Giáo trình giải tích 1 part 1 pdf

TRƯỜNG ĐẠI HỌC ĐÀ LẠT
KHOA TOÁN - TIN HỌC
Y  Z



TẠ LÊ LI






GIẢI TÍCH 1

(Giáo Trình)

Lưu hành nội bộ
Y Đà Lạt 2008 Z
R
q
q
2
=2
√
2
R
R
• R
+:R × R → R, (x, y) → x + y
· : R × R → R, (x, y) → xy
∀x, y x + y = y + x
∀x, y, z (x + y)+z = x +(y + z)
∃0, ∀x, x +0 = x (0 )
∀x, ∃−xx+(−x)=0 (−x x)
∀x, y xy = yx
∀x, y, z (xy)z = x(yz)
∃1 =0, ∀x 1x = x (1 )
∀x =0, ∃x
−1
xx
−1
=1 (x
−1

x)
∀x, y, z x(y + z)=xy + xz
• R ≤
∀x, y x ≤ y y ≤ x
∀xx≤ x
∀x, y x ≤ y, y ≤ x ⇒ x = y
∀x, y, z x ≤ y, y ≤ z ⇒ x ≤ z
∀x, y, z x ≤ y ⇒ x + z ≤ y + z
∀x, y 0 ≤ x, 0 ≤ y ⇒ 0 ≤ xy
• R
R
R
R

R R

n

i=1
x
i
= x
1
+ ···+ x
n
n

i=1
x
i

= x
1
···x
n
x − y = x +(−y)
x
y
= xy
−1
x ≤ y y ≥ x x y y x
x<y y>x x ≤ y x = y x y y x
0 <x x x<0 x
(a, b)={x ∈ R : a<x<b}
[a, b]={x ∈ R : a ≤ x ≤ b}
[a, b), (a, b]
R
O 0 1 =0 1
0 1 M
OM M 1 0
✲
0
t
1 M>0M

< 0
A ⊂ R
b ∈ R x ≤ b, ∀x ∈ A
b
A
A ⊂ R

a ∈ R a ≤ x, ∀x ∈ A
a
A
b
∗
A b
∗
=supA b
∗
A
a
∗
A a
∗
=infA a
∗
A
A = {
1
2
,
3
4
, ··· ,
2
n
−1
2
n
, ···} sup A =1, inf A =

1
2
A = {q : q q
2
< 2}
a
∗
=infA b
∗
=supA R A
a
∗
b
∗
q q
2
=2
R
sup A ∈ A inf A ∈ A A
M
A M =maxA M =supA M ∈ A
m
A m = min A m =infA m ∈ A
A ⊂ R a =supA
a A ∀>0, ∃x

∈ A : a − <x

N, Z, Q
N = {n : n =0 n =

n
  
1+···+1}
Z = {p : p ∈ N − p ∈ N }
Q = {
p
q
: p ∈ Z,q∈ N,q=0}/ ∼,
p

q

∼
p
q
⇔ pq

− qp

=0
x ∈ R
x
|x| =

x x ≥ 0
−x x<0
x, y
|x|≥0, |xy| = |x||y|, |x + y|≤|x|+ |y|
x ∈ R n ∈ N x<n
n ≤ x, ∀n ∈ N

a =supN
sup n
0
∈ N a−1 <n
0
a<n
0
+1 ∈ N 
x, y > 0 n ∈ N x<ny
x>0 n ∈ N 0 <
1
n
<x
x>0 n ∈ N n ≤ x<n+1
x ∈ R
[x]= n n ≤ x<n+1
[0, 5], [−2, 5], [0, 0001]
R
x, y ∈ R x<y r ∈ Q x<r<y
x ∈ R >0 r ∈ Q |x − r| <
n ∈ N 0 <
1
n
<y−x
m ∈ N m ≤ nx < m +1
m
n
≤ x<
m +1
n

r =
m +1
n
∈ Q x<r=
m +1
n
=
m
n
+
1
n
<x+(y − x)=y 
R
n x>0 n ∈ N \{0}
y>0 y
n
= x
y
n x y =
n
√
x
A = {t ∈ R : t
n
≤ x} A = ∅ t =0
1+x y =supA
y
n
= x

