Math 3310 Theoretical Concepts of Calculus Solutions to selected review problems for Exam 2 Exam 2: Wednesday, 4/10/2013 Section 13 Exercise 13:3 Find the interior of each set. a) 1 n : n 2 N Solution: We see that int 1 n : n 2 N = ;; because for any x 2 R and > 0 we have N (x; ) * 1 n : n 2 N : b) [0; 3] [ (3; 5) Solution: We have [0; 3] [ (3; 5) = [0; 5) : Therefore, we have that int ([0; 3] [ (3; 5)) = int ([0; 5)) = (0; 5) : c) r 2 Q :0 <r< p 2 Solution: We have int n r 2 Q :0 <r< p 2 o = ; since for any x 2 R and > 0: N (x; ) * n r 2 Q :0 <r< p 2 o : d) r 2 Q : r p 2 Solution: We have int n r 2 Q : r p 2 o = ; since for any x 2 R and > 0: N (x; ) * n r 2 Q : r p 2 o : e) [0; 2] \ [2; 4] Solution: We have [0; 2] \ [2; 4] = f2g ; therefore, we have int ([0; 2] \ [2; 4]) = int (f2g)= ; since for any x 2 R and > 0: N (x; ) * f2g Exercise 13:4 Find the boundary of each set in Exercise 13:3. 1
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Math 3310 Theoretical Concepts of CalculusSolutions to selected review problems for Exam 2
Exam 2: Wednesday, 4/10/2013
Section 13Exercise 13:3 Find the interior of each set.
a)�1n : n 2 N
Solution: We see that int
��1n : n 2 N
�= ;; because for any x 2 R and � > 0 we have
N (x; �) *�1
n: n 2 N
�:
b) [0; 3] [ (3; 5)Solution: We have
[0; 3] [ (3; 5) = [0; 5) :
Therefore, we have thatint ([0; 3] [ (3; 5)) = int ([0; 5)) = (0; 5) :
c)�r 2 Q : 0 < r <
p2
Solution: We haveint�nr 2 Q : 0 < r <
p2o�
= ;
since for any x 2 R and � > 0 :
N (x; �) *nr 2 Q : 0 < r <
p2o:
d)�r 2 Q : r �
p2
Solution: We haveint�nr 2 Q : r �
p2o�
= ;
since for any x 2 R and � > 0 :N (x; �) *
nr 2 Q : r �
p2o:
e) [0; 2] \ [2; 4]Solution: We have
[0; 2] \ [2; 4] = f2g ;
therefore, we haveint ([0; 2] \ [2; 4]) = int (f2g) = ;
since for any x 2 R and � > 0 :N (x; �) * f2g
Exercise 13:4 Find the boundary of each set in Exercise 13:3.
1
a)�1n : n 2 N
Solution: We see that
bd
��1
n: n 2 N
��=
�1
n: n 2 N
�[ f0g :
Then 0 2 bd��
1n : n 2 N
�; because for any � > 0 we have
N (0; �) \�1
n: n 2 N
�6= ; and N (0; �) \ Rn
�1
n: n 2 N
�6= ;:
Hence, we conclude that
0 2 bd��
1
n: n 2 N
��:
If x 2�1n : n 2 N
: Then x = 1
n for some n 2 N. Let � > 0. We have
N
�1
n; �
�\�1
n: n 2 N
�6= ;
and clearly
N
�1
n; �
�\ Rn
�1
n: n 2 N
�6= ;:
Thus, we conclude (after considering all of the above cases) that
bd
��1
n: n 2 N
��=
�1
n: n 2 N
�[ f0g :
b) [0; 3] [ (3; 5)Solution: Since [0; 3] [ (3; 5) = [0; 5) ; we clearly have
bd ([0; 3] [ (3; 5)) = bd ([0; 5)) = f0; 5g :
To see that, we observe that, for any � > 0 :
N (0; �) \ [0; 5) 6= ; and N (0; �) \ Rn [0; 5) 6= ;
andN (5; �) \ [0; 5) 6= ; and N (5; �) \ Rn [0; 5) 6= ;:
Exercise 13:5 Classify each of the following sets as open, closed, neither, or both.
a)�1n : n 2 N
Solution: As we observed in Exercise 13:3 a);
int
��1
n: n 2 N
��= ; 6=
�1
n: n 2 N
�;
therefore,�1n : n 2 N
is not open. As we see from Exercise 13:4 a);
bd
��1
n: n 2 N
��=
�1
n: n 2 N
�[ f0g *
�1
n: n 2 N
�;
so�1n : n 2 N
is not closed. Therefore, the set
�1n : n 2 N
is neither open nor closed.
b) N.Solution: As one verify
int (N) = ; 6= Nthus, N is not open. However, we see that
bd (N) = N;
so, in particular,bd (N) � N;
thus N is closed.
3
c) Q:Solution: As one can verify
int (Q) = ;
(since Q being countable cannot contain open interval), thus
int (Q) 6= Q;
so Q is not open. Moreover,bd (Q) = R;
(since every open interval (x� �; x+ �) intersects both Q and RnQ), thus
bd (Q) * Q:
Therefore, Q is neither open nor closed.
d)1\n=1
�0; 1n
�Solution: As one can verify
1\n=1
�0;1
n
�= ;;
therefore,1\n=1
�0; 1n
�is both open and closed.
e)�x : jx� 5j � 1
2
Solution: As one can verify�
x : jx� 5j � 1
2
�=
�x : �1
2� x� 5 � 1
2
�=
�x :
9
2� x � 11
2
�=
�9
2;11
2
�Clearly, we have
bd
��9
2;11
2
��=
�9
2;11
2
���9
2;11
2
�Therefore,
�92 ;
112
�is closed. However, we also can verify that
int
��9
2;11
2
��=
�9
2;11
2
�6=�9
2;11
2
�;
therefore�92 ;
112
�is not open.
f)�x : x2 > 0
:
Clearly, we have �x : x2 > 0
= (�1; 0) [ (0;1) ;
andint��x : x2 > 0
�= int ((�1; 0) [ (0;1)) = (�1; 0) [ (0;1) =
�x : x2 > 0
;
hence�x : x2 > 0
is open. However,
bd��x : x2 > 0
�= f0g "
�x : x2 > 0
;
so�x : x2 > 0
is not closed.
