1. No category

Math 131 A, Lecture 3
Real Analysis
Sample Final Exam
Instructions: You have three hours to complete the exam. There are ten problems, worth a
total of one hundred points. Write your solutions in the space below the questions. If a question
is in multiple parts, clearly label each part. If you need more space use the back of the page. Do
not forget to write your name in the space below.
Name:
Question Points Score
1
10
2
10
3
10
4
10
5
10
6
10
7
10
8
10
9
10
10
10
Total:
100
Problem 1.
(a) [5pts.] Define f 0 (a), the derivative of f (x) at x = a.
Solution: We say that f 0 (a) = limx→a
f (x)−f (a)
x−a
where this limit exists.
(b) [5pts.] Let f (x) = 1/g(x). If g(a) 6= 0 and g is differentiable at a, use the definition
−g 0 (a)
of the derivative to prove that f 0 (a) = (g(a))
2.
Solution: Since g is differentiable, hence continuous, at a, there is a small
interval around a on which g(x) 6= 0, so f is defined. Notice that f (x) − f (a) =
1
1
− g(a)
= g(a)−g(x)
. Therefore we have
g(x)
g(x)g(a)
f (x) − f (a)
x→a
x−a
1
−(g(x) − g(a))
= lim
·
x→a g(x)g(a)
x−a
1
· (−g 0 (a))
=
(g(a))2
−g 0 (a)
=
(g(a))2
f 0 (a) = lim
Here in the third step we have used continuity of g to show that g(x) → g(a)
as x → a, and differentiability of g to show that limx→a g(x)−g(a)
= g 0 (a).
x−a
Problem 2.
(a) [5pts.] Define: the sequence (sn ) diverges to −∞.
Solution: We say a sequence (sn ) diverges to −∞ if, for every M < 0, there
exists some N such that n > N implies that sn < M .
(b) [5pts.] Prove that if (sn ) is an decreasing unbounded sequence then (sn ) diverges
to −∞.
Solution: Since (sn ) is decreasing, (sn ) is bounded above, so (sn ) must be
unbounded below. Let M < 0. Since (sn ) is unbounded, there is some N
such that sN < M . But since (sn ) is decreasing, n > N then implies that
sn < sN < M . Ergo (sn ) diverges to −∞.
Problem 3.
(a) [5pts.] State the Mean Value Theorem.
Solution: Let f be a continuous function on [a, b] which is differentiable on
(a, b). Then there is some x ∈ (a, b) such that
f 0 (x) =
(f (b) − f (a)
.
b−a
(b) [5pts.] Let f be a differentiable function on (a, b). Prove iff 0 (x) > 0 for all x ∈ (a, b),
f is strictly increasing.
Solution: For any x1 , x2 ∈ (a, b), by MVT, there is some y ∈ (x1 , x2 ) such that
0 < f 0 (y) =
f (x2 ) − f (x1 )
x2 − x1
Since x2 − x1 > 0, we must have f (x2 ) − f (x1 ) > 0, or f (x2 ) > f (x1 ). So since
x1 and x2 were arbitrary, f is strictly increasing.
Problem 4.
(a) [5pts.] State the Bolzano-Weierstrauss Theorem.
Solution: Every bounded sequence has a convergent subsequence.
(b) [5pts.] Prove that a uniformly continuous function f on a open interval (a, b) (not
closed!) is bounded.
Solution: Suppose not. Wlog, let us assume f is unbounded above. Then for
every n ∈ N, there is some xn such that f (xn ) > n, and (f (xn )) diverges to
infinity. Now, (xn ) has a subsequence (xnk ) which is convergent in R (although
not necessarily in (a, b), hence is Cauchy. But uniformly continuous functions
take Cauchy sequences to Cauchy sequences, so this implies (f (xnk )) is Cauchy,
which is nonsense since this sequence diverges to infinity. Contradiction. Ergo
f is bounded.
Problem 5.
(a) [5pts.] Define limx→a+ f (x) = L.
Solution: We say that limx→a+ f (x) = L if there is an interval (a, b) such that
if (xn ) is any sequence of points in (a, b) converging to a, then lim f (xn ) = L.
(b) [5pts.] Prove that if f f1 (x) ≤ f2 (x) for all x ∈ (a, b) and limx→a+ f1 (x) =
L1 , limx→a+ f2 (x) = L2 , then L1 ≤ L2 .
