+

2(´ b

a(f(x))2dx

)1/2 (´ ba

(g(x))2dx)1/2

+´ ba

(g(x))2dx=

((´ ba

(f(x))2dx)1/2

+(´ b

a(g(x))2dx

)1/2)2

.


Take the square root of both sides to see the inequality.

Chapter 6

Infinite Series

6.1 Convergence of Infinite Series

1. Let {an}∞n=1 be a sequence of real numbers. Prove that∑∞n=1(an− an+1)

converges iff {an}∞n=1converges. If∑∞n=1(an − an+1) converges, what is the

sum?

(⇒) Suppose∑

(an − an+1) < ∞. Then, we must have (an − an+1) → 0(test for divergence). Let m > n. Then, for any ε > 0, there is N such that|an − an+1| < ε

m−n for all n > N . Thus, for m > n > N , |an − am| =|an−an+1 +an+1−an+2 + · · ·+am−1−am| ≤ |an−an+1|+ · · ·+ |am−1−am| <(m− n) ε

(m−n) = ε. Thus, {an} is Cauchy and thus convergent.

(⇐) Suppose an → L. Then, {an} is Cauchy, so an − an+1 → 0. Then,∑(an − an+1) = a1 − a2 + a2 − a3 + · · · = lim(a1 − an) = a1 − L.

2. Let∑∞n=1 an converge. Let {nk}∞k=1 be a subsequence of the sequence of

positive integers.For each k, define bk = ank−1+1 + · · ·+ ank where n0 = 0.Prove that

∑∞k=1 bk converges and that

∑∞k=1 bk =

∑∞n=1 an.

This is the identical sum. All that is happening in∑bk is that we choose

terms of an and add the terms between them (including the an terms them-selves). Thus, there is really no analysis to perform here.

3. Prove that∑∞n=1 2nrn converges if |r| < 1/2 and find the sum.

For a finite series,∑Nn=0(2r)n = 1 + 2r + 4r2 + · · · . Multiplying both sides

by (1 − 2r), we see that (1 − 2r)∑Nn=0(2r)n = (1 − 2r)(1 + 2r + 4r2 + · · · )=

1 − 2r + 2r − 4r2 + 4r2 − · · ·= 1 − (2r)N+1. Thus,∑Nn=0(2r)n = 1−(2r)N+1

1−(2r) .

63

CHAPTER 6. INFINITE SERIES 64

(We will use this for Q4.) If we instead set the lower bound as n = 1, we find∑Nn=1(2r)2 = 2r−(2r)N+1

1−2r . Thus,∑∞n=1(2r)n = lim 2r−(2r)N+1

1−2r

2r<1= 2r

1−2r .

4. Prove that∑∞n=0 3−n converges and find the limit.

Q3 shows that the sum is finite. The sum is 11−1/3 = 1

2/3 = 32 .

5. Determine whether the series∑∞n=1(√n+ 1−

√n)converges or diverges.

Justify your conclusion.∑Nn=1(√n+ 1−

√n) =

√2−√

1+√

3−√

2+ · · ·+√N−√N − 1+

√N + 1−√

N=√N + 1−

√1. Taking N →∞, we see that the sum is ∞.

6. Use induction to show that 1 + 1√2

+ · · ·+ 1√n≥√n for n ≥ 1. Can you

use this fact to determine whether the series∑∞n=1

1√n

converges or diverges?

For n = 1, the inequality is true. Suppose it is true for n = N . Then,1 + 1√

2+ · · · + 1√

N+ 1√

N+1≥√N + 1√

N+1. This inequality holds if and only

if√N + 1

(1 + 1√

2+ · · ·+ 1√

N

)+ 1 ≥

√N + 1 holds. By subtracting 1 from

each side, we obtain√N + 1

(1 + 1√

2+ · · ·+ 1√

N

)≥√N which is immediately

apparent since 1 + 1√2

+ · · ·+ 1√N≥√N by the inductive hypothesis. Thus, the

inequality has been proven by induction.Now,

∑Nn=0

1√n≥√N , so as N →∞, we see that

∑1√n

=∞.

7. Use induction to show that 2(1 + 1

8 + 127 + · · ·+ 1

n3

)< 3− 1

n2 for n ≥ 2.Does the series

∑∞n=0

1n3 converge? Justify your conclusions.

