Giáo trình giải tích 2 part 4 pdf

lim
(x,y)→(0,0)
xy(x + y)
x
2
+ y
2
=0




xy(x + y)
x
2
+ y
2




≤
1
2
|x
2
+ y
2
||x + y|
|x
2

+ y
2
|
≤|x + y|→0 (x, y) → (0, 0).
lim
(x,y)→(0,0)
sin xy
x
= lim
(x,y)→(0,0)
sin xy
xy
x =1.0=0.
lim
(x,y)→(0,0)
x − y
x + y
(x
k
,y
k
)=(
1
k
,
1
k
) (x

k

,y

k
)=(
1
k
, 0) (0, 0) f(x
k
,y
k
) → 0
f(x

k
,y

k
) → 1.
f(x, y)
(x
0
,y
0
) f
a
12
= lim
y→y
0
lim

x→x
0
f(x, y),a
21
= lim
x→x
0
lim
y→y
0
f(x, y),a= lim
(x,y)→(x
0
,y
0
)
f(x, y).
x
0
=0,y
0
=0
f(x, y)=(x + y)sin
1
x
sin
1
y
. a
12

,a
21
a =0
f(x, y)=
x
2
− y
2
x
2
+ y
. a
12
=0,a
21
=1 a
f(x, y)=
xy
x
2
+ y
2
. a
12
= a
21
=0 a
f(x, y)=x sin
1
y

. a
12
=0 a
21
a =0
a = a
12
= a
21
f : X ×Y → R
m
x
0
,y
0
X, Y
lim
y→y
0
f(x, y)=g(x), ∀x ∈ X.
lim
x→x
0
f(x, y)=h(y) y
∀>0, ∃δ>0:x ∈ X, d(x, x
0
) <⇒ d(f(x, y),h(y)) <, ∀y ∈ Y.
lim
(x,y)→(x
0

,y
0
)
f(x, y) = lim
x→x
0
lim
y→y
0
f(x, y) = lim
y→y
0
lim
x→x
0
f(x, y).
x
lim
x→∞
f(x)=L, lim
x→a
f(x)=∞, lim
x→∞
f(x)=∞.
R
n
a ∈ R
n
a = ∞ F
a

(R
n
, R
m
)
a R
n
R
m
a
f,ψ ∈ F
a
(R
n
, R
m
)
f = o(ψ) x → a ⇔ lim
x→a
f(x)
ψ(x)
=0.
f,g,ψ ∈ F
a
(R
n
, R
m
)
f = o(ψ) g = o(ψ) x → a f + g = o(ψ) x → a

f = o(ψ) x → a g <f,g>= o(ψ) x → a
f,ψ ∈ F
a
(R
n
, R
m
)
f = O(ψ) x → a ⇔∃C>0,r >0:f (x) ≤Cψ(x), ∀x ∈ B(a, r).
f,g,ψ ∈ F
a
(R
n
, R
m
)
f = O(ψ) g = O(ψ) x → a f + g = O(ψ) x → a
f = O(ψ) x → a g <f,g>= O(ψ) x → a
o(ψ),O(ψ)
f = o(ψ) g = o(ψ) f = g
f,g ∈ F
a
(R
n
, R)
f ∼ g x → a ⇔ lim
x→a
f(x)
g(x)
=1.

∼
n →∞
P (n)=a
p
n
p
+ a
p−1
n
p−1
+ ···+ a
0
∼ a
p
n
p
(a
p
=0)
1+2+···+ n =
n(n +1)
2
= O(n
2
)
1
2
+2
2
+ ···+ n

2
=
n(2n +1)(n +2)
6
= O(n
3
)
n! ∼

n
e

n
√
2πn = O


n
e

n+
1
2

2
n
,n
p
, ln
q

n, n
p
ln
q
n n → +∞
p ∈ N 1
p
+2
p
+ ···+ n
p
= O(n
p+1
) n →∞
f : X → R
m
X ⊂ R
n
a ∈ X
lim
x→a
f(x)=f(a).
(, δ)
f a
∀>0, ∃δ>0: B(a, δ) ⊂ f
−1
(B(f (a),)
f : R
n
→ R

m
f R
n
V ⊂ R
m
f
−1
(V )
F ⊂ R
m
f
−1
(F )
C(X, R
m
) f : X → R
m
X
f
a f a
m =1
f(x
1
, ··· ,x
n
)=x
i
(i =1, ··· ,n) e
x
ln x sin x arcsin x

f(x
1
, ··· ,x
n
)=

0≤i
1
···i
n
≤N
a
i
1
,···,i
n
x
i
1
1
···x
i
n
n
R
n
x → x
i
,x→ a
T : R

n
−→ R
m
T (αx + βy)=αT (x)+βT(y), ∀x, y ∈ R
n
,α,β ∈ R.
T m n (a
ij
)
m×n
T (e
j
)=
m

i=1
a
ij
e
i
,j=1, ··· ,m.
y = Tx



y
1
y
m




=



a
11
a
12
··· a
1n
a
m1
a
m2
··· a
mn






x
1
x
n




1
T
∃M>0: Tx≤Mx, ∀x ∈ R
n
.
T  =max
x=1
Tx , T
f : K −→ R
m
f K
f(K)
f : K → R K ⊂ R
n
f
K a,b ∈ K f(a)=sup
x∈K
f(x),f(b)= inf
x∈K
f(x)
(y
k
) f(K) x
k
∈ K, y
k
= f(x
k
) K

x
σ(k)
) x ∈ K f
(y
σ(k)
= f(x
σ(k)
)) f(x) ∈ f(K) f(K)
m =1 f(K)
M =supf(K) m =inff(K) f(K) a, b ∈ K
f(a)=M,f(b)=m 
f : K −→ R
m
f K f
K
∀>0, ∃δ>0:x, x