y
n
<x 0 <h<1
(y + h)
n
≤ y
n
+ h(
n

k=1
C
k
n
y
n−k
)=y
n
+ h((y +1)
n
− y
n
)
0 <h<
x − y
n
(y +1)
n
− y
n

h<1 (y + h)
n
<x y + h ∈ A
y + h>y=supA
y
n
>x k>0 (y −k)
n
>x y −k
A y =supA 
R
√
2,
√
3,
3
√
5,
4
√
16
R
∞
R = R ∪{+∞, −∞}
x ∈ R −∞ <x<+∞
x +(+∞)=+∞,x+(−∞)=−∞
x(+∞)=+∞ x>0,x(+∞)=−∞ x<0
x
+∞
=

x
−∞
=0
∞−∞, 0 ∞,
∞
∞
A inf A = −∞ sup A =+∞
1 3
00, 30, 33 0, 333 0, 3333 ···
1
1
2
1
2
2
1
2
3
1
2
4
···
122
2
2
3
2
4
···
010101 ···

•
•
1
3
0
X ⊂ R X
(x
n
)
n∈N
= x
0
,x
1
,x
2
,x
3
, ···
X x : N → X, n → x
n
= x(n)
R
2
{ (n, x
n
): n ∈ N }
0
s
1

s
2
s
3
s
q
s
q
s
n
sx
q
s
q
s
q
s
s
s
s
s
✲
+∞
✻
x
N = {0, 1, 2, ···} n ∈ N n +1∈ N
0 < 1 < 2 < 3 < ···
• Σ={0, 1, ··· ,N}
(x
0

,x
1
,x
2
, ···) x
n
∈ Σ
• x
n
=3.10
−1
+3.10
−2
+ ···+3.10
−n
,
x
n
=
1
2
n
,x
n
=2
n
x
n
=1−(−1)
n

• x
n
= n! x
0
=1,x
n+1
=(n +1)x
n
(n ≥ 1)
x
0
∈ R x
n+1
= f(x
n
) n =0, 1, ··· f
x
0
=0,x
1
=1,x
n+1
= x
n
+ x
n−1
(n ≥ 2)
f(x)=
√
1+x f(x)=4λx(1 − x) λ ∈{0.7, 0.8, 0.9}

x
n+1
= f(x
n
) x
0
=1
x
n
=
n
{x
n
: n ∈ N} (x
n
)
n∈N
a ∈ R
(x
n
)
n∈N
>0
N

 n>N

|x
n
− a| <

∀>0, ∃N : n>N ⇒|x
n
− a| <
(x
n
) a
lim
n→∞
x
n
= a lim x
n
= a x
n
→ a, n →∞
0
s
1
s
2
s
3
s
q
s
q
s
N
s
n

s
q
s
q
s
q
s
q
s
q
s
q
s
a + 
a − 
a
✲
+∞
✻
x
•
• lim
n→∞
x
n
= a lim
n→∞
|x
n
− a| =0

•
{(x, y): y = a } R
2
• (x
n
) a b
(x
n
) |a −b|≤|a −x
n
|+|x
n
−b|→0 n →∞ |a −b| =0 a = b
x
n
=
1
√
n
n =1, 2, ···
lim
n→∞
x
n
=0

1
1
10
1

100
1
1.000
1
1.000.000
N

1 100
 N

0 <
1
<
2
⇒ N

1
≥ N

2
lim
n→∞
x
n
= a |x
n
− a|
|x
n
− a|≤f(N) n>N N

 f(N) <
>0 N n>N |x
n
−a|≤f(N) <
lim
n→∞
1
n
p
=0 p>0
n>N |
1
n
p
− 0| =
1
n
p
<
1
N
p
>0 N>
p

1

N =[
p


1

]+1
n>N |
1
n
p
− 0| <
1
N
p
<
(x
n
)=0 0, 30, 33 0, 333 0, 3333 ···→
1
3
>0 N =[3/] n>N
|x
n
−
1
3
| = |0, 33 ···3
  
n
−
1
3
| <

3
10
n
<
3
10
N
<
3
N
<
•
(2
n
)
lim
n→∞
x
n
=+∞ ∀E>0, ∃N : n>N⇒ x
n
>E
lim
n→∞
x
n
= −∞ ∀E>0, ∃N : n>N⇒ x
n
< −E
•

0 1
lim
n→+∞
a
n
=









0 |a| < 1
1 a =1
+∞ a>1
a ≤−1
(x
n
)
n
0
<n
1
< ···<n
k
< ··· (x
n

k
)
k∈N
(x
n
)
(x
n
)
N −→ N −→ R
k → n(k)=n
k
→ x
n
k
= x
n(k)
a ∈ R
a ((−1)
n
)
(1) (−1)
1 −1
• (x
n
) a a
• a (x
n
) >0
n ∈ N |x

n
− a| <
lim sup
n→∞
x
n
= lim
n→∞
x
n
=sup{a : a (x
n
)}
lim inf
n→∞
x
n
= lim
n→∞
x
n
=inf{a : a (x
n
)}
x
n
=(−1)
n
lim sup x
n

