math 3310 theoretical concepts of calculus section...

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 <

Solution: We haveint�nr 2 Q : 0 < r <

p2o�

since for any x 2 R and � > 0 :

N (x; �) *nr 2 Q : 0 < r <

d)�r 2 Q : r �

Solution: We haveint�nr 2 Q : r �

p2o�

since for any x 2 R and � > 0 :N (x; �) *

nr 2 Q : r �

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.

a)�1n : n 2 N

Solution: We see that

��1

n: n 2 N

��=

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

n: n 2 N

�6= ;:

Hence, we conclude that

0 2 bd��

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: n 2 N

�6= ;

and clearly

n; �

�\ Rn

n: n 2 N

�6= ;:

Thus, we conclude (after considering all of the above cases) that

��1

n: n 2 N

��=

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= ;:

Therefore, we havebd ([0; 3] [ (3; 5)) = bd ([0; 5)) = f0; 5g :

c)�r 2 Q : 0 < r <

Solution: We clearly have

bd�nr 2 Q : 0 < r <

p2o�

=h0;p2i

since for any x 2�0;p2�and � > 0 :

N (x; �) \nr 2 Q : 0 < r <

p2o6= ;

Moreover, we have

N (x; �) \ Rnnr 2 Q : 0 < r <

p2o6= ;:

It follows thatbd�nr 2 Q : 0 < r <

p2o�

=h0;p2i

d)�r 2 Q : r �

Solution: We havebd�nr 2 Q : r �

p2o�

=hp2; 1

�since for any x 2

�p2; 1

�; we have: If � > 0;

N (x; �) \nr 2 Q : r �

p2o6= ;:

Moreover, analogously, we have

N (x; �) \ Rnnr 2 Q : r �

p2o6= ;:

It follows thatbd�nr 2 Q : r �

p2o�

=hp2; 1

e) [0; 2] \ [2; 4] :Solution: We clearly have

[0; 2] \ [2; 4] = f2g :Therefore, we have

bd ([0; 2] \ [2; 4]) = bd (f2g) = f2g :Obviously, for any � > 0;

N (2; �) \ f2g 6= ; and N (2; �) \ Rn f2g 6= ;:

Thus, it follows thatbd ([0; 2] \ [2; 4]) = bd (f2g) = f2g :

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);

��1

n: n 2 N

��= ; 6=

n: n 2 N

therefore,�1n : n 2 N

is not open. As we see from Exercise 13:4 a);

��1

n: n 2 N

��=

n: n 2 N

�[ f0g *

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.

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

�0;1

�= ;;

therefore,1\n=1

�0; 1n

�is both open and closed.

e)�x : jx� 5j � 1

Solution: As one can verify�

x : jx� 5j � 1

�x : �1

2� x� 5 � 1

�x :

2� x � 11

�Clearly, we have

��9

��=

��9

�Therefore,

�92 ;

�is closed. However, we also can verify that

��9

��=

�6=�9

therefore�92 ;

�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.

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

2; 3� 1

�; 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.

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

�n� 1

�; 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

n: n 2 N

�is not closed,

since as we saw in Exercise 13:4 a) we have

��1

n: n 2 N

��=

n: n 2 N

�[ f0g *

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.

Solution: Let

n� 1

�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

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

�qn �

2n; qn +

�; 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 �

2n; qn +

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

�= 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�� < �

< �; hence

n >jkj�

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

�= 0:

(b) For any real number k > 0

limn!1

�= 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�� < �;

limn!1

�= 0:

(c) Show that

limn!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)

n+ 2< �; hence

�� 2

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�

8�>09N0�N8n>N0

��3n+ 1n+ 2� 3�� < �;

limn!1

�3n+ 1

�= 3:

(d) Show that

limn!1

�sin (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

n< �

Let N0 =�1�

�; then for n > N0; we have, since

�1�

�� 1

� :�� sin (n)n� 0�� = jsin (n)j

n� 1

N0� 1

8�>09N0�N8n>N0

�� sin (n)n� 0�� < �;

limn!1

�sin (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

n2 � 3 �2n12n

Therefore, it is su¢ cient to solve the inequality

n< �;

Let N0 = max��

�; 2; then for n > N0; we have, since max

��1�

�; 2� 1

� :�� n+ 2n2 � 3 � 0�� 2n

max��

�; 2 � 1

= �:

8�>09N0�N8n>N0

�� n+ 2n2 � 3 � 0�� < �;

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 + 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

In order to �nd N0; you just need to solve the following inequality

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 �

Thus to �nd N0; you just need to solve the following inequality

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

�� = 9n+ 5

2n2 � 3n �10n

n< �;

(d) Show that

limn!1

� pn

�= 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

1pn< �; so n >

thus take for N0 = 1�2 :

(e) Show that

limn!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 � 6; we have n! � n3; thus��n2n! � 0

�� < n2

n3� 1

n< �; so n >

(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:

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

thus we take N0 = 12M; and then for all n > N0; we have

2n � 2N0 > 2�1

�=M:

Thus, we just showed thatlimn!1

(2n) = +1

(b) Show that bn = (�1)n is divergent.

