xercise - hong kong university of science and technologymamyan/ma1023/homework.pdf · logx = 1 2...
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Math1023 Selected Answers to Section 2.5
Exercise 2.5.1By
(1� x)(1 + x+ x
2 + · · ·+ x
n) = 1� x
n+1
,
we get
1 + x+ x
2 + · · ·+ x
n =1
1� x
� x
n+1
1� x
.
For any ✏ > 0, we have
|x| < min{1, ✏}2
=)����
1
1� x
� (1 + x+ x
2 + · · ·+ x
n)
���� =|x|
|1� x| |x|n ✏|x|n.
Therefore1
1� x
is n-th order di↵erentiable at 0, with the n-th order approximation 1 + x +
x
2 + · · ·+ x
n.
Exercise 2.5.2Similar to Exercise 2.5.1, we have
1
1 + x
= 1� x+ x
2 � · · ·+ (�1)nxn + (�1)n+1
x
n+1
1 + x
.
Substituting x by x
2, we also have
1
1 + x
2
= 1� x
2 + x
4 � · · ·+ (�1)nx2n + (�1)n+1
x
2(n+1)
1 + x
.
For any ✏ > 0, we have
|x| < min{1, ✏}2
=)����
1
1 + x
� (1� x+ x
2 + · · ·+ (�1)nxn)
���� =|x|
|1 + x| |x|n ✏|x|n,
and
|x| < ✏ =)����
1
1 + x
2
� (1� x
2 + x
4 + · · ·+ (�1)nx2n)
���� =|x|
1 + x
2
|x|2n+1 ✏|x|2n+1
.
Therefore1
1 + x
= 1� x+ x
2 + · · ·+ (�1)nxn + o(xn),
and1
1 + x
2
= 1� x
2 + x
4 + · · ·+ (�1)nx2n + o(x2n+1).
Exercise 2.5.3For n 100, the n-th order approximation of 1 + 2x+ 3x2 + · · ·+ 100x100 at 0 is 1 + 2x+
3x2 + · · ·+ nx
n.For n > 100, the n-th order approximation of 1 + 2x+ 3x2 + · · ·+ 100x100 at 0 is 1 + 2x+
3x2 + · · ·+ 100x100.
1
Exercise 2.5.4 (1)
limx!0
sin x� x
x
3
= limx!0
cos x� 1
3x2
= limx!0
� sin x
6x= �1
6.
This implies limx!0
sin x� x+ 1
6
x
3
x
3
= 0, or
sin x = x� 1
6x
3 + o(x3).
Exercise 2.5.4 (2)
limx!0
sin x2 � x
2
x
6
= limx!0
2x cos x2 � 2x
6x5
= limx!0
cos x2 � 1
3x4
= limx!0
�2x sin x2
12x3
= limx!0
�sin x2
6x2
= �1
6.
This implies limx!0
sin x2 � x
2 + 1
6
x
6
x
6
= 0, or
sin x2 = x
2 � 1
6x
6 + o(x6).
Exercise 2.5.4 (4)
limx!0
1
x
5
✓sin x� x+
1
6x
3
◆= lim
x!0
1
5x4
✓cos x� 1 +
1
2x
2
◆= lim
x!0
� sin x+ x
20x3
= limx!0
� cos x+ 1
60x2
= limx!0
� sin x
120x= � 1
120.
This means
sin x = x� 1
6x
3 +1
120x
5 + o(x5).
Exercise 2.5.4 (5)
limx!0
sin x� tan x
x
3
= limx!0
sin x
x
· 1
cos x· cos x� 1
x
2
= 1 · 1 ·✓�1
2
◆= �1
2.
This means
sin x� tan x = �1
2x
3 + o(x3).
Exercise 2.5.4 (6)
limx!0
1
x
3
✓e
x � 1� x� 1
2x
2
◆= lim
x!0
e
x � 1� x
3x2
= limx!0
e
x � 1
6x= lim
x!0
e
x
6=
1
6.
This means
e
x = 1 + x+1
2x
2 +1
6x
3 + o(x3).
2
Exercise 2.5.4 (8)
limx!1
2 log x+ (x� 1)(x� 3)
(x� 1)3= lim
x!1
2
x
+ 2x� 4
3(x� 1)2= lim
x!1
(x� 1)2
3x(x� 1)2=
1
3.
This means
log x = �1
2(x� 1)(x� 3) +
1
3(x� 1)3 + o((x� 1)3)
=1
2(x� 1)(2� (x� 1)) +
1
3(x� 1)3 + o((x� 1)3)
= (x� 1)� 1
2(x� 1)2 +
1
3(x� 1)3 + o((x� 1)3).
Exercise 2.5.5First we need the function to be continuous. This means a + b + c = 1, and implies
f(x) = a+ bx+ cx
2 for x 1.For x � 1, we have
f(x) = x
4 = (1 + (x� 1))4 = 1 + 4(x� 1) + 6(x� 1)2 + o((x� 1)2).
For x 1, we have
f(x) = a+bx+cx
2 = a+b(1+(x�1))+c(1+(x�1))2 = (a+b+c)+(b+2c)(x�1)+c(x�1)2.
Therefore the function to be second order di↵erentiable at 1 is and only if
a+ b+ c = 1, b+ 2c = 4, c = 6.
The solution is a = 3, b = �8, c = 6.
Exercise 2.5.6The assumption tells us
f(x) = a+ bx+ cx
2 + o(x2).