4
Section 14Exercise 14:3 Show that each subset of R is not compact by describing an open cover for it that has no�nite subcover.
a) [1; 3)
Notice: Using Heine-Borel theorem, a subset S of R is compact i¤ S is closed and bounded, we seethat the set
S = [1; 3) is not closed,
since 3 2 S0 (S0 is the set of accumulation points), but 3 =2 S. Therefore, using Heine-Borel theorem,[1; 3) cannot be compact.
Solution: Let
An =
�1
2; 3� 1
n
�; n 2 N
Obviously for all n 2 N; An is open interval, hence it is open. The family
A = fAngn2N
is an open covering of [1; 3) ; however it has no �nite subcover. This shows that [1; 3) is not compact.
b) N
Notice: Using Heine-Borel theorem, a subset S of R is compact i¤ S is closed and bounded, we see thatthe set
S = N is not bounded,
therefore, using Heine-Borel theorem, N is not compact.
Solution: Let
An =
�n� 1
3; n+
1
3
�; n 2 N:
Each An is open interval, so it is an open set in R. The family
A = fAngn2N
is an open covering of N. However, no �nite sub-collection
fAn1 ; An2 ; :::; Ankg
of A = fAngn2N covers N. Hence, N is not compact.
c)�1n : n 2 N
:
Notice: Using Heine-Borel theorem, a subset S of R is compact i¤ S is closed and bounded, we see thatthe set
S =
�1
n: n 2 N
�is not closed,
since as we saw in Exercise 13:4 a) we have
bd
��1
n: n 2 N
��=
�1
n: n 2 N
�[ f0g *
�1
n: n 2 N
�= S
Thus,�1n : n 2 N
is not closed. Thus, by Heine-Borel theorem,
�1n : n 2 N
cannot be compact.
5
Solution: Let
An =
�1
n� 1
2n;1
n+1
2n
�Each An is open interval, so it is an open set in R. The family
A = fAngn2N
is an open covering of�1n : n 2 N
. However, no �nite sub-collection
fAn1 ; An2 ; :::; Ankg
of A = fAngn2N covers�1n : n 2 N
. Therefore
�1n : n 2 N
is not compact.
d) fx 2 Q : 0 � x � 2g
Notice: Using Heine-Borel theorem, a subset S of R is compact i¤ S is closed and bounded, we see thatthe set
S = fx 2 Q : 0 � x � 2g is not closed,since as we saw in Exercise 13:4 a) we have
bd (fx 2 Q : 0 � x � 2g) = [0; 2] * fx 2 Q : 0 � x � 2g = S
Thus, fx 2 Q : 0 � x � 2g is not closed. Thus, by Heine-Borel theorem, fx 2 Q : 0 � x � 2g cannotbe compact.
Solution: Since fx 2 Q : 0 � x � 2g is countable, let fq1; q2; q3; :::g = fx 2 Q : 0 � x � 2g ; be enumerationof all elements of fx 2 Q : 0 � x � 2g : De�ne
An =
�qn �
1
2n; qn +
1
2n
�; n 2 N:
We observe that An is open as it is just an open interval. Moreover, we have that
qn 2 An; n 2 N:
Therefore, we see that
fq1; q2; q3; :::g =1[n=1
fqng �1[n=1
�qn �
1
2n; qn +
1
2n
�=
1[n=1
An;
hence the familyA = fAngn2N
is an open covering of fq1; q2; q3; :::g = fx 2 Q : 0 � x � 2g. However, no �nite sub-collection of Acovers fx 2 Q : 0 � x � 2g. Thus, fx 2 Q : 0 � x � 2g is not compact.
Section 16Exercise 16:6: Using the De�nition 16:2, prove that:
a) For any real number k
limn!1
�k
n
�= 0
Solution: Let � > 0 be given. We need to �nd N0 2 N; such that, for n > N0 we have�� kn � 0
�� < �.Since � > 0 is given, we solve the following inequality:����kn � 0
���� =
����kn���� < �
jkjn
< �; hence
n >jkj�
6
Let N0 =hjkj�
i+ 1; then for n > N0; we have:����kn � 0
���� = ����kn���� = jkj
n<jkjN0
� jkjjkj�
= �
Therefore, we showed that,
8�>09N0�N8n>N0
����kn � 0���� < �;
thus it follows, by De�nition 16:2; that
limn!1
�k
n
�= 0:
(b) For any real number k > 0
limn!1
�1
nk
�= 0
Solution: We �rst observe that1
nk= n�k = e�k ln(n):
Let � > 0 be given. We need to �nd N0 2 N; such that, for n > N0 we have�� 1nk� 0�� < �.
