Math 3210, Fall 2013
Instructor: Thomas Goller
16 December 2013
Final Learning Celebration Solutions!
Example (20 points)
Give an example of each of the following or state that no example exists:
(1) A bounded subset of R with no maximum. (2 points)
[0, 1)
(2) A continuous function on [ 1, 1] that has no antiderivative on [ 1, 1]. (2 points)
No example exists
(3) A power series with center 7 and radius of convergence 13 . (2 points)
1
X
3k (x
7)k
k=0
(4) A sequence of real numbers that has minimum
99 and converges to ⇡. (2 points)
99, ⇡, ⇡, ⇡, . . .
(5) Two functions f and g on R such that f 0 = g 0 on R but f 6= g. (2 points)
f (x) = x, g(x) = x + 2
(6) A sequence of real numbers {an } that is monotone and has lim sup an 6= lim inf an . (2
points)
No example exists
(7) A sequence of functions {fn (x)} on [0, 1] such that {fn (0)} converges but {fn (1)} diverges. (2 points)
fn (x) = nx
(8) A function on R that is analytic at 3 and not continuous at
(
x
if x 6= 2
f (x) =
700 if x = 2
2. (2 points)
(9) An alternating series of real numbers whose partial sums are all positive. (2 points)
1
X
( 1)k+1
k=1
k
(10) A function f on the interval [3, 5] such that limx!4 f (x) = limx!4+ f (x) but f is not
continuous at 4. (2 points)
(
0
if x 2 [3, 4) [ (4, 5]
f (x) =
1 if x = 4
Math 3210, Fall 2013
Instructor: Thomas Goller
16 December 2013
Proof (60 points)
1
(11) Prove that lim p = 0 using the definition of limit. (6 points)
n
Proof. Given ✏ > 0, let N =
1
.
✏2
Then for all n > N ,
1
p
n
1
1
0 = p < p = ✏.
n
N
(12) Let A and B be non-empty subsets of R. Assume that A ⇢ B. Prove that sup A  sup B.
(6 points)
Proof. If sup B = 1, then sup A  sup B is trivially true, so assume sup B 2 R.
Given a 2 A, since a 2 B, a  sup B. Thus sup B is an upper bound for A, hence
sup A  sup B since sup A is the least upper bound for A.
(13) Let f (x) = e2x . Prove that the Taylor series for f centered at 0 converges to f on all of
R. (6 points)
P
2k k
Proof. The Taylor series for f centered at 0 is 1
k=0 k! x . Fix x 2 R and n 2 N. By
Taylor’s formula, there is c between 0 and x such that
f (x)
n
X
2k
k=0
k!
xk =
2n+1 e2c n+1
2n+1 e2|x| n+1
x

|x| .
(n + 1)!
(n + 1)!
Since x is fixed,
2n+1 e2|x| n+1
(2|x|)n+1
2|x|
lim
|x|
= e lim
= 0.
(n + 1)!
(n + 1)!
Thus f (x) =
P1
(14) Let f (x) = x4
2k k
k=0 k! x .
3x + 1. Prove that f has a real root. (6 points)
Proof. f is a polynomial, so f is continuous on R. Since f (0) = 1, f (1) = 1, and
1 < 0 < 1, the Intermediate Value Theorem implies that there is c 2 [0, 1] such that
f (c) = 0.
(15) Let {an } be a sequence of real numbers that converges to a 2 R, and let {bn } be a
sequence of real numbers that converges to b 2 R. Prove that {an 2bn } converges to
a 2b using the definition of convergence. (6 points)
Math 3210, Fall 2013
Instructor: Thomas Goller
16 December 2013
Proof. Let ✏ > 0 be given. Since an ! a, there is N1 2 R such that for all n > N1 ,
|an a| < 2✏ . Since bn ! b, there is N2 2 R such that for all n > N2 , |bn b| < 4✏ . Let
N = max{N1 , N2 }. Then for all n > N ,
|(an
2bn )
2b)| = |(an
(a
a) + 2(b
bn )|  |an
a| + 2|bn
b| <
✏
✏
+ 2 · = ✏.
2
4
(16) Prove that f (x) = x2 is continuous at 2 using the definition of continuity. (6 points)
= min{1, 5✏ }. Then for all x 2 R satisfying |x
Proof. Given ✏ > 0, choose
f (2)| = |x2
|f (x)
4| = |x + 2| · |x
2|  5|x
2| < ,
2| < 5  ✏.
(17) Let f be a function that is di↵erentiable at a 2 R. Prove that the function g(x) = 3·f (x)
is di↵erentiable at a using the definition of the derivative. (6 points)
Proof. Since f is di↵erentiable at a, the definition of the derivative for g at a is
g(x)
x!a
x
lim
g(a)
3 · f (x)
= lim
x!a
a
x
Thus g 0 (a) = 3 · f 0 (a).
(18) Let p 2 (0, 1). Prove that
Z
3 · f (a)
f (x)
= 3 · lim
x!a
a
x
f (a)
= 3 · f 0 (a).
a
2
x
p
dx converges. (6 points)
0
Proof. By definition of improper integrals and the First Fundamental Theorem of Calculus,
✓ p+1
◆
Z 2
Z 2
2
r p+1
p
p
x dx = lim+
x dx = lim+
.
r!0
r!0
p+1
p+1
0
r
Since
p + 1 > 0, limr!0+ r
(19) Prove that the series
1
X
100k
k=0
k!
Proof. Since
lim
the series
P1
k=0
100k
k!
p+1
= 0, so
Z 2
2 p+1
x p dx =
.
p+1
0
converges. (6 points)
100k+1
k!
100
·
= lim
= 0 < 1,
k
(k + 1)! 100
k+1
converges by the ratio test.
Math 3210, Fall 2013
Instructor: Thomas Goller
16 December 2013
x2
(20) Let f (x) = 2
. Prove that limx!1 f (x) = 1 using the definition of limit. (6 points)
x +1
Proof. Given ✏ > 0, choose m =
x2
x2 + 1
p1 .
✏
Then for all x 2 R satisfying x > m,
1 =
x2
1
1
1
< 2 < 2 = ✏.
+1
x
m