One can easily find that the statement is true for n = 2. Suppose it is true

for n = N . Then, 2(

1 + 18 + 1

27 + · · ·+ 1N3 + 1

(N+1)3

)< 3− 1

N2 + 2(N+1)3 which

is equivalent to (N+1)3 ·2(1 + 1

8 + 127 + · · ·+ 1

N3

)+2 < 3(N+1)3− (N+1)3

N2 +2.As expected, we subtract 2 from both sides and find that this is equivalent to(N + 1)3 · 2

(1 + 1

8 + 127 + · · ·+ 1

N3

)< (N + 1)3

[3− 1

N2

]which is immediately

apparent from the inductive hypothesis. Thus, the inequality has been provenby induction.

Now,∑Nn=0

1n3 < 1/2

(3− 1

N2

). Taking N → ∞, we find that

∑1n3 < 3

2 ,proving convergence.

8. Write an infinite series for the repeating decimal for the rational number5/9 and prove that it converges to 5/9.

We claim that 59 =

∑n=1 5(10)−n. That

∑n=1 5(10)−n = 5/9 can be proven

by the division algorithm or it can be proven by a method similar to Q9.


9. Find the rational number that is the limit of the repeating decimal 0.15.

We have x = .151515..., so 100x = 15.1515.... thus, 100x− x = 15 implying99x = 15 or x = 15

99 .

10. Prove Theorem 6.3. [Theorem 6.3 is∑

(αan+βbn) = α∑an+β

∑bn.]

We already have the result that lim(cxn+dyn) = c limxn+d lim yn, so sincean infinite series is the limit to the sequence of partial sums, we have the desiredresult.

11. Prove that the series∑∞n=2

(5

2n −3

5n

)converges and find the limit.

By Q10, the sum can be written as 5 ·∑n=2(1/2)n − 3 ·

∑(1/5)n. The sum

converges by an argument similar to Q3. The limit is 5 · 1/41−1/2 − 3 · 1/25

1−1/5 .

12. Show that the sequence∑∞n=1

(2n + 1

2n

)diverges. Use Theorem 6.3.

We have seen that∑

1n diverges, so the sequence in question must diverge

by Theorem 6.3.

6.2 Absolute Convergence and the ComparisonTest

13. Suppose∑∞n=1 an converges absolutely and {bn}∞n=1 is bounded. Prove

that∑∞n=1 anbn converges absolutely.

Let M = sup{bn}. Then,∑|anbn| ≤

∑|anM | = |M | ·

∑|an| <∞.

14. If∑∞n=1 anconverges absolutely, prove that

∑∞n=1 a

2n converges. Is the

statement true if∑∞n=1 an converges conditionally?

Let M = sup{an} (which must be finite since an → 0 as a necessary con-dition for convergence. We then have

∑|a2n| =

∑|an||an| <

∑|an||M | <

|M |∑|an| <∞.

15. Determine which of the following infinite series converge.a.∑∞n=1

1(n+1)(2n−1)

b.∑∞n=1

nn+1

c.∑∞k=1

23k

d.∑∞m=1

√m+1−

√m

m

e.∑∞m=1

3m

5m+1

f.∑∞m=1

1m2

(100m+1

m

)


a,e, and f converge; the rest diverge.

16. (Limit-comparison test.) Prove the following generalization of Theorem6.5. Suppose that

∑∞n=0 an and

∑∞n=0 bn are series of positive terms such that{

anbn

}∞n=1

converges to L 6= 0. Then∑∞n=0 an and

∑∞n=0 bn either both diverge

or both converge. What can be concluded if L = 0?

We have lim anbn

= L, so lim an = L · lim bn. Suppose∑an <∞. Then, (by

definition of convergence) there is N for which 0 =∑n=N an =

∑n=N L · bn =

L ·∑n=N bn. Since L 6= 0, we conclude that

∑n=N bn = 0 and so

∑bn <∞.

If∑an = ∞, then the proof is very similar. (Just change the first equality to

<.)We can see from the proof that if L = 0, than we cannot determine if the tail

sum of∑bn goes to 0. Thus, we cannot conclude convergence or divergence.

17. Apply the limit-comparison test (Exercise 16) to the following series.

a.∑∞n=1

√n+17

n2−64n−112

b.∑∞n=1

1√n2+64n+

√n2+3

We see that the quotient of the sequences is√n+17

(n2−54n−112)(√n2+64n+

√n2+3)

which clearly tends to 0. We cannot conclude anything about the series. (Forthe good of the order, (a) converges while (b) doesn’t since (a) is of the order∑

1np with p > 1 while (b) is of the order

∑1np with p = 1.)