Since � > 0 is given, we solve the following inequality:���� 1nk � 0���� =
���� 1nk���� < �; since 1
nk= e�k ln(n):
e�k ln(n) < �; hence
e�k ln(n) < eln(�); thus � k ln (n) < ln (�) ; son > e�
1k ln(�)
Let N0 =he�
1k ln(�)
i+ 1; then for n > N0; since nk > (N0)
k; we have:���� 1nk � 0
���� = ���� 1nk���� < 1
(N0)k� 1�
e�1k ln(�)
�k = eln(�) = �Therefore, we showed that,
8�>09N0�N8n>N0
���� 1nk � 0���� < �;
thus it follows, by De�nition 16:2; that
limn!1
�1
nk
�= 0:
(c) Show that
limn!1
3n+ 1
n+ 2= 3
Solution: Let � > 0 be given. We need to �nd N0 2 N; such that, for n > N0 we have��� 3n+1n+2 � 3
��� < �.Since � > 0 is given, we solve the following inequality:����3n+ 1n+ 2
� 3���� =
����3n+ 1� 3 (n+ 2)n+ 2
���� = j�5jn+ 2
< � (since n � 1)
5
n+ 2< �; hence
n >5
�� 2
7
Let N0 =�5�
�; then for n > N0; we have, since
�5�
�� 5
� :����3n+ 1n+ 2� 3���� = ����3n+ 1� 3 (n+ 2)n+ 2
���� = 5
n+ 2<5
n<55�
= �
Therefore, we showed that,
8�>09N0�N8n>N0
����3n+ 1n+ 2� 3���� < �;
thus it follows, by De�nition 16:2; that
limn!1
�3n+ 1
n+ 2
�= 3:
(d) Show that
limn!1
�sin (n)
n
�= 0
Solution: Let � > 0 be given. We need to �nd N0 2 N; such that, for n > N0 we have��� sin(n)n � 0
��� < �.Since � > 0 is given, we solve the following inequality:���� sin (n)n
� 0���� =
jsin (n)jn
� 1
n< �
n >1
�
Let N0 =�1�
�; then for n > N0; we have, since
�1�
�� 1
� :���� sin (n)n� 0���� = jsin (n)j
n� 1
n<
1
N0� 1
1�
= �
Therefore, we showed that,
8�>09N0�N8n>N0
���� sin (n)n� 0���� < �;
thus it follows, by De�nition 16:2; that
limn!1
�sin (n)
n
�= 0:
(e) Show that
limn!1
�n+ 2
n2 � 3
�= 0
Solution: Let � > 0 be given. We need to �nd N0 2 N; such that, for n > N0 we have��� n+2n2�3 � 0
��� < �.Since � > 0 is given, we solve the following inequality:���� n+ 2n2 � 3 � 0
���� = n+ 2
n2 � 3 < �
However, since n+ 2 � 2n (for n > 1) and n2 � 3 � n2 � 12n
2 = 12n
2 (for n > 1), then
n+ 2
n2 � 3 �2n12n
2=4
n:
Therefore, it is su¢ cient to solve the inequality
4
n< �;
n >4
�
8
Let N0 = max��
1�
�; 2; then for n > N0; we have, since max
��1�
�; 2� 1
� :���� n+ 2n2 � 3 � 0���� � 2n
12n
2=4
n<
1
N0=
1
max��
1�
�; 2 � 1
1�
= �:
Therefore, we showed that,
8�>09N0�N8n>N0
���� n+ 2n2 � 3 � 0���� < �;
thus it follows, by De�nition 16:2; that
limn!1
�n+ 2
n2 � 3
�= 0:
Exercise 16:7: Using any results from section 16 solve the following problems:,
(a) Show that
limn!1
�1
1 + 3n
�= 0
Solution: Hint : Use the same structure of the proof as in the solutions to problems 16:6 (a� e) ; andobserve that ���� 1
1 + 3n� 0���� = 1
1 + 3n<1
n
In order to �nd N0; you just need to solve the following inequality
1
n< �;
thus take for N0 = 1� :
(b) Show that
limn!1
�4n2 � 72n3 � 5
�= 0
Solution: Hint : Use the same structure of the proof as in the solutions to problems 16:6 (a� e) ; andobserve that for n � 2 ����4n2 � 72n3 � 5 � 0
���� = 4n2 � 72n3 � 5 �
4n2
n3=4
n
Thus to �nd N0; you just need to solve the following inequality
4
n< �;
thus take for N0 = max�4� ; 2
:
(c) Show that
limn!1
�6n2 + 5
2n2 � 3n
�= 3
Solution: Hint : Use the same structure of the proof as in the solutions to problems 16:6 (a� e) ; andobserve that for n � 3���� 6n2 + 52n2 � 3n � 3
���� =�����6n2 + 5� 3
�2n2 � 3n
�2n2 � 3n
����� = 9n+ 5
2n2 � 3n �10n
n2=10
n
Thus to �nd N0; you just need to solve the following inequality
10
n< �;
thus take for N0 = max�10� ; 3
:
9
(d) Show that
limn!1
� pn
n+ 1
�= 0
Solution: Hint : Use the same structure of the proof as in the solutions to problems 16:6 (a� e) ; andobserve that for n � 1 ���� pnn+ 1
� 0���� < p
n
n=
1pn
Thus to �nd N0; you just need to solve the following inequality
1pn< �; so n >
1
�2
thus take for N0 = 1�2 :
(e) Show that
limn!1
�n2
n!
�= 0
Solution: Hint : Use the same structure of the proof as in the solutions to problems 16:6 (a� e) ; andobserve that for n � 6; we have n! � n3; thus����n2n! � 0
���� < n2
n3� 1
n
Thus to �nd N0; you just need to solve the following inequality
1
n< �; so n >
1
�
thus take for N0 = max�1� ; 6
:
(f) Show that if jxj < 1 thenlimn!1
xn = 0:
Solution: Suppose that 0 < jxj < 1 and let � > 0 be given. We need to �nd N0; such that, for n > N0 wehave jxn � 0j < �.Since � > 0 is given, we solve the following inequality:
jxn � 0j = jxjn < �n ln (jxj) < ln (�) :
Now, we observe that since 0 < jxj < 1; then ln (jxj) < 0: Thus, we have
n >ln (�)
ln (jxj)
Let N0 = maxln(�)ln(jxj) : Then for n > N0; we have, since 0 < jxj < 1:
jxn � 0j � jxjn < jxjN0 < jxjln(�)ln(jxj) =
�eln(jxj)
� ln(�)ln(jxj)
= eln(�) = �:
Therefore, we showed that, if 0 < jxj < 1; then
8�>09N0�N8n>N0jxn � 0j < �;
thus it follows, by De�nition 16:2; thatlimn!1
(xn) = 0:
10
Now, we consider separately the case when x = 0: We have:
limn!1
xn = limn!1
(0)n= lim
n!10 = 0:
Thus, we showed that, if jxj < 1;limn!1
(xn) = 0:
Exercise 16:8: Show that each of the following sequences is divergent.
(a) Show thatlimn!1
(2n) = +1
Solution: Recall thatlimn!1
an = +1
if and only if8M>0; M2R9N0
8n>N0an �M:
Let M > 0; M 2 R; then we need to show that there is N0; such that, for n > N0; we have:
2n > M:
Since, the last inequality implies that
n >1
2M;
thus we take N0 = 12M; and then for all n > N0; we have
2n � 2N0 > 2�1
2M
�=M:
Thus, we just showed thatlimn!1
(2n) = +1
(b) Show that bn = (�1)n is divergent.