Then a = f(0) and
3f(x)� 3f(2x) + f(3x) = [3a+ 3bx+ 3cx2 + 3o(x2)]
� [3a+ 3b(2x) + 3c(2x)2 + 3o((2x)2)]
+ [a+ b(3x) + c(3x)2 + o((3x)2)]
= (3a� 3a+ a) + (3b� 6b+ 3b)x+ (3c� 12c+ 9c)x2
+ 3o(x2)� 3o(4x2) + o(9x2)
= a+ o(x2).
Exercise 2.5.7Suppose
f(x) = a
0
+ a
1
x+ a
2
x
2 + · · ·+ a
n
x
n + o(xn).
3
Thenf(�x) = a
0
� a
1
x+ a
2
x
2 + · · ·+ (�1)nan
x
n + o(xn).
By the uniqueness of high order approximation (see Exercise 2.5.8), we find that f(�x) = f(x)(f is even) implies a
1
= a
3
= · · · = a
2k+1
= 0, we also find that f(�x) = �f(x) (f is odd)implies a
0
= a
2
= · · · = a
2k
= 0.
Exercise 2.5.8
If A0
6= 0, then limx!a
P (x)
(x� a)2= A
0
· (+1) = 1. So for limx!a
P (x)
(x� a)2to converge, we
must have A
0
= 0.
If A0
= 1 and A
1
6= 0, then limx!a
P (x)
(x� a)2= lim
x!a
A
1
+ A
2
(x� a)
(x� a)= A
1
·1 = 1. So
for limx!a
P (x)
(x� a)2to converge, we must also have A
1
= 0.
If A0
= A
1
= 0, then limx!a
P (x)
(x� a)2= lim
x!a
A
2
= A
2
. By the assumption, we have
A
2
= 0.Suppose both P (x) = A
0
+A
1
(x�a)+A
2
(x�a)2 and Q(x) = B
0
+B
1
(x�a)+B
2
(x�a)2 arequadratic approximations of f(x) at a, then P (x)�Q(x) = (A
0
�B
0
)+(A1
�B
1
)(x�a)+(A2
�
B
2
)(x�a)2 is a quadratic approximations of f(x)�f(x) = 0. Therefore limx!a
P (x)�Q(x)
(x� a)2=
0. By what we proved before, we have A
0
� B
0
= A
1
� B
1
= A
2
� B
2
= 0.
Exercise 2.5.9 (1)
(ax)0 = a
x log a,
(ax)00 = a
x(log a)2,
(ax)(n) = a
x(log a)n.
Exercise 2.5.9 (6)
(log(ax+ b))0 = (ax+ b)�1
,
d
n log(ax+ b)
dx
n
= a
n�1(�1)(�2) . . . (�n+ 1)(ax+ b)�n = (�a)n�1(n� 1)!(ax+ b)�n
,
d
n
dx
n
✓log
ax+ b
cx+ d
◆=
d
n
dx
n
(log(ax+ b)� log(cx+ d))
= (�a)n�1(n� 1)!(ax+ b)�n � (�c)n�1(n� 1)!(cx+ d)�n
.
Exercise 2.5.9 (8)
1
(ax+ b)(cx+ d)=
1
ad� bc
✓a
ax+ b
� c
cx+ d
◆,
d
n
dx
n
✓a
ax+ b
◆=
✓x+
b
a
◆�1
!(n)
= (�1)nn!
✓x+
b
a
◆�n�1
=(�1)nn!an+1
(ax+ b)n+1
,
d
n
dx
n
✓1
(ax+ b)(cx+ d)
◆=
(�1)nn!
ad� bc
✓a
n+1
(ax+ b)n+1
� c
n+1
(cx+ d)n+1
◆.
4
Exercise 2.5.10 (1)
(tan x)0 = sec2 x,
(tan x)00 = 2 secx(sec x tan x) = 2 sec2 x tan x,
(tan x)000 = 2(2 sec2 x tan x) tan x+ 2 sec2 x(sec x tan x)
= 4 sec2 x(sec2 x� 1) + 2 sec3 x tan x
= 4 sec4 x+ 2 sec3 x tan x� 4 sec2 x.
Exercise 2.5.10 (2)
(sec x)0 = secx tan x,
(sec x)00 = (secx tan x) tan x+ secx sec2 x = 2 sec3 x� sec x,
(sec x)000 = 6 sec2 x(sec x tan x)� sec x tan x = 6 sec3 x tan x� sec x tan x.
Exercise 2.5.10 (5)
(arctan x)0 =1
x
2 + 1,
(arctan x)00 =�2x
(x2 + 1)2.
Exercise 2.5.10 (6)
(xx)0 = x
x(log x+ 1),
(xx)00 = x
x(log x+ 1)2 + x
x
1
x
= x
x(log x+ 1)2 + x
x�1
,
(xx)000 = x
x(log x+ 1)3 + x
x2(log x+ 1)1
x
+ x
x�2(x log x+ x� 1)
= x
x(log x+ 1)3 + 3xx�1(log x+ 1) + x
x�3
.