6.3 Ratio and Root Tests

18. Give an example of an infinite series for which Theorem 6.8 yields a

conclusion, but Theorem 6.9 does not. [Theorem 6.8 is∣∣∣an+1

an

∣∣∣ < 1 for n > N

for some N⇒∑an is absolutely convergent, and Theorem 6.9 is lim

∣∣∣an+1

an

∣∣∣ <1⇒

∑an is absolutely convergent.]

Examine the sum∑

1n2 . Now, lim

1/(n+1)2

1/n2 = lim n2

n2+2n+1 = 1 and so the

traditional ratio test is inconclusive. However, the fraction on the RHS of thefirst equality is always less than 1 (top<bottom), so Theorem 6.8 allows us toconclude that

∑1n2 is absolutely convergent.

19. Use Theorem 6.9 to determine the values of r for which∑∞n=0 nr

n

converges.

lim∣∣∣ (n+1)rn+1

nrn

∣∣∣ = lim∣∣∣nrn+1

nrn + rn+1

nrn

∣∣∣= lim∣∣r · nrnnrn + r · r

n

nrn

∣∣= lim |r + r/n| =r. Thus, the sum converges if r < 1.


20. Prove that {nxn}∞n=1 converges to zero if |x| < 1.

By Q19, we have shown the sequence of partial sums of this sequence con-verges. A necessary condition for the convergence of the series is a convergenceof the sequence to 0.

21. Prove the following version of the root test. If∑∞n=0 an is an infinite

series and{√|an|

}∞n=1

converges to L, then

1. if L < 1, the series converges absolutely; and2. if L > 1, the series diverges.

There must be a typo in the book, since if lim√|an| 6= 0, the series must

diverge. Thus, we will assume that we should replace√|an| with n

√an in the

problem statement.If n√|an| → L, then |an| → Ln. Immediately, we see that

∑an can only

hope to converge if L < 1 (otherwise an 9 0). Next, if |an| → Ln with L < 1,

then there is N for which∑n=N |an|

±ε=∑n=N L

n <∞. (See Q3 for the reasonthat this converges.) Thus,

∑an is absolutely convergent.

22. If {an}∞n=1 is a sequence of positive real numbers such that{an+1

an

}∞n=1

converges to L, prove that{n√an}∞n=1

converges to L.

Suppose that an+1/an±ε= L for all n > N for some N . This implies that

an+1±ε= L · an. For any k, we see that aN+k

±ε= L · aN+k−1

±ε= L2 · aN+k−2

±ε=

· · · ±ε= LkaN using the previous formula. Substituting n = N + k (n > N), we

see that an±ε= Ln−N · aN which is the same as an

1/n ±ε= Ln−Nn aN

1/n. Takingn→∞, lim n

√an = L1 · a0

N = L.

23. Prove that{

n√n!n

}∞n=1

converges and find the limit. You might want to

look at Example 6.12.

n+1√

(n+1)!

n+1 · nn√n!

= (n+1)1/(n+1)(n)1/(n+1)···(1)·n(n+1)(n)1/n(n−1)1/n···(1)

. We see that for all n, the

denominator is greater than the numerator, so the ratio is less than 1. Thus,∑ n√n!n <∞ by the Ratio Test. Thus,

n√n!n → 0.

24. Test the following series for convergence.a.∑∞n=1 n

ppn, p > 0b.∑∞n=1( n

√n− 1)n

c.∑∞n=1 n

−1−(1/n)

d.∑∞n=2

1pn−qn , 0 < q < p


(a) Suppose that lim (n+1)ppn+1

nppn = L. Then, doing some algebra, we see that

lim(n+1n

)p · p = L since n+1n → 1, we find that L = p and so the sum is finite

for 0 < p < 1.(b) Suppose that lim( n

√n− 1)n)1/n = L. This means that limn1/n = L+ 1.

Taking the natural log of both sides, lim lnnn = ln(L + 1). By l’Hospital’s rule,

the LHS is equal to lim 1/n1 = 0. Hence, 0 = ln(L + 1) and so 1 = L + 1 or

L = 0 < 1. Thus, the series is absolutely convergent.(c) Using algebra, the sum becomes

∑1

n·n1/n . Using the ratio test, we see

that n·n1/n

(n+1)(n+1)1/n< 1 for all n. Thus, the sum is absolutely convergent.

(d)∣∣∣ 1pn−qn

∣∣∣1/n ≤ ∣∣∣ 1(pn)1/n−(qn)1/n

∣∣∣ = 1|p−q| < 1 iff |p−q| < 1. (The inequality

is a simple case of Minkowski’s inequality [see Q5.42].) Thus,∑n=2

1pn−qn <∞

if |p− q| < 1.