Solution: Since, we have
bn =
��1 if n = 2k � 11 if n = 2k
By Theorem 16:14; the limit of every convergent sequence is unique, however, in our case we clearly observethat we must have one of the following possibilities:
limn!1
bn = �1 or limn!1
bn = 1
Suppose, by contradiction, thatlimn!1
bn = 1:
Then, if 0 < � < 1; then there is N0; such that, for n > N0 :
jbn � 1j < �
However, we have
jbn � 1j =�2 if n = 2k � 10 if n = 2k
in particular, if n = 2k � 1 > N0; then we have
� > jbn � 1j = jb2k�1 � 1j = 2:
11
We have a contradiction, since we assumed that 0 < � < 1: Therefore, we showed that
limn!1
bn 6= 1:
Similar argument shows thatlimn!1
bn 6= �1
Suppose (by contradiction) thatlimn!1
bn = L; where L 6= �1:
and let� =
1
2min fjL� 1j ; jL+ 1jg > 0:
Then according to the de�nition of convergence, we have: there is N0 2 N; such that for n > N0 :
jbn � Lj < �
However, we have
jbn � Lj =�j�1� Lj if n = 2k � 1j1� Lj if n = 2k
thus� > jbn � Lj � min fjL� 1j ; jL+ 1jg ;
since � = 12 min fjL� 1j ; jL+ 1jg ; we have a contradiction.
Therefore, the sequence fbng is divergent.
(c) Show that cn = cos�n�3
�is divergent.
Solution: Since we havecn = cos
�n�3
�2 f�1; 1;�1
2;1
2g;
we can show that
limn!1
cn 6= �1 and limn!1
cn 6= 1 and limn!1
cn 6= �1
2; and lim
n!1cn 6=
1
2:
If L =2��1; � 1
2 ;12 ; 1
; then take
� =1
2min
�jL+ 1j ; jL� 1j ;
����L� 12���� ; ����L+ 12
����� > 0and use the argument (by contradiction) as in (b) :
(d) Show that dn = (�n)2 is divergent.
Solution: The argument is similar to one used for (a) ; since we have
dn = (�n)2 = n2
then show thatlimn!1
dn = +1;
thus we have that the sequencefdng is divergent.Section 17Exercise 17:5: For sn given by the following formulas, determine the convergence or divergence of the sequence(sn) : Find any limits that exist.
(a) sn =3�2n1+n
12
Solution: The sequence (sn) is convergent to �2 since
limn!1
sn = limn!1
3� 2n1 + n
= limn!1
n�3n � 2
�n�1n + 1
� = limn!1
3n � 21n + 1
Since
limn!1
�3
n� 2�= �2 and lim
n!1
�1
n+ 1
�= 1;
therefore, by theorem
limn!1
sn = limn!1
3n � 21n + 1
=limn!1
�3n � 2
�limn!1
�1n + 1
� = �21= �2:
(b) sn =(�1)nn+3
Solution: The sequence (sn) is convergent to 0 since
jsn � 0j =���� (�1)nn+ 3
���� = 1
n+ 3
Sincelimn!1
1
n+ 3= 0;
then by Theorem 16:8; it follows that
limn!1
(�1)n
n+ 3= 0:
(c) sn =(�1)nn2n�1
Solution: The sequence (sn) is divergent. Since
sn =
�� n2n�1 if n = 2k � 1n
2n�1 if n = 2k
and
limn!1
�� n
2n� 1
�= �1
2and lim
n!1
�n
2n� 1
�=1
2:
Now, you can apply arguments as in 16:8 to show that (sn) is divergent.
(d) sn =23n
32n
The sequence (sn) is convergent to 0 since
limn!1
sn = limn!1
23n
32n= lim
n!1
�23
32
�n= lim
n!1
�8
9
�n= 0; since
8
9< 1:
(e) sn =n2�2n+1
Solution: The sequence (sn) is divergent. Since, for n > 2
n2 � 2n+ 1
�n2 � 1
2n2
n+ 1�
12n
2
n+ 1=
n2
2 (n+ 1)� n2
4n=1
4n
and the, as we know
limn!1
1
4n = +1:
(f) sn =3+n�n21+2n
13
Solution: The sequence (sn) is divergent. Since, for n > 2
3 + n� n21 + 2n
� � n2
1 + 2n� �n
2
4n= �1
4n
and
limn!1
��14n
�= �1
(g) sn =1�n2n
Solution: The sequence (sn) is convergent to 0 since
jsn � 0j =����1� n2n � 0
���� = n� 12n
� n
2n
Now, by ratio test
limn!1
n+12n+1
n2n
= limn!1
1
2n(n+ 1) =
1
2< 1;
thuslimn!1
n
2n= 0:
From Theorem 16:8; it follows that
limn!1
1� n2n
= 0:
(h) sn =3n
n3+5
Solution: The sequence (sn) is divergent. Since, for n > 4; we have 3n > n4; and therefore
3n
n3 + 5� n4
2n3=1
2n
andlimn!1
1
2n = +1:
(i) sn =n!2n
Solution: The sequence (sn) is divergent. To show this, we observe that, n! > 3n; for n > 6: Thus, we have:
n!
2n� 3n
2n=
�3
2
�n:
Since
limn!1
�3
2
�n= +1;
we have
limn!1
n!
2n= +1
(j) sn =n!nn
Solution: The sequence (sn) is convergent to 0. To show this, we observe that, n! < nn�1; for n > 3: Thus,we have: ���� n!nn � 0
���� � nn�1
nn=1
n:
Sincelimn!1
1
n= 0;
by Theorem 16:8; we have
limn!1
n!
nn= 0:
14
(k) sn =n2
2n
Solution: The sequence (sn) is convergent to 0. To show this, we observe that, 2n > n3; for n > 6: Thus,we have: ����n22n � 0
���� � n2
n3=1
n:
Sincelimn!1
1
n= 0;
by Theorem 16:8; we have
limn!1
n2
2n= 0:
(l) sn =n2
n!