Exercise 2.5.10 (7)
d
dx
✓1 +
1
x
◆x
=
✓1 +
1
x
◆x�1
✓✓1 +
1
x
◆log
✓1 +
1
x
◆� 1
x
2
x
◆
=
✓1 +
1
x
◆x
✓log
✓1 +
1
x
◆� 1
1 + x
◆
=
✓1 +
1
x
◆x
✓log(1 + x)� log x� 1
1 + x
◆
d
2
dx
2
✓1 +
1
x
◆x
=
✓1 +
1
x
◆x
✓log(1 + x)� log x� 1
1 + x
◆2
+
✓1 +
1
x
◆x
✓1
1 + x
� 1
x
� 1
(1 + x)2
◆
=
✓1 +
1
x
◆x
✓(log(1 + x)� log x)2 � 2
log(1 + x)� log x
1 + x
� 1
x(1 + x)
◆.
5
Exercise 2.5.11 (6)
e
1
2
= e,
(ex2
)0 = 2xex2
, (ex2
)0x=1
= 2e,
(ex2
)00 = 2(1 + 2x2)ex2
, (ex2
)00x=1
= 6e,
(ex2
)000 = 4(3x+ 2x3)ex2
, (ex2
)000x=1
= 20e,
(ex2
)(4) = 4(3 + 12x2 + 4x4)ex2
, (ex2
)(4)x=1
= 4 · 19e,(ex
2
)(5) = 8(15x+ 20x3 + 4x5)ex2
, (ex2
)(5)x=1
= 8 · 39e,(ex
2
)(6) = 8(15 + 90x2 + 60x4 + 8x6)ex2
, (ex2
)(6)x=1
= 8 · 173e.
The 6-th order approximation is
e
x
2
= e+2e(x�1)+e(x�1)2+10
3e(x�1)3+
19
6e(x�1)4+
13
5e(x�1)5+
173
90e(x�1)6+o((x�1)6).
Exercise 2.5.13We can use the usual chain rule twice
(g(f(x)))00 = (g0(f(x))f 0(x))0 = (g0(f(x)))0f 0(x) + g
0(f(x))(f 0(x))0
= g
00(f(x))f 0(x)2 + g
0(f(x))f 00(x).
Exercise 2.5.14 (2)
((x2 + 1) sin x)(n) = (x2 + 1)(sin x)(n) + n(x2 + 1)0(sin x)(n�1) +n(n� 1)
2(x2 + 1)00(sin x)(n�2)
= (x2 + 1)(sin x)(n) + 2nx(sin x)(n�1) + n(n� 1)(sin x)(n�2)
.
For n = 4k, ((x2 + 1) sin x)(n) = (x2 � n
2 + n+ 1) sin x� 2nx cos x.For n = 4k + 1, ((x2 + 1) sin x)(n) = (x2 � n
2 + n+ 1) cosx+ 2nx sin x.For n = 4k + 2, ((x2 + 1) sin x)(n) = �(x2 � n
2 + n+ 1) sin x+ 2nx cos x.For n = 4k + 3, ((x2 + 1) sin x)(n) = �(x2 � n
2 + n+ 1) cosx� 2nx sin x.
Exercise 2.5.15
The formula holds for n = 0. Suppose (xn�1
e
1
x )(n) =(�1)n
x
n+1
e
1
x . Then
(xn
e
1
x )(n+1) = (x · xn�1
e
1
x )(n+1) = x(xn�1
e
1
x )(n+1) + (n+ 1)x0(xn�1
e
1
x )(n)
= x
✓(�1)n
x
n+1
e
1
x
◆0
+ (n+ 1)(�1)n
x
n+1
e
1
x
= x
✓�(�1)n(n+ 1)
x
n+2
e
1
x � (�1)n
x
n+1
1
x
2
e
1
x
◆+ (n+ 1)
(�1)n
x
n+1
e
1
x
= x
✓�(�1)n
x
n+1
1
x
2
e
1
x
◆=
(�1)n+1
x
n+2
e
1
x
.
6
Exercise 2.5.16We have
(eax sin(bx+ c))0 = e
ax(a sin(bx+ c) + b cos(bx+ c))
=pa
2 + b
2(cos ✓ sin(bx+ c) + sin ✓ cos(bx+ c))
=pa
2 + b
2 sin(bx+ c+ ✓),
This shows that the e↵ect of taking the derivative of eax sin(bx+c) means multiplyingpa
2 + b
2
and adding ✓ to c. Therefore taking the derivative n times means multiplyingpa
2 + b
2
n timesand adding ✓ to c n times. The result is
(eax sin(bx+ c))(n) = (a2 + b
2)n
2
e
ax sin(bx+ c+ n✓).
Similarly, we have
(eax cos(bx+ c))0 = e
ax(a cos(bx+ c)� b sin(bx+ c))
=pa
2 + b
2(cos ✓ cos(bx+ c)� sin ✓ sin(bx+ c))
=pa
2 + b
2 cos(bx+ c+ ✓).
This implies(eax cos(bx+ c))(n) = (a2 + b
2)n
2
e
ax cos(bx+ c+ n✓).
Exercise 2.5.17Since f(x) has second order derivative near x
0
, we have
f(x) = f(x0
) + f
0(x0
)(x� x
0
) +1
2f
00(x0
)(x� x
0
)2 +R(x� x
0
), limh!0
R(h)
h
2
= 0.
Then
f(x0
+ h) = f(x0
) + f
0(x0
)h+1
2f
00(x0
)h2 +R(h),
f(x0
� h) = f(x0
)� f
0(x0
)h+1
2f
00(x0
)h2 +R(�h),
and
limh!0
f(x0
+ h) + f(x0
� h)� 2f(x0
)
h
2
= limh!0
f
00(x0
)h2 +R(h) +R(�h)
h
2
= f
00(x0
) + limh!0
R(h)
h
2
+ limh!0
R(�h)
(�h)2
= f
00(x0
) + 0 + 0 = f
00(x0
).
