York University
Faculty of Arts, Faculty of Science
Math 3210
Midterm Test 2
SOLUTIONS
Instructions:
1. There are 4 questions on 4 pages.
2. Answer all questions.
3. Your work must justify the answer you give.
Question
1
2
3
4
Total
Points
10
16
8
11
45
M
Math 3210 M
Quiz 2 January 31, 2003
SOLUTIONS
Each of the following statements is true or false. Establish which. If the statement is true, give a
formal proof. If the statement is false, then explain why it is false. If you give a counterexample,
prove t
Math 3210 M
Quiz 3 March 7, 2003
SOLUTIONS
This quiz has three questions. Each question will be scored from 10. The grade for the quiz will
be, best score + half the sum of the two lower scores, for a maximum of 20.
1. Let f : R R with
f (x) =
x for x rat
Math 3210 M
Quiz 4 March 26, 2003
SOLUTIONS
1. Let f : (0, 1) R.
(a) (2 points) Explain what is meant by the statement, f is uniformly continuous on (0, 1).
Answer:
Given any > 0, there exists > 0, such that if x, y (0, 1) with |x y | < , then
|f (x) f (y
York University
Faculty of Arts, Faculty of Science
Math 3210
Midterm Test 1
SOLUTIONS
Instructions:
1. There are 4 questions on 4 pages.
2. Answer all questions.
3. Your work must justify the answer you give.
Question
1
2
3
4
Total
Points
13
12
11
9
45
M
Math 3210 M
Quiz 1 Version 1 January 17, 2003
SOLUTIONS
1. (a) (2 points) What is meant by the statement, (xn ) converges with limit L.
Answer:
For any given > 0, there exists N , such that, if n > N then |xn L| < .
(b) (8 points) Using the denition in (a
Math 3210 Homework 5 due April 5 at Noon
SOLUTIONS
p.166: 1b
For c = 0, we have f (c) = lim
xc
p.166: 1c
1
x
1
cx
1
1
1
c
= lim
= lim
= 2 . Then f (x) = 2 .
xc xc(x c)
xc xc
xc
c
x
For c > 0, we have f (c) = lim
xc
x c
xc
1
1
= . Thenf (x) = .
= lim
x
Math 3210 Homework 1 due January 11 at Noon
SOLUTIONS
p.38: 9
Let al be a lower bound for A and au an upper bound. Let bl be a lower bound for B and bu
an upper bound. Clearly inf(al , bl ) = min(al , bl ) is a lower bound for A B and sup(al , bl ) =
max
Math 3210 Homework 2 due January 25 at Noon
SOLUTIONS
p.74: 2
We have x1 > 1. If xn > 1, then x1 < 1, from which 2 x1 > 2 1 = 1. By Mathematical
n
n
Induction, it follows that xn > 1 for all n, so in particular, (xn ) is bounded below. To show
that (xn )
Math 3210 Homework 3 due February 22 at Noon
SOLUTIONS
p.118: 5
x
x1
1
1
For a) and b) note that
=
+
=1+
. Then
x1
x1 x1
x1
x
a) lim
= +.
+ x1
x1
x
b) lim
does not exist.
x1 x 1
2
c) lim x + = +.
x
x0+
2
d) lim x + = +.
x
x
x+1
x+1
x+1
x+1
For e) and f)
Math 3210 Homework 4 due March 8 at Noon
SOLUTIONS
p.128: 7
Set
f (x) =
1
1
if x is rational
if x is irrational
.
p.128: 8
Let c R. As Q is dense in R, there exists a sequence (xn ), xn Q for all n, such that
xn c. By the continuity of f and g we have f