HOMEWORK 8
Solution
Exercise 1. Consider the function h : R → R defined by
x x∈Q
h(x) =
0 x 6∈ Q
(1) Show that for any c 6= 0 you can find three sequences (xn )n∈N (yn )n∈N
and (zn )n∈N with lim xn = lim yn = lim zn = c such that lim h(xn ) = 0
lim h(yn ) = c and the series (h(zn )) does not converge.
Proof. Let c 6= 0. For each n ∈ N consider the open interval In = (c− n1 , c+
1
n ). Since the irratinas and ratinal numbers are both dense in R there is an
iraional number xn ∈ In and a rational number yn ∈ In . We thus get two
seqeuences (xn ) and (yn ) and since |xn − c| < n1 we get that xn → c and
similarly yn → c. Now h(xn ) = 0 for all n so lim h(xn ) = 0 and h(yn ) = yn
xn n is odd
so lim h(yn ) = lim yn = c. Finally let zn =
. Then
yn n is even
zn → c as well but lim h(zn ) does not exiset since it has two subsequences
(h(z2n )) and (h(z2n+1 )) converging to different limits.
(2) On the other hand, show that any sequence (xn )n∈N with lim xn = 0 satisfies that limn→∞ h(xn ) = 0.
Proof. Assume that xn → 0. Then for any > 0 there is N ∈ N such
that |xn | < for all n > N . But then |h(xn )| < for all n > N (since
if xn is rational then |h(xn )| = |xn | < and if xn is irrationhal then
|h(xn )| = 0 < ).
Exercise 2. In each of the following cases prove the following limit formulas directly
from the definition
(1) limx→2 (2x + 4) = 8
For any > 0 let δ = 2 then if |x − 2| < δ then |2x + 4 − 8| = |2(x − 2)| <
2δ = .
(2) limx→0 x3 = 0
For any > 0 let δ = 1/3 . Then if |x| < δ then |x3 − 0| = |x|3 < δ 3 = .
(3) limx→2 x3 = 8
For any > 0 let δ = min{1, /12}. Using the identity x3 − 8 = (x −
2)(x2 +2x+4) and noting that if |x−2| < 1 then 1 < x < 3 and x2 +2x+4 <
12 we get that if |x − 2| < δ then |x2 − 8| < 12δ = .
(4) limx→π bxc = 3 where the floor function bxc denotes the greatest integer
smaller or equal to x (for example b2.2c = 2).
Recall that 3.14 < π < 3.15 in particular, for any > 0 if |x−π| < δ = 0.1
then 3.04 < x < 3.05 and bxc = 3 so |bxc − 3| = 0 < .
1
2
HOMEWORK 8
(5) limx→0 h(x) = 0 where h(x) is the function from exe 1
For any > 0 let δ = and note that if |x| < δ then either |h(x)| = 0 < or |h(x)| = |x| < .
Exercise 3. Use the sequential criterion for functional limits to show that following
limits do not exist
(1) limx→0 |x|
x
|xn |
Let xn = n1 and yn = −1
n then lim xn = lim yn = 0 but lim xn = 1 and
lim |yynn | = −1 so the limit limx→0 |x|
x does not exist.
(2) limx→1 h(x) where h(x) is the function from exe 1.
This follows from the fact that there is a sequence zn → 1 where lim h(zn )
does not exist.
Exercise 4. Let f, g : A → R be two functions and let c ∈ A0 a limit point
of A. Assume that f (x) ≥ g(x) for all x ∈ A and that limx→c f (x) = L and
limx→c g(x) = M . Show that L ≥ M .
Proof. Let (xn ) be a sequence in A with xn 6= c and lim xn = x. Then lim f (xn ) = L
and limx→c g(xn ) = M . Since f (xn ) ≥ g(xn ) for all n we get that L ≥ M .
Exercise 5. Let f, g : A → R be two functions and let c ∈ A0 a limit point of A.
Assume that f (x) is bounded, that is, that there is some M > 0 with |f (x)| ≤ M
for all x ∈ A. Show that if limx→c g(x) = 0 then limx→c (f (x)g(x)) = 0 as well.
Proof. For any > 0 let δ > 0 be such that if |x − c| < δ then |g(x)| < /M . Then
if |x − c| < δ then |f (x)g(x) − 0| ≤ M |g(x)| < .
Exercise 6 (Challenge question). Consider the function t : R → R defined by

 1 x=0
1
x = pq with p ∈ Z, q ∈ N in lowest terms
t(x) =
 q
0 x 6∈ Q
You will now prove that for any x ∈ R one has limx→c t(x) = 0.
(1) For any M > 0 let
p
{QM = { |0 < q < M and |p| < (|c| + 1)q in lowest terms}
q
Show that this set has finitely many elements.
Proof. Since any pq in QM satisfies 0 < q < M and |p| < (|c| + 1)q <
(|c| + 1)M then QM has at most M 2 (|c| + 1) elements.
(2) Let δ1 = min{|r − c| : r ∈ QM , r 6= c} and explain why δ1 > 0 must be
strictly positive.
Proof. This is a finite set of positive elements (since |r − c| > 0 whenever
r 6= c) so their minimum is positive.
HOMEWORK 8
(3) Let δ = min{δ1 , 1} and show that if |x − c| < δ and x 6= c then |t(x)| <
3
1 1
M
Proof. Let x ∈ R and assume that 0 < |x − c| < δ. If x is irrational then
1
so it is enough to consider the case where x ∈ Q. For
|t(x)| = 0 < M
x ∈ Q write x = pq with p ∈ Z and q ∈ N in reduced form. If q > M then
t(q) = 1q < M and we are done. We will now show that the assumption
0 < q ≤ M leads to a contradiction. Indeed since | pq − c| < δ ≤ 1 we have
p
that |p|
q < |c| + 1 so if 0 < q < M then q ∈ QM . But this contradicts the
p
bound | q − c| < δ ≤ δ1 = min{|r − c| : r ∈ QM , r 6= c}.
(4) Show that indeed limx→c t(x) = 0 (using the , δ definition).
Proof. Given > 0 let M > 0 with
|x − c| < δ we showed that |t(x)| <
1
M
1
M
< . Let δ > 0 be as above, then if
< so we are done.
(5) At what points is the function t(x) continuous?
We have that limx→c t(x) = 0 at all points so the function is continuous
only at points for which t(x) = 0, that is it is continuous at all x ∈ R \ Q.
1Hint:Consider the cases when x = p is rational and x is irrational separately
q