Exercise 2.5.19 (1)
Suppose limx!0
R(x)
x
5
= 0. Then
limx!0
R(x)
x
3
= limx!0
x
2
R(x)
x
5
= limx!0
x
2 limx!0
R(x)
x
5
= 0 · 0 = 0.
7
Exercise 2.5.19 (4)
Suppose limx!0
R(x)
x
5
= 0. Then
limx!0
x
3
R(x)
x
8
= limx!0
R(x)
x
5
= 0.
Exercise 2.5.19 (6)
Suppose limx!0
R(x)
x
3
= 0. Then
limx!0
R(x) + x
5
x
3
= limx!0
R(x)
x
3
+ limx!0
x
2 = 0 + 0 = 0.
Exercise 2.5.21By Example 2.5.8, we have
d
n arctan x
dx
n
����x=0
=d
n�1
dx
n�1
����x=0
✓1
1 + x
2
◆=
(0, if n = 2k,
(�1)k(2k)!, if n = 2k + 1.
Then the Taylor expansion of arctanx at 0 is
arctan x =nX
k=0
1
(2k + 1)!
d
2k+1 arctan x
dx
2k+1
����x=0
x
2k+1 + o(x2n+2)
=nX
k=0
(�1)k
2k + 1x
2k+1 + o(x2n+2)
= x� 1
3x
3 +1
5x
5 + · · ·+ (�1)n
2n+ 1x
2n+1 + o(x2n+2).
Exercise 2.5.22We have
1p1� x
2
= (1� x
2)�1
2
= 1 +
��1
2
�
1!x
2 +
��1
2
� ��3
2
�
2!x
4 + · · ·+��1
2
� ��3
2
�. . .
��2n�1
2
�
n!x
2n + o(x2n+1)
= 1� 1
2x
2 +3
8x
4 + · · ·+ (�1)n(2n)!
4n(n!)2x
2n + o(x2n+1).
Then
d
n
dx
n
����x=0
✓1p
1� x
2
◆=
8><
>:
0, if n = 2k � 1,
(�1)k(2k)!
4k(k!)2(2k)! = (�1)k
✓(2k)!
2kk!
◆2
, if n = 2k.
8
We further get
d
n arcsin x
dx
n
����x=0
=d
n�1
dx
n�1
����x=0
✓1p
1� x
2
◆=
8><
>:
0, if n = 2k,
(�1)k✓(2k)!
2kk!
◆2
, if n = 2k + 1.
The Taylor expansion is then
arcsin x =nX
k=0
(�1)k
(2k + 1)!
✓(2k)!
2kk!
◆2
x
2k+1 + o(x2n+2)
= x� 1
6x
3 +3
40x
5 + · · ·+ (�1)n(2n)!
4n(2n+ 1)(n!)2x
2n+1 + o(x2n+2).
Exercise 2.5.24 (5)
log(1 + 3x+ 2x2) = log(1 + x)(1 + 2x) = log(1 + x) + log(1 + 2x)
= �x+1
2x
2 � 1
3x
3 + · · ·+ (�1)n�1
1
n
x
n + o(xn)
� 2x+1
2(2x)2 � 1
3x
3 + · · ·+ (�1)n�1
1
n
(2x)n + o((2x)n)
= �(1 + 2)x+1 + 22
2x
2 � 1 + 23
3x
3 + · · ·+ (�1)n�1
1 + 2n
n
x
n + o(xn),
(log(1 + 3x+ 2x2))(n)x=0
= (�1)n�1(1 + 2n)(n� 1)!.
Exercise 2.5.24 (9)
sin x cos 2x =1
2(sin 3x� sin x)
=1
2
✓3x� 1
3!(3x)3 + · · ·+ (�1)n
(2n+ 1)!(3x)2n+1 + o(x2n+2)
◆
� 1
2
✓x� 1
3!x
3 + · · ·+ (�1)n
(2n+ 1)!x
2n+1 + o(x2n+2)
◆
= x� 13
6x
3 +121
120x
5 + · · ·+ (�1)n(32n+1 � 1)
2(2n+ 1)!x
2n+1 + o(x2n+2),
(sin x cos 2x)(n)x=0
=(�1)n(32n+1 � 1)n!
2(2n+ 1)!.
Exercise 2.5.25 (1)
x+ 3
x+ 1= 1 +
1
1 + 1
2
(x� 1)
= 2� 1
2(x� 1) +
1
22(x� 1)2 � 1
23(x� 1)3 + · · ·+ (�1)n
1
2n(x� 1)n + o((x� 1)n),
✓x+ 3
x+ 1
◆(n)
x=1
= (�1)nn!
2n.
9
Exercise 2.5.25 (8)
sin x = sin⇣⇣
x� ⇡
2
⌘+
⇡
2
⌘= cos
⇣x� ⇡
2
⌘
= 1� 1
2!
⇣x� ⇡
2
⌘2
+1
4!
⇣x� ⇡
2
⌘4
+ · · ·+ (�1)n
(2n)!
⇣x� ⇡
2
⌘2n
+ o
✓⇣x� ⇡
2
⌘2n+1
◆,
(sin x)(2n)x=
⇡
2
= (�1)n,
(sin x)(2n+1)
x=
⇡
2
= 0.