25. Determine those values of p for which∑∞n=2

1n(log n)p converges.

Using the ratio test, examine lim n(logn)p

(n+1)(log[n+1])p = lim[

nn+1 ·

(logn

log(n+1)

)p].

If p ≥ 0, both factors have the numerator is greater than the denominator,which would imply convergence. If p < 0, then the second factor has a quotientgreater than 1. Thus, for p < 0, the sum is infinite.

26. For each positive integer n, define γn = 1 + 12 + · · ·+ 1

n − log n. Prove

that {γn}∞n=1 converges. (Use the fact that log x =´ x

1dtt for x > 0.)

We could visualize 1+ 12 + · · ·+ 1

n as the area of n rectangles, each with base1 and height 1

n . As mentioned in the hint, we can visualize log(n) as the areaunder the curve 1

t from 1 to n. Thus, the difference 1 + 12 + · · · + 1

n − log(n)can be visualized by the area between the afformentioned rectangles and thecurve 1

t from 1 to n. The following is an illustration of the first three rectanglesdescribed along with the curve 1

t . The area that {γn} represents is equal to thesum of areas of the rectangles above 1

t .


The area mentioned at the end of the previous paragraph is surely less than

the sum of areas of rectangles whose base is 1 and height is(

1n −

1n+1

). In

the figure, these rectangles are the “sub-rectangles” defined by the portionof the whole rectangles above the dashed lines. Thus,

∑n=2

1n − log n <∑

n=2

(1n −

1n+1

)= 1

2 − lim 1n+1 = 1

2 <∞. Thus, {γn} converges.

6.4 Conditional Convergence

27. If∑∞n=1 an converges and {bn}∞n=1 is monotone and bounded, prove that∑∞

n=1 anbn converges.

Since {bn} is monotone and bounded, bn → L for some L. Thus, for any

ε there is N1 for which bn±ε= L for n > N1. Also, since

∑an < ∞, there is

N2 for which∑n=N2

an±ε= 0 for the same ε. Let N = max(N1, N2). Then,∑

n=N anbn <∑n=N (L + ε)an = L

∑n=N an + ε

∑n=N an. We see that the

RHS becomes L(0 + ε) + ε(0 + ε) = ε(L + ε). Since ε is arbitrary, we can takeε→ 0 to conclude that RHS→ 0, or, in other words,

∑anbn <∞.

28. Suppose {an}∞n=1 is a sequence of positive real numbers convergingto 0 such that an ≥ an+1 for all n. Then, by the alternating series test,∑∞n=1(−1)nan converges; call the sum S. Let Sn be the nth partial sum of∑∞n=1(−1)nan. Prove that |Sn − S| ≤ an+1.

Since an ≥ an+1, we can easily find that |Sn − Sn+1| = |Sn − (Sn +(−1)n+1an+1)| = |an+1|. Similarly, |Sn−Sn+2| = |Sn−(Sn+1+(−1)n+2an+2)| =|Sn − (Sn + (−1)n+1an+1 + (−1)n+2an+2)| = |an+1 − an+2| < an+1. In general,

|Sn − Sn+k| =∣∣∣∑n+k

`=n+1(−1)`a`

∣∣∣ TI<∑n+k` odd |a` − a`+1|. Now, taking a random

term of the series on the RHS of the inequality, say |ak − ak+1|, we know that


the next term is |ak+2 − ak+3| < ak+2 < ak+1. Thus, what is being addedin subsequent terms is never more than what is initially subtracted. Hence,{|Sn − Sk|}∞k=n+1 is a decreasing sequence. Thus, |Sn − S| ≤ an+1.

29. Let∑∞n=1 an be an infinite series and {nk}∞k=1 a subsequence of the

sequence of positive integers. Prove that if∑∞n=1 an converges absolutely, then∑∞

k=1 ank converges absolutely. What can be concluded if∑∞n=1 an converges

conditionally?

We see that∞ >∑|an| ≥

∑|ank | since all terms are positive. Thus,

∑ank

is absolutely convergent.If∑an is conditionally convergent, then we cannot say anything about the

convergence of subseries a priori. For example, by the alternating series test,∑(−1)n 1

n <∞. However, we have∑n even(−1)n 1

n =∞ and∑n odd(−1)n 1

n <∞.

30. Prove that∑∞k=1

sin kxk converges.