=1 lim inf x
n
= −1
x
n
=(−1)
n
n lim sup x
n
=+∞ lim inf x
n
= −∞
x
n
=sin
nπ
2
lim sup x
n
= lim inf x
n
=
x
0
∈ R n ≥ 1
x
n
=

3x

n−1
+1 x
n−1
1
2
x
n−1
x
n−1
x
0
=17 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ···
1 1, 4, 2, 1, 4, 2, 1, ···
x
0
n x
n
=1
• lim sup x
n
lim inf x
n
∞
• lim inf x
n
≤ lim sup x
n
• (x
n
) lim inf x

n
= lim sup x
n
• lim sup x
n
= M >0 x
n
>M−
x
n
>M+ 
• lim inf x
n
= m >0 x
n
<m+ 
x
n
<m−
(x
n
) M |x
n
| <M,∀n
(x
n
) (y
n
)
(x

n
+ y
n
) (x
n
y
n
)

x
n
y
n

lim
n→∞
y
n
=0
lim
n→∞
(x
n
+y
n
) = lim
n→∞
x
n
+ lim

n→∞
y
n
, lim
n→∞
(x
n
y
n
) = lim
n→∞
x
n
lim
n→∞
y
n
, lim
n→∞
x
n
y
n
=
lim
n→∞
x
n
lim
n→∞

y
n
(x
n
) (y
n
) n
x
n
≤ y
n
lim
n→∞
x
n
≤ lim
n→∞
y
n
n x
n
≤ y
n
≤ z
n
lim
n→∞
x
n
=

lim
n→∞
z
n
= a lim
n→∞
y
n
= a
lim
n→∞
x
n
= a lim
n→∞
y
n
= b
 =1 N |x
n
− a| < 1, ∀n>N
M =max{|x
0
|, ··· , |x
N
|, |a| +1} |x
n
| <M,∀n
|(x
n

+ y
n
) − (a + b)|≤|x
n
− a| + |y
n
− b|
|x
n
y
n
− ab|≤|x
n
y
n
− x
n
b + x
n
b − ab|≤M|y
n
− b| + |b||x
n
− a|
b =0  = |b| /2 N |y
n
− b| < |b|/2 ∀n>N
n>N |y
n
| = |b − b + y

n
|≥|b|−|y
n
− b| > |b|/2




x
n
y
n
−
a
b




=




x
n
b − y
n
a
by

n




=




x
n
b − ab
by
n
+
ab − y
n
a
by
n




≤
|x
n
− a|
|y

n
|
+
|a||b −y
n
|
|by
n
|
≤
|x
n
− a|
|b|/2
+
|a||b −y
n
|
|b||b|/2
n → +∞ → 0
n x
n
≤ y
n
a>b
 =
a−b
2
> 0 n |x
n

− a| < |y
n
− b| <
y
n
<b+  =
a+b
2
= a − <x
n
>0 lim x
n
= lim z
n
= a N
1
|x
n
− a| <,|z
n
− a| < n>N
1
N
2
x
n
≤ y
n
≤ z
n

, ∀n ≥ N
2
n ≥ max(N
1
,N
2
)
−<x
n
− a ≤ y
n
− a ≤ z
n
− a< |y
n
− a| <
lim y
n
= a 
• ((−1)
n
)
• (x
n
), (y
n
) x
n
<y
n

, ∀n lim
n→∞
x
n
≤ lim
n→∞
y
n
lim
n→∞
x
n
= a lim
n→∞
|x
n
| = |a| lim
n→∞
p

|x
n
| =
p

|a|
lim
n→∞
n
2

− 3n +6
3n
2
+4n +2
lim
n→∞
√
n(
√
n +2−
√
n +1)
n
2
n
n
2
lim
n→∞
n
2
− 3n +6
3n
2
+4n +2
= lim
n→∞
n
2
(1 − 3/n +6/n

2
)
n
2
(3 + 4/n +2/n
2
)
= lim
n→∞
1 − 3/n +6/n
2
3+4/n +2/n
2
=
1 − lim 3/n + lim 6/n
2
3 + lim 4/n + lim 2/n
2
=
1 − 0+0
3+0+0
=
1
3