Exercise 2.5.25 (11)
sin2
x = 1� 2 cos 2x = 1� 2 cos(2(x� ⇡) + 2⇡) = 1� 2 cos 2(x� ⇡)
= �1 +2
2!(x� ⇡)2 � 2
4!(x� ⇡)4 + · · ·+ (�1)n+12
(2n)!(x� ⇡)2n + o((x� ⇡)2n+1),
(sin2
x)(2n)x=⇡
= (�1)n+12,
(sin2
x)(2n+1)
x=⇡
= 0.
Exercise 2.5.27 (3)From the Taylor expansions of log(1 + x) and e
x at 0, we get
(1 + x)x = e
x log(1+x) = e
x
⇣x�x
2
2
+
x
3
3
�x
4
4
+o(x
4
)
⌘
= e
x
2�x
3
2
+
x
4
3
�x
5
4
+o(x
5
)
= 1 +
✓x
2 � x
3
2+
x
4
3� x
5
4+ o(x4)
◆+
1
2!
✓x
2 � x
3
2+
x
4
3� x
5
4+ o(x4)
◆2
+1
3!
✓x
2 � x
3
2+
x
4
3� x
5
4+ o(x4)
◆3
+ o
✓x
2 � x
3
2+
x
4
3� x
5
4+ o(x4)
◆3
!
= 1 +
✓x
2 � x
3
2+
x
4
3� x
5
4
◆+
1
2!
✓x
4 � 2x2 · x3
2
◆+ o(x5)
= 1 + x
2 � 1
2x
3 +5
6x
4 � 3
4x
5 + o(x5).
The derivatives of the function f(x) = (1 + x)x at 0 are
f
0(0) = 1! · 0 = 0, f
00(0) = 2! · 1 = 2, f
000(0) = 3! ·✓�1
2
◆= �3,
f
(4)(0) = 4! · 56= 20, f
(5)(0) = 5! ·✓�3
4
◆= �90.
Exercise 2.5.27 (4)
logsin x
x
= log1
x
✓x� x
3
3!+
x
5
5!+ o(x6)
◆= log
✓1� x
2
3!+
x
4
5!+ o(x5)
◆
= �✓x
2
3!� x
4
5!+ o(x5)
◆� 1
2
✓x
2
3!� x
4
5!+ o(x5)
◆2
+ o
✓x
2
3!� x
4
5!+ o(x5)
◆3
!
= �x
2
3!+
x
4
5!� 1
2
✓x
2
3!
◆2
+ o(x5) = �x
2
6� x
4
180+ o(x5).
10
Exercise 2.5.28 (2)By
sin x� tan x =sin x(cos x� 1)
cos x=
1
cos x(x+ o(x))
✓�1
2x
2 + o(x2)
◆= �
1
2x
3 + o(x3)
cos x
and limx!0
cos x = 1, we get limx!0
sin x� tan x
x
3
= �1
2.
Exercise 2.5.28 (4)We have
1
x
2
� 1
tan2
x
=1
x
2
✓1�
⇣x cos x
sin x
⌘2
◆
=1
x
2
0
BBB@1�
0
BB@
x
✓1� x
2
2+ o(x3)
◆
x� x
3
6+ o(x4)
1
CCA
2
1
CCCA=
1
x
2
0
BB@1�
0
B@1� x
2
2+ o(x3)
1� x
2
6+ o(x3)
1
CA
2
1
CCA
=1
x
2
1�
✓✓1� x
2
2+ o(x3)
◆✓1 +
x
2
6+ o(x3)
◆◆2
!
=1
x
2
1�
✓1� x
2
3+ o(x3)
◆2
!=
1
x
2
✓1�
✓1� 2x2
3+ o(x3)
◆◆
=2
3+ o(x).
Therefore
limx!0
✓1
x
2
� 1
tan2
x
◆=
2
3.
Exercise 2.5.28 (6)We have
1
x
log(cosx+ sin x) =1
x
log(1 + x+ o(x)) =1
x
[(x+ o(x)) + o(x+ o(x))] = 1 + o(1).
Therefore
limx!0
1
x(x+ 1)log(cosx+ sin x) = lim
x!0
1
x
log(cosx+ sin x) = 1,
andlimx!0
(cos x+ sin x)1
x(x+1) = e
lim
x!0
1
x(x+1)
log(cosx+sinx) = e.
Exercise 2.5.28 (8)
limx!1
x
x � x
log x� x+ 1= lim
x!0
(1 + x)1+x � 1� x
log(1 + x)� x
.
11
We have
(1 + x)1+x � 1� x = e
(1+x) log(1+x) � 1� x = e
(1+x)(x� 1
2
x
2
+o(x
2
)) � 1� x
= e
x+
1
2
x
2
+o(x
2
) � 1� x = 1 +
✓x+
1
2x
2
◆+
1
2!x
2 + o(x2)� 1� x
= x
2 + o(x2),
log(1 + x)� x = �1
2x
2 + o(x2).
Therefore
limx!0
(1 + x)1+x � 1� x
log(1 + x)� x
= limx!0
x
2 + o(x2)
�1
2x
2 + o(x2)= �2.
Exercise 2.5.28 (13)
limx!1
(x� 1) log x
1 + cos ⇡x= lim
x!0
x log(x+ 1)
1 + cos(⇡x+ ⇡)= lim
x!0
x log(x+ 1)
1� cos ⇡x.
