2. For each of the following, either give an example and prove why it is an
example or prove why such an example is impossible.
(a) An innite set A with no limit points.
(b) A family of compact sets (Kn) contained in [0; 1] such that
S1
n=1 Kn is not compact.
(c) A sequence (xn) for which the subsequences (x2n) and (x2n+1) converge, but (xn)
does not converge.
(d) A sequence (xn) of positive terms for which the series
P1
n=1 xn converges and the
series
P1
n=1
nxn
n+1 diverges.
(e) A continuous function f : [0; 2] ! R that has no maximum value.
(f) A continuous function on [a; b] with range a subset of Q.
(g) A function f : [a; b] ! R which is continuous on [a; b] and dierentiable on (a; b) for
which there exits c 2 (a; b) such that f0(c) = 0 and f does not have an extremum
at c.
(h) A function f : [a; b] ! R for which jfj is Riemann integrable on [a; b], but f is not
Riemann integrable on [a; b].
3. Dene the function
ha(x) =
(
xa if x > 0;
0 if x 0:
where a is a real constant.
(a) (3 points) For what values of a is ha continuous at 0?
(b) (4 points) For what values of a is ha dierentiable at 0?
(c) (3 points) For what values of a is the derivative function h0
a continuous on R?
Make sure to show all your work.
4. Let K R be compact and f : K ! R be continuous. In the following problem, we will
show that for every ” > 0, there exists M(“) > 0 such that for all x; y 2 K we have
jf(x) .. f(y)j M(“)jx .. yj + “:
(a) (3 points) We will argue by contradiction. Show that if the above statement does
not hold, there exists “0 > 0 and sequences (xn); (yn) in K such that
jf(xn) .. f(yn)j > njxn .. ynj + “0:
Hint: Proceed inductively.
(b) (2 points) Given your sequences (xn) and (yn) as above, show that there are sub-
sequences (xnk) and (ynk ), such that xnk ! x0 2 K and ynk ! y0 2 K.
(c) (5 points) Using parts (a) and (b), arrive at a contradiction by considering the
cases when x0 = y0 or x0 6= y0.
5. (a) (3 points) Using the identity
log(1 + x) =
x
0
1
1 + t
dt for all x 0;
show that there exists ` : [0;1) ! [0;1) such that, for every 0 x 1,
log(1 + x) = x ?
x2
2
+ `(x);
where
`(x)
x3
3
:
Hint: You do not need to use Taylor series (which we never covered). Try exploiting
that 1
1+t = 1 ? t
1+t .
(b) (3 points) For n 2 N, let (xn) denote the sequence
xn = 1 +
1
2
+ : : : +
1
n
? log n:
Using part (a), or otherwise, show that for any n 2 N, we have
jxn+1 ? xnj
2
n2 +
1
3n3 :
(c) (3 points) Using part (b), or otherwise, show that the sequence (xn) converges.
(d) (3 points) Show that
lim
n!1
1
log n
Xn
k=1
1
k
= 1:
6. (9 points) Prove using the “? denition of continuity that f(x) = x3?3×2+x?1
is continuous on R.
Hint: You may nd the following inequality from helpful: for any a; b 2 R,
we have ab 1
2a2 + 1
2b2.
7. Given a function f : [a; b] ! R and a partition P = fa = t0 < : : : < tn = bg of [a; b], let
V(f;P) :=
Xn
j=1
jf(tj) ? f(tj?1)j:
We then dene
V (f) := sup
P
V(f;P);
where supremum is over all partitions P of [a; b]. In the following, we further assume
that f is continuously dierentiable on [a; b] (i.e. f0 exists and is a continuous function
on [a; b])
(a) (4 points) Using the Fundamental Theorem of Calculus, show that
V (f)
b
a
jf0(x)jdx:
(b) (5 points) Using the Mean Value Theorem (for derivatives), establish the reverse
inequality:
V (f)
b
a
jf0(x)jdx:
Remark: Combining parts (a) and (b), we get the equality
V (f) =
b
a
jf0(x)jdx:
8. (a) (5 points) Let A R be a non-empty, bounded set. Show that
supA .. inf A = sup
x;y2A
jx .. yj;
where the supremum is taken over all pairs x; y 2 A. You need to argue why both
sides of the above inequality exist and are nite to begin with.
(b) (3 points) Show that f : [a; b] ! R is Riemann integrable on [a; b] if and only if for
every ” > 0, there exists a partition P = fa = t0 < < tn = bg of [a; b] such that
Xn
j=1
sup
x;y2[tj..1;tj ]
jf(x) .. f(y)j(tj .. tj..1) < “:
(c) (5 points) Let f : [c; d] ! R be continuous on [c; d] and g : [a; b] ! [c; d] be Riemann
integrable on [a; b]. Show that the composition f g is Riemann integrable on [a; b].
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