Solution: Since, we have

��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:

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� =

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

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=

If L =2��1; � 1

2 ;12 ; 1

; then take

� =1

�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

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

limn!1

n� 2�= �2 and lim

�= 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

Sincelimn!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

�� n2n�1 if n = 2k � 1n

2n�1 if n = 2k

limn!1

�� n

2n� 1

�= �1

2and lim

2n� 1

Now, you can apply arguments as in 16:8 to show that (sn) is divergent.

(d) sn =23n

The sequence (sn) is convergent to 0 since

limn!1

sn = limn!1

32n= lim

�n= lim

�n= 0; since

(e) sn =n2�2n+1

Solution: The sequence (sn) is divergent. Since, for n > 2

n2 � 2n+ 1

�n2 � 1

n+ 1�

2 (n+ 1)� n2

and the, as we know

limn!1

4n = +1:

(f) sn =3+n�n21+2n

Solution: The sequence (sn) is divergent. Since, for n > 2

3 + n� n21 + 2n

� � n2

1 + 2n� �n

4n= �1

limn!1

��14n

�= �1

(g) sn =1�n2n

Solution: The sequence (sn) is convergent to 0 since

jsn � 0j =��1� n2n � 0

�� = n� 12n

Now, by ratio test

limn!1

n+12n+1

= limn!1

2n(n+ 1) =

thuslimn!1

2n= 0:

From Theorem 16:8; it follows that

limn!1

1� n2n

(h) sn =3n

Solution: The sequence (sn) is divergent. Since, for n > 4; we have 3n > n4; and therefore

n3 + 5� n4

andlimn!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:

2n� 3n

limn!1

�n= +1;

we have

limn!1

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

Sincelimn!1

by Theorem 16:8; we have

limn!1

nn= 0:

(k) sn =n2

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

Sincelimn!1

limn!1

2n= 0:

(l) sn =n2

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

Sincelimn!1

limn!1

n!= 0:

Exercise 17:15: Prove that

(a) limn!1

�pn+ 1�

pn�= 0

Solution: We have:

limn!1

�pn+ 1�

pn�= lim

�pn+ 1�

pn� �p

n+ 1 +pn�

pn+ 1 +

= limn!1

n+ 1� npn+ 1 +

= limn!1

1pn+ 1 +

Since1p

n+ 1 +pn� 1

andlimn!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

�pn2 + n� n

� �pn2 + n+ n

�pn2 + n+ n

= limn!1

n�pn2 + n+ n

� = limn!1

n�qn2�1 + 1

�+ n

�= lim

n�q�

1 + 1n

�+ 1� = lim

1�q�1 + 1

�+ 1�

Since limn!1

�q�1 + 1

�+ 1�= 2, we have

limn!1

�pn2 + n� n

�= lim

1�q�1 + 1

�+ 1� = 1

limn!1

�q�1 + 1

�+ 1� = 1

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) �

4(2 + 5) =

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) �

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:

limn!1

sn = limn!1

sn+1 =1

4limn!1

(sn + 5) =1

�limn!1

sn + 5�

4(s+ 5) : Thus, we have:

4(s+ 5)

Thus, we have

limn!1

(d) s1 = 2 and sn+1 =p2sn + 1 for n 2 N:

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 =

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:

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

��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

(d) vn = n sin�n�2

�Solution: We observe that

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

� �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

�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+3)(x�1)(x�1)(x+1) = lim

(x+3)(x+1) =

limx!1

(x+1) =42 = 2

(c) limx!1

px�1x�1 = lim

(d) limx!0

x2+4xx2+2x = lim

x+4x+2 = 2:

(e) limx!0

x2+3xx2+1 = 0:

(f) limx!0

p4+x�2x = lim

(p4+x�2)(

p4+x�2)

x(p4+x+2)

= limx!0

4+x�4x(p4+x+2)

= limx!0

x(p4+x+2)

= limx!0

1p4+x+2

= 14 .