We havex log(x+ 1)
1� cos ⇡x=
x(x+ o(x))
1�✓1� (⇡x)2
2+ o(x3)
◆ =1 + o(1)
⇡
2
2+ o(x)
.
Therefore
limx!1
(x� 1) log x
1 + cos ⇡x= lim
x!0
x log(x+ 1)
1� cos ⇡x
Exercise 2.5.30 (1)
limx!0
+
x
p
e
�x
q
= 0 · e�0 = 0 · 1 = 0.
Exercise 2.5.30 (2)Let r = p
q
. Then by Exercise 1.6.15 and 0 < e
�1
< 1, we have
limx!+1
x
p
e
�x
q
= limx!+1
x
r
e
�x = limx!+1
x
r(e�1)x = 0.
Exercise 2.5.31 (1)We have (see Example 2.5.12)
tanp
x� x
p =
✓x+
1
3x
3 + o(x4)
◆p
� x
p = x
p
✓1 +
1
3x
2 + o(x3)
◆p
� 1
�
= x
p
1 + p
✓1
3x
2 + o(x3)
◆+ o
✓1
3x
2 + o(x3)
◆� 1
�= x
p
⇣p
3x
2 + o(x3)⌘,
sinp
x� x
p =
✓x� 1
3x
3 + o(x4)
◆p
� x
p = x
p
✓1� 1
3x
2 + o(x3)
◆p
� 1
�
= x
p
1 + p
✓�1
3x
2 + o(x3)
◆+ o
✓�1
3x
2 + o(x3)
◆� 1
�= x
p
⇣�p
3x
2 + o(x3)⌘.
12
Therefore
limx!0
+
tanp
x� x
p
sinp
x� x
p
= limx!0
+
x
p
⇣p
3x
2 + o(x3)⌘
x
p
⇣�p
3x
2 + o(x3)⌘ = �1.
Exercise 2.5.31 (2)We have (see Example 2.5.12)
tanp
x� sinp
x =
✓x+
1
3x
3 + o(x4)
◆p
�✓x� 1
3x
3 + o(x4)
◆p
= x
p
✓1 +
1
3x
2 + o(x3)
◆p
�✓1� 1
3x
2 + o(x3)
◆p
�
= x
p
⇣1 +
p
3x
2 � 1 +p
3x
2 + o(x3)⌘=
2p
3x
p+2(1 + o(x)).
Therefore
limx!0
+
sinp
x� tanp
x
x
q
= limx!0
+
2p
3x
p+2�q(1 + o(x)) =
8><
>:
0, if p+ 2 > q,
2p
3
, if p+ 2 = q,
+1, if p+ 2 < q.
Exercise 2.5.31 (3)
limx!⇡
4
tan x� cot x
4x� ⇡
= limx!⇡
4
sin2
x� cos2 x
(4x� ⇡) sin x cos x= lim
x!⇡
4
� cos 2x
(4x� ⇡) 1p2
1p2
= limx!⇡
2
2 cosx
(⇡ � 2x)= lim
x!0
2 sin x
2x= 1.
Exercise 2.5.31 (4)
a tan bx� b tan ax
a sin bx� b sin ax=
a
�bx+ 1
3
(bx)3 + o(x4)�� b
�ax+ 1
3
(ax)3 + o(x4)�
a
�bx� 1
3
(bx)3 + o(x4)�� b
�ax� 1
3
(ax)3 + o(x4)�
=(ab3 � a
3
b)x3 + o(x4)
�(ab3 � a
3
b)x3 + o(x4)=
ab
3 � a
3
b+ o(x)
�(ab3 � a
3
b) + o(x).
Therefore
limx!0
a tan bx� b tan ax
a sin bx� b sin ax= �1.
Exercise 2.5.32 (1)
limx!1
x
3
✓sin
1
x
� 1
2sin
2
x
◆= lim
x!0
1
2x3
(2 sin x� sin 2x)
= limx!0
1
2x3
✓2x� 2
1
6x
3 � 2x+1
6(2x)3 + o(x3)
◆
=1
2
✓�2
1
6+
1
623◆
=1
2.
13
Exercise 2.5.32 (2)
x
✓2
x(2 + x)� 1
e
x � 1
◆= x
2
x(2 + x)� 1
x+ x
2
2
+ x
3
6
+ o(x3)
!
=1
1 + x
2
� 1
1 + x
2
+ x
2
6
+ o(x2)
=
1� x
2+
x
2
4+ o(x2)
��1�
✓x
2+
x
2
6
◆+
x
2
4+ o(x2)
�
=x
2
6+ o(x2)
Therefore
limx!0
1
x
✓2
x(2 + x)� 1
e
x � 1
◆= lim
x!0
1
x
2
✓x
2
6+ o(x2)
◆=
1
6.
Exercise 2.5.32 (3)
1
log(1 + x)�1
x
=1
x� x
2
2
+ o(x2)�1
x
=1
x
✓1
1� x
2
+ o(x)� 1
◆=
1
x
⇣1 +
x
2+ o(x)� 1
⌘=
1
2+o(1).
Therefore limx!0
✓1
log(1 + x)� 1
x
◆=
1
2.
Exercise 2.5.32 (4)We have
x
log(1 + x)=
x
x� x
2
2
+ o(x2)=
1
1� x
2
+ o(x)= 1 +
x
2+ o(x),
x
log(x+p1 + x
2)=
x
log(x+ 1 + 1
2
x
2 + o(x2))=
x
(x+ 1
2
x
2)� 1
2
x
2 + o(x2)=
1
1 + o(x)= 1 + o(x).