Let ak = sin kx. Then, sin kx =cos(n−1

2 )x−cos(n+12 )x

2·sin( x2 )(a formula usually

derived in a trigonometry class), so the partial sums are always telescoping andthus, since the trig functions always give outputs less than or equal to 1, we canbound each sum. Next, set bk = 1

k , which we know to be decreasing toward 0.By Theorem 6.12,

∑anbn <∞.

31. Test each of the following series for absolute convergence, conditionalconvergence, or absolute convergence.

a.∑∞n=1

(−1)nnn2−5n+1

b.∑∞n=1

(−1)n(n−5)n3−7n−9

c.∑∞n=1(√n+ 1 −

√n)an where an = (−1)f(n) and f(n) =

[n2

]; that is,

f(n) is the integer part of n2 .

d.∑∞n=1(−1)n n

n+1

(a)∑ (−1)nn

n2−5n+1 <∑ (−1)nn

n2−5n =∑ (−1)n

n−5 <∞. Thus, the sum is conditionallyconvergent by the alternating series test.

(b)∑ (−1)n(n−5)

n3−7n−9 <∑n=9

(−1)nnn3−8n < ∞ since the sum is of the form

∑1np

for p > 1. Thus, the sum is absolutely convergent.(c)∑

(√n+ 1−

√n)an is a telescoping series which is convergent since 0 <

an < 1. Since all terms are positive, it is absolutely convergent.(d)

∑(−1)n n

n+1 is conditionally convergent by the ratio test.

6.5 Power Series


32. Suppose that∑∞n=0 an diverges and that {an}∞n=0is bounded. Prove that

the radius of convergence of∑∞n=0 anx

n is equal to 1.

Let M = max{an}. Then,∑anx

n < M ·∑xn implying that the radius of

convergence is 1.

33. Suppose∑∞n=1 an converges conditionally. Prove that the radius of

convergence of∑∞n=1 anx

n is equal to 1.

lim an+1xn+1

anxn= lim an+1

an· x

n+1

x . Since∑an <∞, we have an+1

an≤ 1. Hence,

lim an+1

an· x

n+1

x < limx. Thus, the radius of convergence is 1.

34. Show that∑∞n=1 n!xn converges only for x = 0.

lim (n+1)!xn+1

n!x = lim (n+1)x1 =∞ unless x = 0.

35. Show that∑∞n=1

xn

n converges iff −1 ≤ x < 1.

limxn+1

n+1xn

n

= lim xn+1nxn(n+1) < lim xn

n = x. Thus, the radius of convergence is 1.

If x = −1, it is an alternating series whose terms are decreasing, so the serieswould be convergent.

If x = 1, then the sum would be∑

1/n =∞.

36. Determine the radius of convergence of the power series∑∞n=1

2nn!nn x

n.

lim2n+1(n+1)!

(n+1)n+1 xn+1

2nn!nn xn

= lim 2n+1(n+1)!·nn2nn!(n+1)n+1 x = lim 2(n+1)·nn

(n+1)n+1 x < lim 2(n+1)n+1

(n+1)n+1 x <

2x. Thus, the radius of convergence is 1/2.

37. Determine the interval of convergence of the power series∑∞n=1

1npx

n

for different values of p.

Let lim n

√xn

np = lim xn√np = x

limnp/n= L. Then, log(L) = log(x)−lim p/n log(n).

Thus, L = exp[log(x) − lim p/n log(n)]H= exp[log(x) − 0] = x. Thus, radius of

convergence is 1, regardless of the value of p.If x = 1, then the sum is

∑1np which is convergent if p > 1.

If x = −1, then the sum is∑ (−1)n

np which is convergent for p > 0.

38. Show that the power series∑∞n=0 anx

n and∑∞n=1 nanx

n−1 either bothconverge for all x, both converge for x = 0, or both have the same finite nonzeroradius of convergence.

Suppose that lim an+1xn+1

anxn= lim an+1

an· x = L. Then, by l’Hospital’s rule,

we must have lim (n+1)an+1xn

nanxn−1 = lim (n+1)an+1

nan· x = L. Thus,

∑n=1 nanx

n−1


converges or diverges with∑n=0 anx

n. Also, by the factorization of the limit,we see that the radii of convergence must coincide.

39. Let {an}∞n=1 be a sequence of real numbers bounded from below, andlet A = {p : there is a subsequence of {an}∞n=1 converging to p}. Suppose A isnonvoid. Define a = inf A. Prove that a ∈ A and that, for each ε > 0, thereis N such that for all n ≥ N, a− ε < an and there are infinitely many m suchthat am < a+ ε.