Solution: The sequence (sn) is convergent to 0. To show this, we observe that, n! > n3; for n > 5: Thus,we have: ����n2n! � 0
���� � n2
n3=1
n:
Sincelimn!1
1
n= 0;
by Theorem 16:8; we have
limn!1
n2
n!= 0:
Exercise 17:15: Prove that
(a) limn!1
�pn+ 1�
pn�= 0
Solution: We have:
limn!1
�pn+ 1�
pn�= lim
n!1
�pn+ 1�
pn� �p
n+ 1 +pn�
pn+ 1 +
pn
= limn!1
n+ 1� npn+ 1 +
pn
= limn!1
1pn+ 1 +
pn
Since1p
n+ 1 +pn� 1
2pn
andlimn!1
1
2pn= 0;
thus by Theorem 16:8; we havelimn!1
�pn+ 1�
pn�= 0
(b) Analogous proof as for (a) :
(c) We have
limn!1
�pn2 + n� n
�= lim
n!1
�pn2 + n� n
� �pn2 + n+ n
�pn2 + n+ n
= limn!1
n�pn2 + n+ n
� = limn!1
n�qn2�1 + 1
n
�+ n
�= lim
n!1
n
n�q�
1 + 1n
�+ 1� = lim
n!1
1�q�1 + 1
n
�+ 1�
15
Since limn!1
�q�1 + 1
n
�+ 1�= 2, we have
limn!1
�pn2 + n� n
�= lim
n!1
1�q�1 + 1
n
�+ 1� = 1
limn!1
�q�1 + 1
n
�+ 1� = 1
2
Section 18Exercise 18:3: Prove that each sequence is monotone and bounded. Then �nd the limit.
(a) s1 = 1 and sn+1 = 14 (sn + 5) for n 2 N:
Solution: We show that the sequence fsng is bounded by 2; that is:
8n�N sn � 2
To show this, we apply induction on n.Base: For n = 1; we have s1 = 1 � 2:Induction hypothesis: Suppose that sn � 2 for a �xed number n 2 N:We show that sn+1 � 2. We observe that, by the induction hypothesis, we have:
sn+1 =1
4(sn + 5) �
1
4(2 + 5) =
7
4� 2:
By induction, it follows that8n�N sn � 2:
Now, we show that fsng is increasing, that is sn � sn+1. Again, we apply induction on n.Base: For n = 1; we have s1 = 1 � s2 = 1
4 (1 + 5) =32 :
Induction hypothesis: Suppose that sn � sn+1 for a �xed number n 2 N:We show that sn+1 � sn+2. We observe that, by the induction hypothesis, we have:
sn+2 =1
4(sn+1 + 5) �
1
4(sn + 5) = sn+1
By induction, it follows that8n�N sn � sn+1:
Thus, the sequence fsng is increasing. Using Theorem 18:3, we have that fsng is convergent. Let
limn!1
sn = s:
Since
limn!1
sn = limn!1
sn+1 =1
4limn!1
(sn + 5) =1
4
�limn!1
sn + 5�
=1
4(s+ 5) : Thus, we have:
s =1
4(s+ 5)
s =5
3:
Thus, we have
limn!1
sn =5
3
(d) s1 = 2 and sn+1 =p2sn + 1 for n 2 N:
16
Solution: We show that the sequence fsng is bounded by 3; that is:
8n�N sn � 3
To show this, we apply induction on n.Base: For n = 1; we have s1 = 2 � 3:Induction hypothesis: Suppose that sn � 3 for a �xed number n 2 N:We show that sn+1 � 2. We observe that, by the induction hypothesis, we have:
sn+1 =p2sn + 1 �
p2� 3 + 1 =
p7 � 3:
By induction, it follows that8n�N sn � 3:
Now, we show that fsng is increasing, that is sn � sn+1. Again, we apply induction on n.Base: For n = 2; we have s1 = 2 � s2 =
p2s1 + 1 =
p4 + 1 =
p5:
Induction hypothesis: Suppose that sn � sn+1 for a �xed number n 2 N:We show that sn+1 � sn+2. We observe that, by the induction hypothesis, we have:
sn+2 =p2sn+1 + 1 �
p2sn + 1 = sn+1
By induction, it follows that8n�N sn � sn+1:
Thus, the sequence fsng is increasing. Using Theorem 18:3, we have that fsng is convergent. Let
limn!1
sn = s:
Since
s = limn!1
sn = limn!1
sn+1 = limn!1
p2sn + 1 =
qlimn!1
2sn + 1
=p2s+ 1: Thus, we have:
s =p2s+ 1
s2 = 2s+ 1
s2 � 2s� 1 = 0
s =p2 + 1 > 0 or s = 1�
p2 < 0
Thus, we havelimn!1
sn =p2 + 1
Section 19Exercise 19.3: For each sequence, �nd the set S of subsequential limits, the limit superior, and the limitinferior.
(a) sn = (�1)n
Solution: We observe that
sn =
��1 if n = 2k � 11 if n = 2k
Therefore,S = f�1; 1g
sincelimk!1
s2k�1 = �1 and limk!1
s2k = 1:
Moreover, since
lim supn!1
sn = supS = 1
lim infn!1
sn = inf S = �1
17
(d) vn = n sin�n�2
�Solution: We observe that
vn =
8<: 0 if n = 2kn if n = 2k � 1
�n if n = 2k + 1
Therefore,S = f�1; 0; 1g
sincelimk!1
s2k�1 =1 and limk!1
s2k = 0 and limk!1
s2k+1 = �1:
Moreover,
lim supn!1
sn = 1;
lim infn!1
sn = �1:
Exercise 19.4: For each sequence, �nd the set S of subsequential limits, the limit superior, and the limitinferior.
(a) wn =(�1)nn
Solution: We observe that
wn =
� �1n if n = 2k + 11n if n = 2k + 2
Therefore,S = f0g
sincelimk!1
w2k+1 = 0 and limk!1
w2k+2 = 0:
Moreover,lim supn!1
wn = lim infn!1
wn = limn!1
wn = 0:
(d) zn = (�n)n
Solution: We observe that
zn =
�nn if n = 2k
�nn if n = 2k + 1
Therefore,S = f�1; 1g
sincelimk!1
s2k =1 and limk!1
s2k+1 = �1:
Moreover,
lim supn!1
sn = 1;
lim infn!1
sn = �1:
Section 20Exercise 20.3
(a) limx!1
x3+5x2+2 =
limx!1(x3+5)
limx!1
(x2+2) =63 = 2:
(b) limx!1
x2+2x�3x2�1 = lim
x!1
(x+3)(x�1)(x�1)(x+1) = lim
x!1
(x+3)(x+1) =
limx!1
(x+3)
limx!1
(x+1) =42 = 2
(c) limx!1
px�1x�1 = lim
x!1
1px+1
= 12
18
(d) limx!0
x2+4xx2+2x = lim
x!0
x+4x+2 = 2:
(e) limx!0
x2+3xx2+1 = 0:
(f) limx!0
p4+x�2x = lim
x!0
(p4+x�2)(
p4+x�2)
x(p4+x+2)
= limx!0
4+x�4x(p4+x+2)
= limx!0
x
x(p4+x+2)
= limx!0
1p4+x+2
= 14 .