Then
limx!0
✓1
log(x+p1 + x
2)� 1
log(1 + x)
◆= lim
x!0
1
x
✓x
log(x+p1 + x
2)� x
log(1 + x)
◆
= limx!0
1
x
⇣�x
2+ o(x)
⌘= �1
2.
Exercise 2.5.33 (1)By Example 2.4.10, we have lim
y!0
+
y log y = 0. This further implies limy!0
+
o(y) log y = 0.Then we get
limx!1
�log x log(1� x) = lim
y!0
+
log(1 + y) log y = limy!0
+
(y + o(y)) log y
= limy!0
+
y log y + limy!0
+
o(y) log y = 0.
14
Exercise 2.5.33 (2)By Example 2.4.10, we have lim
x!0
+
x log x = 0. By the composition rule, we then get
limx!0
+
x
x � 1
x log x= lim
x!0
+
e
x log x � 1
x log x= lim
y!0
e
y � 1
y
= limy!0
y + o(y)
y
= 1.
Exercise 2.5.33 (3)
limx!0
+
x
x � 1� x log x
x
2(log x)2= lim
x!0
+
e
x log x � 1� x log x
x
2(log x)2= lim
y!0
e
y � 1� y
y
2
=1
2.
Exercise 2.5.33 (4)By Example 2.4.10, we have lim
x!0
+
x log x = limx!0
+
x(log x)2 = 0. Then
x
x
x
= x
e
x log x
= x
1+x log x+o(x log x) = x · xx log x+o(x log x) = xe
x(log x)
2
+o(x(log x)
2
)
= x(1 + x(log x)2 + o(x(log x)2)) == x+ x
2(log x)2 + o(x2(log x)2).
Then
limx!0
+
x
(x
x
) � x
x
2(log x)2= lim
x!0
+
x
2(log x)2 + o(x2(log x)2)
x
2(log x)2= 1.
Exercise 2.5.33 (5)
limx!1
x
log x � 1
(log x)2= lim
x!1
e
(log x)
2 � 1
(log x)2= lim
y!0
+
e
y � 1
y
= 1.
Exercise 2.5.33 (6)Let x = 1 + y. Then x ! 1 means y ! 0, and
x
x = (1+y)1+y = e
(1+y) log(1+y) = e
(1+y)(y� 1
2
y
2
+o(y
2
)) = e
y+
1
2
y
2
+o(y
2
) = 1+
✓y +
1
2y
2
◆+1
2y
2+o(y2).
Then
limx!1
x
x � x
log x� x+ 1= lim
y!0
(1 + y)1+y � 1� y
log(1 + y)� y
= limy!0
y
2 + o(y2)
�1
2
y
2 + o(y2)= �2.
Exercise 2.5.34 (2)
(1 + ax+ cx
2)b � (1 + bx+ dx
2)a
x
2
= x
�2
✓1 + b(ax+ cx
2) +b(b� 1)
2(ax)2 � 1� a(bx+ dx
2)� a(a� 1)
2(bx)2 + o(x2)
◆
=
✓(bc� ad) +
1
2(b(b� 1)a2 + a(a� 1)b2)
◆+ o(1).
Then
limx!0
(1 + ax+ cx
2)b � (1 + bx+ dx
2)a
x
2
= bc� ad+ a
2
b
2 +1
2ab(a+ b).
15
Exercise 2.5.34 (3)
limx!a
a
x � x
a
x� a
= limx!a
a
x � a
a
x� a
� limx!a
x
a � a
a
x� a
= (ax)0x=a
�(xa)0x=a
= a
a log a�a·aa�1 = a
a(log a�1).
Exercise 2.5.34 (4)For x = ay, we have
limx!0
(a+ x)x � a
x
x
2
= limy!0
a
x
(1 + y)ay � 1
(ay)2= a
�2 limy!0
(1 + y)ay � 1
y
2
.
By(1 + y)ay = e
ay log(1+y) = e
ay(y+o(y)) = e
ay
2
+o(y
2
) = 1 + ay
2 + o(y2),
we get
limx!0
(a+ x)x � a
x
x
2
= a
�2 limy!0
1 + ay
2 + o(y2)� 1
y
2
= a
�2
.
Exercise 2.5.34 (6)We have
log(sin ax) = log(ax+ o(x)) = log x+ log(a+ o(1)).
Note that log(a+ o(1)) converges and is therefore bounded, while limx!0
+ log x = �1. There-fore lim
x!0
+
log(a+o(1))
log x
= 0, and
limx!0
+
log(sin ax)
log(sin bx)= lim
x!0
+
log x+ log(a+ o(1))
log x+ log(b+ o(1))= lim
x!0
+
1 + log(a+o(1))
log x
1 + log(b+o(1))
log x
= 1.
Exercise 2.5.34 (7)We have
log(cos ax) = log(1� 1
2(ax)2 + o(x3)) = �1
2(ax)2 + o(x3).
Then
limx!0
+
log(cos ax)
log(cos bx)= lim
x!0
+
�1
2
(ax)2 + o(x3)
�1
2
(bx)2 + o(x3)=
a
2
b
2
.
Exercise 2.5.34 (8)We have
cos(sin x)� cos x = cos
✓x� 1
6x
3 + o(x4)
◆� cos x
= 1� 1
2!