(g) Since x! 0�; then x < 0 and jxj = �x; therefore

limx!0�

jxj = limx!0�

�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; �

; we have: for x 2 R such that 0 < jx� 5j < �:��(x2 � 3x+ 1)� 11�� jx� 5jjx+ 2j < � jx+ 2j

Since for 0 < jx� 5j < � = min�1; �

; we have: jx+ 2j < 8; thus:��(x2 � 3x+ 1)� 11�� jx� 5jjx+ 2j < � jx+ 2j < 8� < 8 � �

8= �:

8�>09�>08x2R(0 < jx� 5j < � ) j(x2 � 3x+ 1)� 11j < �;

limx!5

�x2 � 3x+ 1

�= 11:

Exercise 20.9 Determine whether or not the following limits exist. Justify your answers.

(a) limx!0+

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

sn= lim

= 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

�; 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

�= 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

�= lim

n!1sin

�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:

8� > 0 9� > 0 8x 2 R (jx� 2j < � ) jf(x)� f(2)j =��(x2 � 3x+ 5)� (22 � 3 � 2 + 5)�� < �):

Since��(x2 � 3x+ 5)� (22 � 3 � 2 + 5)�� = jx2 � 3x+ 2j = j(x� 1)(x� 2)j < 2� = � for � = 1, � = �

2 .Now let � > 0 be given. We take � = min

�1; �

: 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) :

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+

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)

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+

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+

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)

�1 � sin�1

�� 1

Therefore, if x > 0; we have

�x � x sin�1

�� x

and if x < 0; we have

�x � x sin�1

�� x

Since limx!0+ x = limx!0� x = 0; it follows that

limx!0

�= 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

�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.

(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

3x� 2� 1x� 1 = lim

3(x� 1)x� 1 = 3

limx!1+

f (x)� f (1)x� 1 = lim

x3 � 1x� 1 = lim

(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

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

3x� 2� 1x� 1 = lim

3(x� 1)x� 1 = 3

limx!1+

f (x)� f (1)x� 1 = lim

x2 � 1x� 1 = lim

(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

(2x+ 7)� (2c+ 7)x� c = lim

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

limx!c

x3 � c3x� c = lim

(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

limx!c

1x �

x� c = limx!c

c�xxc

(x� c) = limx!c

� (x� c)xc

(x� c)

= limx!c

�1xc

= � 1c2:

Therefore, we have, for x 2 R; x 6= 0 :f 0 (x) = � 1

(d) f (x) =px; for x > 0:

Solution: Let c 2 R; c > 0. Using the De�nition 25:1; we need to �nd limx!c

limx!c

x� c = limx!c

(px�

pc) (px+

pc)= lim

Therefore, we have, for x 2 R; x > 0 :f 0 (x) =

(e) f (x) = 1px; x > 0:

Solution: Let c > 0. Using the De�nition 25:1; we need to �nd limx!c

limx!c

1px� 1p

x� c = limx!c

pc�pxp

(px�

pc) (px+

pc)= lim

(pc�

cx (px�

pc) (px+

= limx!c

�1pcx (

pc)= � 1

Therefore, we have, for x > 0 :

f 0 (x) = � 1

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

limx!c

3px� 3

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

3 ( 3pc)2 =

3c�2=3:

Therefore, we have, for x 2 R; x 6= 0 :f 0 (x) =

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= lim

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).

Solution: Using the product and the chain rule, we have, for x 6= 0 :

f 0 (x) =

�x2 sin

��0=�x2�0sin

�+ x2

�sin

��0= 2x sin

�+ x2

�cos

��1

�0!= 2x sin

�+ x2

�cos

�� 1

�= 2x sin

�� cos

Hence, we have:

f 0 (x) = 2x sin

�� cos

(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

x2 sin�1x

�� 0

x= lim

�x sin

��= 0;

because

�1 � sin�1

�� 1;

so it is bounded, andlimx!0

x = 0:

f 0 (0) = limx!0

f (x)� f (0)x� 0 = lim

x!0x sin

�= 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

�� cos

��Now, we observe that

limx!0

�does not exist. To see this, let sn = 1

�n . We have:

limn!1

sn = 0

however:

limn!1

�= 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

�= 0;

limx!0

f 0 (x) = limx!0

�2x sin

�� cos

��DOES NOT EXIST. Therefore, f 0 is not continuous at x = 0.

math 3310 theoretical concepts of calculus section...

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