(g) Since x! 0�; then x < 0 and jxj = �x; therefore
limx!0�
4x
jxj = limx!0�
4x
�x = �4
(h) Since x! 1+; thus x > 1 and jx� 1j = x� 1; so
limx!0�
x2 � 1jx� 1j = lim
x!0�
(x� 1) (x+ 1)(x� 1) = 1.
Exercise: 20.6: Use De�nition 20:1 to prove each limit.
(a) limx!5
�x2 � 3x+ 1
�= 11
Solution: Let � > 0 be given. We need to �nd a positive �; such that, for all x 2 R, we have:
If 0 < jx� 5j < � )��(x2 � 3x+ 1)� 11�� < �:
Since � > 0 is given, and��(x2 � 3x+ 1)� 11�� =��x2 � 3x� 10�� = j(x� 5)(x+ 2)j
= jx� 5jjx+ 2j
Now, we observe that if � < 1; then
jx� 5j < � < 1; so
�1 + 5 < x < 1 + 5
4 < x < 6; so
6 = 4 + 2 < x+ 2 < 6 + 2 = 8; so in particular
jx+ 2j < 8:
Therefore, we have for � < 1; one has:��(x2 � 3x+ 1)� 11�� =��x2 � 3x� 10�� = j(x� 5)(x+ 2)j
= jx� 5jjx+ 2j < � � 8 = 8� < �:
Thus, if � = min�1; �
8
; we have: for x 2 R such that 0 < jx� 5j < �:��(x2 � 3x+ 1)� 11�� � jx� 5jjx+ 2j < � jx+ 2j
Exercise 20.9 Determine whether or not the following limits exist. Justify your answers.
19
(a) limx!0+
1x
Solution: Let f (x) = 1x ; then domain of f is (�1; 0) [ (0; 1) : Since 0 is an accumulation point of
(�1; 0) [ (0; 1) ; Theorem 20:10 applies. We have: Let sn = 1n ; n 2 N. We see that
limn!1
sn = 0 and sn > 0; and
limn!1
f (sn) = limn!1
1
sn= lim
n!1
11n
= limn!1
(n) =1:
Therefore, by theorem 20:10; we have limx!0+
1x does not exist.
(b) limx!0+
sin�1x
�Solution: Let f (x) = sin
�1x
�; then domain of f is (�1; 0) [ (0; 1) : Since 0 is an accumulation point of
(�1; 0) [ (0; 1) ; Theorem 20:10 applies. We have: Let sn = 12�n ; n 2 N. We see that
limn!1
sn = 0 and sn > 0; and
limn!1
f (sn) = limn!1
sin
�1
sn
�= lim
n!1sin
�112�n
�= lim
n!1sin (2�n) = 0:
Moreover, let pn = 1�2+2�n
; n 2 N: We see that
limn!1
pn = 0 and pn > 0; and
limn!1
f (pn) = limn!1
sin
�1
pn
�= lim
n!1sin
11
�2+2�n
!= lim
n!1sin��2+ 2�n
�= 1:
Sincelimn!1
f (pn) 6= limn!1
f (sn)
Therefore, by theorem 20:10; we have limx!0+
sin�1x
�does not exist.
Section 21Exercise 21:3 Let f (x) = (x2 � 4x � 5)=(x � 5) for x 6= 5: How should f(5) be de�ned so that f will becontinuous at 5:Solution: Notice that limx!5 f(x) = limx!5(x
2 � 4x� 5)=(x� 5) = limx!5(x�5)(x+1)
x�5 = limx!5(x+1) = 6:For f to be continuous at 5, we need limx!5 f(x) = f(5), so f(5) = 6:Exercise 21:4 De�ne f : R! R by f (x) = x2 � 3x+ 5: Use De�nition 21:1 to prove that f is continuous at2:Solution: We need to use the de�nition of continuity to show that limx!2 f(x) = f(2), that is we need toshow:
: Then we have for x 2 R such that jx� 2j < �:��(x2 � 3x+ 5)� 3�� � jx� 1jjx� 2j < �:
Section 22Exercises 22:4 : Show that 2x = 3x for some x 2 (0; 1).Solution: Let f (x) = 2x � 3x: Notice
2x = 3x has a solution inside (0; 1) if and only if
f (x) = 0 for x 2 (0; 1) :
20
Since f is continuous on R, and f (0) = 1 and f (1) = 2� 3 = �1, by the Intermediate Value Theorem, wehave that
9c2(0; 1) (f (c) = 0) :
Thus, there is c 2 (0; 1) ; such that 2c = 3c:Exercise 22:5 : Show that 3x = x2 has at least one real solution.Solution: Let f (x) = 3x � x2: Notice
3x = x2 has at least one real solution if and only if
f (x) = 0 for some x 2 R.
Since f is continuous on R, and f (�1) = 13 � 1 = � 2
3 and f (0) = 1 � 0 = 1, by the Intermediate ValueTheorem, we have that
9c2(�1; 1) (f (c) = 0) :
Thus, there is c 2 (�1; 0) ; such that 3c = c2: Therefore, 3x = x2 has at least one real solution.Exercise 22:6 : Show that any polynomial of odd degree has at least one real root.Solution: Let f (x) = anx
n + an�1xn�1 + ::: + a1x + a0; ai 2 R; i = 0; 1; :::; n; be polynomial of degree
n > 0; and n is odd. Assume that an > 0: Then
limx!�1
f (x) = �1 and limx!1
f (x) =1:
It follows, that8M > 0 9a;b2R f (a) < �M and f (b) > M
Thus, in particular, we havef (a) < 0 and f (b) > 0
Since all polynomial functions are continuous, then by the Intermediate Value Theorem, we have that
9c2(a; b) (f (c) = 0) :
Thus, there is c 2 (a; b) ; such that f (c) = 0; that is, f has at least one real root:Section 23Exercise 23:3: Determine which of the following continuous functions are uniformly continuous on the givenset. Justify your answer.
