Name:
SSN:
Exam 1
MAA 4211
Spring 2002
To receive credit you MUST show your work.
1. (15 pts) Prove that
(a1 b1 + a2 b2 )2 (a2 + a2 )(b2 + b2 ), for all a1 , a2 , b1 , b2 R.
1
2
1
2
When does equality hold?
Solution: Expanding both sides, the inequality i

Solution Homework 8:
MAA 4211
Spring 2002
5, page 92. (a) If f is dierentiable at a then f is continuous at a. Since
f (a) = 0, it follows from a result done in Chapter 3 (see lemma 3.28, or Ex.
4, p. 78) that f (x) = 0, for any x in an interval I = (a ,

Solution Homework 6:
MAA 4211
Spring 2002
2 (b), (c), page 78. (b) The functions g (x) = 1 x, h(x) = 1 + x are
continuous on [0, 1] and h(x) = 0 for any x [0, 1]. By Theorem 3.22, it
then follows that f (x) = g (x)/h(x) is continuous on [0, 1]. 2
(c) As i

Solution Homework 5:
MAA 4211
Spring 2002
5, page 64. (i) Well show that limxa h(x) = L using the sequential
characterization of limits (Theorem 3.6).
Let xn I \ cfw_a be an arbitrary sequence such that xn a. Because
limxa f (x) = limxa g (x) = L, from Th

Solution Homework 4:
MAA 4211
Spring 2002
9, page 47. (a) We prove by induction that 0 < yn < xn , n N. The
statement for n = 1 is true, by hypothesis. Now assume 0 < yn < xn . From
the arithmetic/geometric mean inequality of positive numbers we have
yn+1

Solution Homework 3:
MAA 4211
Spring 2002
4, page 37. (a) |xn a| bn for large n means that there exists N0 N
such that
|xn a| bn , n N0 .
Let > 0. Since bn 0, there exists N1 N, N1 N0 , such that |bn | <
, n N1 .
Thus, for any n N1 , we have
|xn a| bn < .

Solution Homework 2:
MAA 4211
Spring 2002
4, page 23. Given a R and n N, apply the density theorem of rational
numbers to a 1/n < a +1/n (the inequality holds because n > 0). It follows
that there exists rn Q such that a 1/n < rn < a + 1/n. This double
in

Name:
SSN:
Exam 2
MAA 4211
Spring 2002
To receive credit you MUST show your work.
1. (15 pts) State the Mean Value Theorem.
Solution: If f is continuous on [a, b] and is dierentiable on (a, b), then there exists c (a, b) such that
f (b) f (a) = f (c)(b a)

Name:
SSN:
Exam 1
MAA 4211
Spring 2002
To receive credit you MUST show your work.
1. (15 pts) Prove that
(a1 b1 + a2 b2 )2 (a2 + a2 )(b2 + b2 ), for all a1 , a2 , b1 , b2 R.
1
2
1
2
When does equality hold?
Solution: Expanding both sides, the inequality i

Name:
SSN:
Quiz 1
MAA 4211
Spring 2002
To receive credit you MUST show your work.
1. (20 pts) (a) (10 pts) Let cfw_xn n be a sequence of positive real numbers (that is, xn > 0, n N) and
assume that xn 0, as n . Show that
lim
n
1
= .
xn
Solution: Need to s

Solution Homework 8:
MAA 4211
Spring 2002
5, page 92. (a) If f is dierentiable at a then f is continuous at a. Since
f (a) = 0, it follows from a result done in Chapter 3 (see lemma 3.28, or Ex.
4, p. 78) that f (x) = 0, for any x in an interval I = (a ,

Solution Homework 6:
MAA 4211
Spring 2002
2 (b), (c), page 78. (b) The functions g (x) = 1 x, h(x) = 1 + x are
continuous on [0, 1] and h(x) = 0 for any x [0, 1]. By Theorem 3.22, it
then follows that f (x) = g (x)/h(x) is continuous on [0, 1]. 2
(c) As i

Solution Homework 5:
MAA 4211
Spring 2002
5, page 64. (i) Well show that limxa h(x) = L using the sequential
characterization of limits (Theorem 3.6).
Let xn I \ cfw_a be an arbitrary sequence such that xn a. Because
limxa f (x) = limxa g (x) = L, from Th

Solution Homework 4:
MAA 4211
Spring 2002
9, page 47. (a) We prove by induction that 0 < yn < xn , n N. The
statement for n = 1 is true, by hypothesis. Now assume 0 < yn < xn . From
the arithmetic/geometric mean inequality of positive numbers we have
yn+1

Solution Homework 3:
MAA 4211
Spring 2002
4, page 37. (a) |xn a| bn for large n means that there exists N0 N
such that
|xn a| bn , n N0 .
Let > 0. Since bn 0, there exists N1 N, N1 N0 , such that |bn | <
, n N1 .
Thus, for any n N1 , we have
|xn a| bn < .

Solution Homework 2:
MAA 4211
Spring 2002
4, page 23. Given a R and n N, apply the density theorem of rational
numbers to a 1/n < a +1/n (the inequality holds because n > 0). It follows
that there exists rn Q such that a 1/n < rn < a + 1/n. This double
in

Name:
SSN:
Exam 2
MAA 4211
Spring 2002
To receive credit you MUST show your work.
1. (15 pts) State the Mean Value Theorem.
Solution: If f is continuous on [a, b] and is dierentiable on (a, b), then there exists c (a, b) such that
f (b) f (a) = f (c)(b a)

Name:
SSN:
Quiz 1
MAA 4211
Spring 2002
To receive credit you MUST show your work.
1. (20 pts) (a) (10 pts) Let cfw_xn n be a sequence of positive real numbers (that is, xn > 0, n N) and
assume that xn 0, as n . Show that
lim
n
1
= .
xn
Solution: Need to s