A is the set of limit points of {an}. Since a = inf A, either a is the smallestlimit point of the sequence, or else we have a sequence of limit points convergingto a. Suppose the latter is true. Now, each limit point has an infinity of terms of{an} in any ε neighborhood of itself. Thus, since a contains an infinity of limitpoints within any ε neighborhood (definition of convergence), a consequentlycontains an infinity of terms of {an} within any ε neighborhood. Thus, a is alimit point of {an}. Thus, a ∈ A.

Since a is itself a limit point with ank → a, we know that for any ε > 0,there is N so that we have −ε < ank − a < ε for all nk > N . As a naturalconsequence, a− ε < ank < a+ ε, and so, obviously, ank < a+ ε for all nk > N .Thus, since {ank} ⊆ {an}, there are infinitely many m such that an < a+ ε.

For the same ε and N as above, we have a−ε < ank for all nk > N . However,since a is the smallest limit point, this must hold true for all n; that is, a−ε < anfor all n > N .

40. State and prove theorems similar to Theorems 6.8 and 6.21 in terms of

limn→∞

inf

∣∣∣∣an+1

an

∣∣∣∣ and limn→∞

sup

∣∣∣∣an+1

an

∣∣∣∣ .(Theorem 6.8) Let

∑an be an infinite series of nonzero terms. Then, if

L = lim sup∣∣∣an+1

an

∣∣∣ < 1, then∑|an| < ∞ and if L = lim inf

∣∣∣an+1

an

∣∣∣ > 1, then∑an =∞.

(Proof) If L < 1, then given ε > 0, there is N for which∣∣∣an+1

an

∣∣∣ < 1− ε for all

N . Thus, |an+1| < (1−ε)|an| and so∑n=N |an+1| = (1−ε)|aN |+(1−ε)2|aN |+

(1− ε)3 + · · · = |aN |∑n=N (1− ε)n. Since 0 < 1− ε < 1 for small ε, the sum is

(absolutely) convergent.

If L > 1, then for the same ε, there is N such that∣∣∣an+1

an

∣∣∣ > 1 + ε for

all n > N . Performing the same steps as above, we see that∑n=N |an+1| >

|an|∑n=N (1 + ε)n =∞. Thus, the sum is divergent.

(Theorem 6.21) Let∑anx

n be a power series with an 6= 0 for all n. Then,

let L = lim sup∣∣∣an+1

an

∣∣∣ and L = lim inf∣∣∣an+1

an

∣∣∣. Then,∑anx

n <∞if |x| < 1L

and∑anx

n =∞ if |x| > 1L .

(Proof) Follows from the altered Theorem 6.8.


41. Find the interval of convergence of the power series∑∞n=1 anx

n wherean is as given below. Be sure to check the endpoints.

a. an = (n+ 1)(n+ 2)b. an = sinnc. an = 3−n

√n

d. an = (n!)2

(2n)!

e. an =(1 + 1

n

)n2

f. an = 3n

n + 2n

n2

(a) lim∣∣∣ (n+2)(n+3)

(n+1)(n+2)

∣∣∣ = lim∣∣∣n+3n+1

∣∣∣ = 1. Thus, interval of convergence is (−1, 1).

The endpont x = −1 allows for convergence, but x = 1 causes the sum todiverge.

(b) The radius of convergence is 1 by Q32. Neither of the endpoints converge.

(c) lim√n+1·3n

3n+1·√n

= lim√n+1

3√n

= L ⇒ L2 = lim 19 ·

n+1n = lim 1/9(1 + 1/n) =

1/9 ⇒ L = 1/3. Thus, the radius of convergence is 3. The negative endpointcauses convergence (AST).

(d) lim [(n+1)!]2·(2n)!(2n+2)!(n!)2 = lim (n+1)2(n!)2·(2n)!

(2n+2)(2n+1)(2n!)(n!)2 = lim (n+1)!(2n+2)(2n+1) = ∞ (by

l’Hospital’s rule, e.g.). Thus, the series converges only for x = 0.

(e) lim(1+ 1

n+1 )(n+1)(n+1)

(1+ 1n )

n·n = lim en+1

en = e > 1. Hence, the radius of conver-

gence is 1/e. The negative endpoint converges.

(f) lim (3n+1(n+1)+2n+1)·n2

(n+1)2·(3nn+2n) = 3. Thus, the radius of convergence is 1/3.

Neither endpoint converges.

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Intro to Analysis Solutions