Solutions, Assignment #6
Section 21.
x2 4x 5
(x + 1)(x 5)
= lim
= lim (x + 1) = 6. Set f (5) = 6.
x5
x5
x5
x5
x5
3. lim
6.
(a) True. Set g(x) k , constant. Then g is continuous on R. By Theorem 21.6 (a)
g f = kf is continuous on D.
(b) True. g = (f + g) f
Assignment #1
Section 10
3. Prove that 12 + 22 + + n2 =
1
6
n(n + 1)(2n + 1) for all n N.
Let S be the set of positive integers for which the equation holds.
Step 1. 1 S :
1
1 = 12 = 6 (1)(2)(3) = 1. Therefore, 1 S .
Step 2. Assume that the positive integ
5333: Analysis
Solutions, Assignment # 10
Section 30
1. (a) True. This is Theorem 30.1.
(b) False. Take, for example, a function that is increasing on [a, b] but is not
continuous. Heres a specic example:
x,
f (x) =
x + 1,
0x<1
1x2
f is increasing on [0,
5333: Analysis
Solutions, Assignment # 9
Section 28.
1. (a) True.
(b) False. pn has the same values at f AT x0 ; pn approximates the values of f
at points near x0 .
4. By Taylors Theorem, |ex pn (x)|
max |f (n+1) (t)| n+1
e2
9 2n+1
|x|
2n+1 <
(n + 1)!
(n
Section 5.2
2a) True, by theorem 5.2.2 part d
2b) True, by theorem 5.1.14 : polynomial, P, and c , lim () = (). Also by 5.2.10 and
5.2.12, if g(x)=x, h(x)= sin(x), and f(x)=1/x, then P=g*h(f(x)=x sin 1/x is continuous 0 on
2c) False, Let f: (0, 1) , when
Section 3.3
2) a. True, Every non-empty set of has a min. Let S be a non-empty set of . If n S, then
S has a least element. Suppose n for every m , when m n-1. If m S, then
there is a least element in S.
b. False; Archimedean Property of states that if we
Section 4.2
2a) (Only if they converge) lim Sn= s lim Sn S= 0 and by the triangle inequality|(sntn)-(st)|=
|(sntn-snt)+( snt-st)| |sntn-snt|+| snt-st|=|Sn|tn-t|+|t|Sn-S| for every s and t
belonging to natural numbers
2b) False; Diverges to + , which only
Section 3.5
2) a) False, a set is compact iff it is bounded and closed
b) True, by theorem 3.5.6 (Bolzano- Weierstrass) If a bounded subset S of contains infinitely
many points, then there exists at least one point in that is an accumulation point of S.
c
Homework 5
Section 4.4
2a) False; By theorem 4.4.8, Every unbounded sequence contains a monotone subsequence that
has either + as a limit. Every sequence does not have a convergent subsequence.
2b) True; By theorem 4.4.7, Every bounded sequence has a conv
Section 3.3
2) a. True, Every non-empty set of has a min. Let S be a non-empty set of . If n S, then S
has a least element. Suppose n for every m , when m n-1. If m S, then there is a least
element in S.
b. False; Archimedean Property of states that if we
Section 4.2
2a) (Only if they converge) lim Sn= s lim Sn S= 0 and by the triangle inequality|(sntn)-(st)|=
|(sntn-snt)+( snt-st)| |sntn-snt|+| snt-st|=|Sn|tn-t|+|t|Sn-S| for every s and t belonging to
natural numbers
2b) False; Diverges to +, which only i
Section 3.5
2) a) False, a set is compact iff it is bounded and closed
b) True, by theorem 3.5.6 (Bolzano- Weierstrass) If a bounded subset S of contains infinitely
many points, then there exists at least one point in that is an accumulation point of S.
c
Homework 5
Section 4.4
2a) False; By theorem 4.4.8, Every unbounded sequence contains a monotone subsequence that
has either + as a limit. Every sequence does not have a convergent subsequence.
2b) True; By theorem 4.4.7, Every bounded sequence has a conv
Section 5.2
2a) True, by theorem 5.2.2 part d
2b) True, by theorem 5.1.14 :
lim P ( x )=P ( c )
polynomial, P, and c R , x c
. Also by
5.2.10 and 5.2.12, if g(x)=x, h(x)= sin(x), and f(x)=1/x, then P=g*h(f(x)=x sin 1/x is continuous
x 0 on R
2c) False,
5333: Analysis
Solutions, Assignment # 8
Section 26
1. (a) False. For example, f (x) = 1/x on (0, 1). (0, 1) is a bounded interval, f has
neither a maximum nor a minimum.
(b) False. You need f dierentiable in order to apply the Mean-Value Theorem.
(c) Fal
Solutions, Assignment 7
Section 23.
3. (d) f (x) = x2 + 3x 5 on D = (1, 3). Dene g on [1, 3] by
f (x),
x (1, 3)
g (x) =
1,
x=1
13,
x=3
Then g (x) is continuous on [1, 3] which implies that f is continuous on (1, 3) by
Theorem 23.9.
4. (a) Let
> 0. For x,
Solutions, Homework #4
Section 17.
6. (a) Counterexample: (sn ) = (1, 1, 1, 1, );
(tn ) = (1, 1, 1, 1, ).
(b) Counterexample: the example in part (a) will work here, too.
(c) Suppose (sn ) A and (sn + tn ) B . Then tn = (sn + tn ) sn B A.
(d) Counterexamp
MATH 5333:
Assignment #3
Section 14
3. (a) Here are two open covers of [1, 3) which do not have nite subcovers.
G=
1
1 n, 3
F=
2, 3
1
n
1
n
, n = 1, 2, 3, . . .
, n = 1, 2, 3, . . .
(b) Here are two open covers of N which do not have nite subcovers.
G =
MATH 5333:
Assignment #2
Section 12
3 Let s denote sup S , and let m denote max S .
(a) s = m = 3.
(c) s = m = 4.
(e) s = m = 1.
(g) s = 1, no max.
(i) None.
(k) s = m = 1.
(m) s = 5, no max
(b) s = m = .
(d) s = 4, no max.
(f) s = 1, no max.
(h) s = m =
Section 5.4
2a) > 0, > 0 |() ()| < | | < ,
. ; (, ). > 0 > 0, > 0, >
0 (, ).
2b) True. By definition 5.4.1, every continuous function is uniformly continuous f is
uniformly continuous on set D. Let (xn) be a Cauchy sequence in D and given > 0, (since f i
Section 5.4
2a)
>0, >0 |f ( x )f ( y )|< whenever| x y|< x , y D . False ; depends on on the point ( x , y ) . If >
2b) True. By definition 5.4.1, every continuous function is uniformly continuous f is uniformly
continuous on set D. Let (xn) be a Cauchy
Solutions, Homework #4
Section 4.2
6. (a) Counterexample: (sn ) = (1, 1, 1, 1, );
(tn ) = (1, 1, 1, 1, ).
(b) Counterexample: the example in part (a) will work here, too.
(c) Proof: Suppose (sn ) A and (sn + tn ) B. Then
tn = (sn + tn ) sn B A
1 1
1
1, ,