(a) f (x) = ex
x on [2; 5]
Solution: Since the set [2; 5] � R is closed and bounded, thus it is compact by Heine-Borel theorem. Since,f is continuous (as quotient of two continuous functions ex and x), it follows from Theorem 23:6 that f isalso uniformly continuous on [2; 5].
(b) f (x) = ex
x on (0; 2)
Solution: We observe thatlimx!0+
ex
x=1:
Therefore, the function f cannot be extended to a function ef that is continuous on [0; 2]. It follows fromTheorem 23:9 that f is not uniformly continuous on (0; 2).
(c) f (x) = x2 + 3x� 5 on [0; 4]
Solution: Since the set [0; 4] � R is closed and bounded, thus it is compact by Heine-Borel theorem. Since,f is continuous (f is a polynomial function), it follows from Theorem 23:6 that f is also uniformly continuouson [0; 4].
(d) f (x) = x2 + 3x� 5 on (0; 4)
21
Solution: We observe that
limx!0+
�x2 + 3x� 5
�= �5 and lim
x!4�
�x2 + 3x� 5
�= 23:
Therefore, the function f can be extended to a function ef that is continuous on [0; 4] ; namely, we de�ne:ef : [0; 4]! Ref (x) = x2 + 3x� 5
It follows from Theorem 23:9 that f is uniformly continuous on (0; 4).
(e) f (x) = 1x2 on (0; 1)
Solution: We observe thatlimx!0+
1
x2=1:
Therefore, the function f cannot be extended to a function ef that is continuous on [0; 1]. It follows fromTheorem 23:9 that f is not uniformly continuous on (0; 1).
(f) f (x) = 1x2 on (0; 1)
Solution: Suppose that f is uniformly continuous on (0; 1). Then its restriction to a smaller subinterval(0; a) ; a > 0 given by:
f : (0; a)! Rf (x) = f (x) ; x 2 (0; a)
is also uniformly continuous. By Theorem 23:9; it follows that f can be extended to the function ef that iscontinuous on [0; a]. However, this is impossible, since
limx!0+
1
x2=1:
We have derived a contradiction with the assumption that f is uniformly continuous on (0; 1). Therefore,f is not uniformly continuous on (0; 1).
(g) f (x) = x sin�1x
�on (0; 1)
Solution: We observe that
�1 � sin�1
x
�� 1
Therefore, if x > 0; we have
�x � x sin�1
x
�� x
and if x < 0; we have
�x � x sin�1
x
�� x
Since limx!0+ x = limx!0� x = 0; it follows that
limx!0
x sin
�1
x
�= lim
x!0+x = lim
x!0�x = 0
Therefore, the function f can be extended to a function ef that is continuous on [0; 1] ; namely, we de�ne:ef : [0; 1]! Ref (x) =
�x sin
�1x
�if x 2 (0; 1]
0 if x = 0
It follows from Theorem 23:9 that f is uniformly continuous on (0; 1).Section 25Exercise 25:3: Determine if each function is di¤erentiable at x = 1: If it is, �nd the derivative. If not, explainwhy not.
22
(a) f (x) =�3x� 2 if x < 1x3 if x � 1
Solution: Notice that f is continuous at 1 because
limx!1+
f(x) = 1 = limx!1�
f(x) = f(1):
Recall, f is di¤erentiable at x0 if
limx!x0
f (x)� f (x0)x� x0
exists. Such limit exists if and only if
limx!x+0
f (x)� f (x0)x� x0
= limx!x�0
f (x)� f (x0)x� x0
We observe that, for x0 = 1 :
limx!1�
f (x)� f (1)x� 1 = lim
x!1
3x� 2� 1x� 1 = lim
x!1
3(x� 1)x� 1 = 3
and
limx!1+
f (x)� f (1)x� 1 = lim
x!1
x3 � 1x� 1 = lim
x!1
(x� 1)�x2 + x+ 1
�x� 1
= limx!1
�x2 + x+ 1
�= 3:
Therefore, we have
limx!x+0
f (x)� f (x0)x� x0
= limx!x�0
f (x)� f (x0)x� x0
;
so the function f is di¤erentiable at x0 = 1, and
f 0 (0) = 3
(b) f (x) =�2x+ 1 if x < 1x2 if x � 1
Solution: Notice that f is discontinuous at 1 because
limx!1+
f(x) = limx!1+
�x2�= 1 = f (1) ;while
limx!1�
f(x) = limx!1�
(2x+ 1) = 3:
Therefore, we havelimx!1
f (x) does not exist.
It follows, f is not continuous at x0 = 1. Thus, using Theorem 25:6; f cannot be di¤erentiable at x0 = 1:
(c) f (x) =�3x� 2 if x < 1x2 if x � 1
Solution: Notice that f is continuous at 1 because
limx!1+
f(x) = 1 = limx!1�
f(x) = f(1):
Recall, f is di¤erentiable at x0 if
limx!x0
f (x)� f (x0)x� x0
23
exists. Such limit exists if and only if
limx!x+0
f (x)� f (x0)x� x0
= limx!x�0
f (x)� f (x0)x� x0
We observe that, for x0 = 1 :
limx!1�
f (x)� f (1)x� 1 = lim
x!1
3x� 2� 1x� 1 = lim
x!1
3(x� 1)x� 1 = 3
and
limx!1+
f (x)� f (1)x� 1 = lim
x!1
x2 � 1x� 1 = lim
x!1
(x� 1) (x+ 1)x� 1
= limx!1
(x+ 1) = 2:
Therefore, we have
limx!x+0
f (x)� f (x0)x� x0
6= limx!x�0
f (x)� f (x0)x� x0
;
so the function f is not di¤erentiable at x0 = 1.Exercise 25:4 : Use De�nition 25:1 to �nd the derivative of each function.