Solution: Let (xn ) be a sequence of points in (a, b) converging to a. Then
by definition, the sequence (f1 (xn )) converges to L1 and the sequence (f2 (xn ))
converges to L2 .
Problem 6.
(a) [5pts.] Define lim inf sn .
Solution: We say lim inf sn = limN →∞ inf{sn : n > N }.
(b) [5pts.] Prove that if lim inf sn = lim sup sn = s then (sn ) converges to s.
Solution: Since s = lim sup sn , there exists a positive number N1 such that
|s − sup{sn : n > N1 }| < . In particular, sup{sn : n > N1 } < s + , so for
n > N1 , sn < s + . Similarly, since lim sup sn = s, there exists N2 such that
n > N2 implies that s− < sn . Therefore if n ≥ max{N1 , N2 }, s− < sn < s+ ,
so |s − sn | < . Ergo sn → s.
Problem 7.
(a) [5pts.] Define: the series
Solution: The series
P
n
k = 0 ak converges.
P
P
an converges.
an converges if the sequence (sn ) of partial sums sn =
(b) [5pts.] Prove that if an ≥ 0 for all n and
all p > 1.
P
an converges, then
P
apn converges for
P
Solution: Since
an converges, lim an = 0. Therefore there is some integer N
such thatPn > N implies that 0 ≤ an < 1. Therefore forP
n > N , 0 < apn < an ,
∞
∞
p
so since
P p n=N +1 an converges, by the Comparison Test, n=N +1 an converges.
Ergo
an converges.
Problem 8.
(a) [5pts.] What does it mean to say that Q is dense in R?
Solution: Every open interval (a, b) in R contains a rational number.
(b) [5pts.] Suppose that f (x) is continuous for all x ∈ R and f (q) = q 2 for all q ∈ Q.
Prove that f (x) = x2 for all x ∈ R.
Solution: Let g(x) = x2 , then f −g(x) is zero on each rational number q. Since
every interval contains a rational, for x irrational, we can choose a sequence of
rational points qn ∈ (x − n1 , x) such that qn → x, so by continuity f − g(qn ) →
f − g(x). But f − g(qn ) = 0 for all n, so f − g(x) = 0, i.e. f (x) − x2 .
Problem 9.
(a) [5pts.] Let f (x) be bounded on [a, b]. Assuming that the upper and lower Darboux
integrals have been defined, what does it mean to say that f (x) is integrable on
[a, b]?
Solution: We say f is integrable if the upper Darboux integral U (f ) and lower
Rb
Darboux integral L(f ) are equal, and we say this number is a f .
(b) [5pts.] Prove that if f (x) is continuous on [a, b] and f (x) ≥ 0 for all x ∈ [a, b] then
Z
b
f (x) dx = 0
a
implies that f (x) = 0 for all x ∈ [a, b]. (Note: You may assume that a function
that is bounded on [a, b] and continuous at all but finitely many points of [a, b] is
integrable on [a, b].)
Solution: Suppose not. Then there is some x0 ∈ [a, b] such that f (x0 ) = α > 0,
Rb
which implies that f (x) > α2 on some interval (c, d) containing x. Then a f ≥
Rd
f ≥ α2 (d − c) > 0. Contradiction.
c
Problem 10.
(a) [5pts.] Part II of the Fundamental
Theorem of Calculus states that if f is integrable
Rx
on [a, b], and F (x) = a f (t)dt, then F (x) is continuous on [a, b], and if f (x) is
continuous at x0 ∈ (a, b), then F (x) is differentiable at x0 with derivative F 0 (x0 ) =
f (x0 ). State Part I.
Solution: If g is a differentiable function on (a, b), then
Rb
a
g 0 = g(b) − g(a).
(b) [5pts.] Prove that if g 0 (x) is continuous, then Part II implies
Part I, that is, prove
Rx
Part RI assuming that Part II is true. (Hint: Let G(x) = a g 0 (t)dt. You may assume
a
that a f (x) dx = 0 for any function f .)
Rx
Solution: Let G(x) = a g 0 (t)dt. RThen G0 (x) = g 0 (x), so G(x) = g(x) + c,
a
where c is a constant. Now, G(a) = a g 0 (t)dt = 0 = g(a) − g(a), so we conclude
Rb 0
c = −g(a). Therefore G(b) = a g (t)dt = g(b) − g(a).