✓x� 1
6x
3
◆2
+1
4!x
4 � 1 +1
2!x
2 � 1
4!x
4 + o(x4)
=1
6x
4 + o(x4).
Therefore
limx!0
cos(sin x)� cos x
x
4
=1
6.
16
Exercise 2.5.34 (9)By Exercise 2.5.22, we have
arcsin 2x� 2 arcsin x =
✓2x� 1
6(2x)3
◆� 2
✓x� 1
6x
3
◆+ o(x4) = �x
3 + o(x4).
Therefore
limx!0
arcsin 2x� 2 arcsin x
x
3
= �1.
Exercise 2.5.34 (11)Let y = x� ⇡
6
. Then x ! ⇡
6
means y ! 0, and
1� 2 sin x = 1� 2 cos⇡
6sin y � 2 sin
⇡
6cos y
= 1�p3(y + o(y2))� (1 + o(y)) = �
p3y + o(y),
cos 3x = � sin 3y = �3y + o(y2).
Then
limx!⇡
6
1� 2 sin x
cos 3x= lim
y!0
�p3y + o(y)
�3y + o(y2)= � 1p
3.
Exercise 2.5.34 (13)We have lim
x!0
+
x log x = limx!0
+
x(log x)2 = 0. Therefore
x
x
x�1 = e
(e
x log x�1)logx = e
(x log x+o(x log x))logx = e
x(log x)
2
+o(x(log x)
2
)
,
andlimx!0
+
x
x
x�1 = e
lim
x!0
+
(x(log x)
2
+o(x(log x)
2
)) = e
0 = 1.
Exercise 2.5.34 (16)We have
log⇣e
�1(1 + x)1
x
⌘ 1
x
=1
x
✓1
x
log(1 + x)� 1
◆=
1
x
✓1
x
✓x� 1
2x
2 + o(x2)
◆� 1
◆= �1
2+ o(1).
Therefore
limx!0
⇣e
�1(1 + x)1
x
⌘ 1
x
= e
lim
x!0
(� 1
2
+o(1)) =1pe
.
Exercise 2.5.35
We have ✓ =1
log(1 + x)� 1
x
. If we use l’Hospital’s rule, then we get
limx!0
✓1
log(1 + x)� 1
x
◆= lim
x!0
x� log(1 + x)
x log(1 + x)= lim
x!0
1� 1
1+x
log(1 + x) + x
1+x
= limx!0
x
(1 + x) log(1 + x) + x
= limx!0
1
log(1 + x) + 1 + 1=
1
2.
17
The calculation using the approximation is given by Exercise 2.5.32 (3).In general, if f(x) has second order derivative at x
0
, then
f(x) = f(x0
) + f
0(x0
)(x� x
0
) +f
00(x0
)
2(x� x
0
)2 + o((x� x
0
)2),
f
0(x) = f
0(x0
) + f
00(x0
)(x� x
0
) + o(x� x
0
).
If f 00(x0
) 6= 0, Then the mean value theorem says the following two are equal
f(x)� f(x0
)
x� x
0
= f
0(x0
) +f
00(x0
)
2(x� x
0
) + o(x� x
0
),
f
0(x0
+ ✓(x� x
0
)) = f
0(x0
) + f
00(x0
)✓(x� x
0
) + o(✓(x� x
0
))
= f
0(x0
) + f
00(x0
)✓(x� x
0
) + o(x� x
0
).
If f 00(x0
) 6= 0, then this implies ✓ = 1
2
+ o(1). Then we conclude limx!x
0
✓ = 1
2
.
Exercise 2.5.37 (1)We have
x� (a+ b cos x) sin x = x�✓a+ b� 1
2!bx
2 + o(x3)
◆✓x� 1
3!x
3 + o(x4)
◆
= (1� a+ b)x+
✓1
3!(a+ b)� 1
2!b
◆x
3 + o(x4).
Therefore x� (a+ b cos x) sin x = o(x4) means
1� a+ b = 0,1
3!(a+ b)� 1
2!b = 0.
The solution is a = 2, b = 1.
Exercise 2.5.38 (2)
Taking derivative of sin y = x in x, we get (cos y)y0 = 1. Therefore y
0 =1
cos y.
Taking derivative of (cos y)y0 = 1 in x, we get �(sin y)y02 + (cos y)y00 = 1. Therefore
y
00 =sin y
cos yy
02 =sin y
cos3 y.
Exercise 2.5.38 (3)
Takingd
dx
of the equation, we get 2x
a
2
+ 2yy
0
b
2
= 0. Then y
0 = � b
2
x
a
2
y
.
Takingd
dx
of 2x
a
2
+ 2yy
0
b
2
= 0, we get 21
a
2
+ 2y
02 + yy
00
b
2
= 0. Then
y
00 = �1
y
✓b
2
a
2
� y
02◆
= �1
y
✓b
2
a
2
� b
4
x
2
a
4
y
2
◆=
b
2(b2x2 � a
2
y
2)
a
4
y
3
.
18
Exercise 2.5.39 (2)
dy
dx
=(a(1� cos t))0
(a(t� sin t))0=
sin t
1� cos t,
d
2
y
dx
2
=d
dx
✓dy
dx
◆=
d
dt
✓dy
dx
◆
dx
dt
=
d
dt
✓sin t
1� cos t
◆
d
dt
(a(t� sin t))
=cos t(1� cos t)� sin t sin t
(1� cos t)2a(1� cos t)= � 1
a(1� cos t)2.
19