(a) f (x) = 2x+ 7; for x 2 R
Solution: Let c 2 R. Using the De�nition 25:1; we need to �nd limx!c
f(x)�f(c)x�c . We have:
limx!c
f (x)� f (c)x� c = lim
x!c
(2x+ 7)� (2c+ 7)x� c = lim
x!c
2 (x� c)(x� c) = 2:
Therefore, we have, for x 2 R :f 0 (x) = 2
(b) f (x) = x3; for x 2 R:
Solution: Let c 2 R. Using the De�nition 25:1; we need to �nd limx!c
f(x)�f(c)x�c . We have:
limx!c
f (x)� f (c)x� c = lim
x!c
x3 � c3x� c = lim
x!c
(x� c)�x2 + cx+ c2
�(x� c)
= limx!c
�x2 + cx+ c2
�= c2 + c2 + c2 = 3c2:
Therefore, we have, for x 2 R :f 0 (x) = 3x2
(c) f (x) = 1x ; for x 6= 0:
Solution: Let c 2 R; c 6= 0. Using the De�nition 25:1; we need to �nd limx!c
f(x)�f(c)x�c . We have:
limx!c
f (x)� f (c)x� c = lim
x!c
1x �
1c
x� c = limx!c
c�xxc
(x� c) = limx!c
� (x� c)xc
� 1
(x� c)
= limx!c
�1xc
= � 1c2:
Therefore, we have, for x 2 R; x 6= 0 :f 0 (x) = � 1
x2
(d) f (x) =px; for x > 0:
24
Solution: Let c 2 R; c > 0. Using the De�nition 25:1; we need to �nd limx!c
f(x)�f(c)x�c . We have:
limx!c
f (x)� f (c)x� c = lim
x!c
px�
pc
x� c = limx!c
(px�
pc)
(px�
pc) (px+
pc)= lim
x!c
1px+
pc
=1p
c+pc=
1
2pc:
Therefore, we have, for x 2 R; x > 0 :f 0 (x) =
1
2px:
(e) f (x) = 1px; x > 0:
Solution: Let c > 0. Using the De�nition 25:1; we need to �nd limx!c
f(x)�f(c)x�c . We have:
limx!c
f (x)� f (c)x� c = lim
x!c
1px� 1p
c
x� c = limx!c
pc�pxp
xpc
(px�
pc) (px+
pc)= lim
x!c
(pc�
px)p
cx (px�
pc) (px+
pc)
= limx!c
�1pcx (
px+
pc)= � 1
2cpc:
Therefore, we have, for x > 0 :
f 0 (x) = � 1
2xpx
Exercise 25:5 : Let f (x) = x1=3 for x 2 R:
(a) Use De�nition 25:1 to prove that
f 0 (x) =1
3x�2=3 for x 6= 0
Solution: Let c 2 R; c 6= 0. Using the De�nition 25:1; we need to �nd limx!c
f(x)�f(c)x�c . We have:
limx!c
f (x)� f (c)x� c = lim
x!c
3px� 3
pc
x� c = limx!c
( 3px� 3
pc)�( 3px)2+ 3px 3pc+ ( 3
pc)2�
(x� c)�( 3px)2+ 3px 3pc+ ( 3
pc)2�
= limx!c
(x� c)(x� c)
�( 3px)2+ 3px 3pc+ ( 3
pc)2� = 1
( 3pc)2+ 3pc 3pc+ ( 3
pc)2
=1
3 ( 3pc)2 =
1
3c�2=3:
Therefore, we have, for x 2 R; x 6= 0 :f 0 (x) =
1
3x�2=3:
(b) Show that f is not di¤erentiable at 0.
Solution: Using the De�nition 25:1; we show that limx!0
f(x)�f(0)x�0 does not exist. We have:
limx!0
f (x)� f (0)x� 0 = lim
x!0
3px
x= lim
x!0
13px2=1
Therefore, limx!0
f(x)�f(0)x�0 does not exist, and f is not di¤erentiable at x = 0:
Exercise 25:6 : Let f (x) = x2 sin�1x
�for x 6= 0 and f (0) = 0:
(a) Use the chain rule and the product rule to show that f is di¤erentiable at each c 6= 0 and �nd f 0 (c).
25
Solution: Using the product and the chain rule, we have, for x 6= 0 :
f 0 (x) =
�x2 sin
�1
x
��0=�x2�0sin
�1
x
�+ x2
�sin
�1
x
��0= 2x sin
�1
x
�+ x2
�cos
�1
x
���1
x
�0!= 2x sin
�1
x
�+ x2
�cos
�1
x
���� 1
x2
�= 2x sin
�1
x
�� cos
�1
x
�:
Hence, we have:
f 0 (x) = 2x sin
�1
x
�� cos
�1
x
�:
(b) Use De�nition 25:1 to show that f is di¤erentiable at x = 0 and �nd f 0 (0).
Solution: By De�nition 25:1 we need to compute:
limx!0
f (x)� f (0)x� 0 :
Since f (0) = 0 and f (x) = x2 sin�1x
�for x 6= 0, we have:
limx!0
f (x)� f (0)x� 0 = lim
x!0
x2 sin�1x
�� 0
x= lim
x!0
�x sin
�1
x
��= 0;
because
�1 � sin�1
x
�� 1;
so it is bounded, andlimx!0
x = 0:
Thus,
f 0 (0) = limx!0
f (x)� f (0)x� 0 = lim
x!0x sin
�1
x
�= 0
(c) Show that f 0 is not continuous at x = 0:
Solution: We show thatlimx!0
f 0 (x) 6= f 0 (0) = 0:
We have:
limx!0
f 0 (x) = limx!0
�2x sin
�1
x
�� cos
�1
x
��Now, we observe that
limx!0
cos
�1
x
�does not exist. To see this, let sn = 1
�n . We have:
limn!1
sn = 0
however:
limn!1
cos
�1
sn
�= lim
n!1cos
�11�n
�= lim
n!1cos (�n) =
�1; n is even�1; n is odd
:
By Theorem 20:10, limx!0
cos�1x
�does not exist. Since, as we showed
limx!0
2x sin
�1
x
�= 0;
limx!0
f 0 (x) = limx!0
�2x sin
�1
x
�� cos
�1
x
��DOES NOT EXIST. Therefore, f 0 is not continuous at x = 0.