∈ K, d(x, x

) <δ =⇒ d(f(x),f(x

)) <.
f
∃>0, ∀k ∈ N, ∃x
k
,x

k
∈ K : d(x
k

,x

k
) <
1
k
, d(f(x
k
),f(x

k
)) ≥ .
K (x
σ(k)
) (x
k
) x ∈ K
d(x
σ(k)
,x

σ(k)
)) <
1
σ(k)
(x

σ(k)
) (x


k
) x
f d(f(x
σ(k)
),f(x

σ(k)
)) d(f(x),f(x)) = 0

f(x)=
1
x
,x∈ (0, +∞)
E
E
E R
N : E → R (N1)(N2)(N3)
R
n
x → max
1≤i≤n
|x
i
| x →

n
i=1
|x
i
|

x
N
1
,N
2
M,m
mN
1
(x) ≤ N
2
(x) ≤ MN
1
(x), ∀x ∈ E.
E f
1
, ··· ,f
n
E
T : E → R
n
,x
1
f
1
+ ···+ x
n
f
n
→ (x
1

, ··· ,x
n
)
N
E
= T
−1
◦ N E N R
n
E
N R
n

R
n
E
S
n−1
= {x ∈ R
n
: x =1} N
S
n−1
M =max
x∈S
n−1
N(x) m = min
x∈S
n−1
N(x)

M,m > 0 x ∈ R
n
\{0}
x
x
∈ S
n−1
(N2)
mx≤N(x) ≤ Mx 
f : C → R
m
f C f(C)
f : C → R f C f(C)
a, b ∈ C µ ∈ R f(a) <µ<f(b) c ∈ C : f (c)=µ
f C f(C)
f f
f(C)
A, B f(C) f U, V
f
−1
(A)=C ∩ U f
−1
(B)=C ∩V U, V C
C
R
1

f :[a, b] → [a, b] x
∗
∈ [a, b]

f(x
∗
)=x
∗
f :[a, b] → R f(b),f(a)
(x
k
) f(x)=0
f : S
n
−→ R,n ≥ 1,
x
0
∈ S
n
f(x
0
)=f(−x
0
)
S
n
= {x ∈ R
n+1
: x =1}
g(x)=f(x) −f(−x) g S
n
g(S
n
) R g(x)g(−x) ≤ 0 g(S

n
)
0 
M ⊂ R
n
f : M → M
d
∃θ, 0 <θ<1: d(f ( x) ,f(y)) ≤ θd(x, y), ∀x, y ∈ M.
f ∃!x
∗
∈ M : f(x
∗
)=x
∗
x
0
∈ M (x
k
) x
1
= f(x
0
),x
k+1
= f(x
k
)(k =2, 3, ···)
(x
k
) x

∗
f
(x
k
)
d(x
k+1
,x
k
)=d(f(x
k
),f(x
k−1
) ≤ θd(x
k
,x
k−1
) ≤···≤θ
k
d(x
1
,x).
m =1, 2, ···
d(x
k+m
,x
k
) ≤ d(x
k+m
,x

k+m− 1
)+···+ d(x
k+1
,x
k
) ≤ (θ
k+m
+ ···θ
k
)d(x
1
,x
0
)
≤
θ
k
1 − θ
d(x
1
,x) → 0, k →∞.
(x
k
) lim x
k
= x
∗
M x
∗
∈ M

f f f(x
∗
)=x
∗
¯x ∈ M f f(¯x)=¯x
d(¯x, x
∗
)=d(f(¯x),f(x
∗
)) ≤ θd(¯x, x
∗
).
θ<1 d(¯x, x
∗
)=0 ¯x = x
∗

f : R → R 0 <θ<1 |f

(x)| <θ,∀x
|f(x) − f(y)| = |f

(c)||x −y|≤θ|x − y|, ∀x, y ∈ R
f
f : M → M d(f(x),f(y)) <d(x, y), ∀x, y ∈ M,x = y
f(x)=x +
1
x
x ∈ M =[1, ∞)
T : R

n
→ R
n
(t
ij
)
T
n

i,j=1
t
2
ij
< 1hay
n

i,j=1
|t
ij
| < 1 n max
1≤i,j≤n
|t
ij
| < 1
(f
k
)
k∈N
f
k

: X → R
m
,X⊂ R
n
(f
k
) f x ∈ X lim f
k
(x)=f(x)
(f
k
) X f
∀>0, ∃N(): k ≥ N ⇒ d(f
k
(x),f(x)) ≤ , ∀x ∈ X,
M
k
=sup
x∈X
d(f
k
(x),f(x)) lim
k→∞
M
k
=0
X
∞

k=0

f
k
= f
0
+ f
1
+ ···+ f
k
+ ··· , f
k
: X → R
m
kS
k
= f
0
+ f
1
+ ···+ f
k
X (S
k
) X
R f
k
(x)=



1 −

1
k
|x| |x|≤k,
0 |x| >k.
(f
k
) f(x) ≡ 1
sup
x∈R
|f
k
(x) − f(x)| =1 → 0, k →∞.
∞

k=0
x
k
f(x)=
1
1 − x
[−1, 1) 0 ≤ r<1
[−r, r]
S
k
(x)=1+x + ···+ x
k
=
1 − x
k+1
1 − x

[−r, r]
sup
|x|≤r
|S
k
(x) − f(x)| =sup
|x|≤r





x
k+1
1 − x





=
r
k+1
1 − r
→ 0, k →∞.
f (−1, 1)
(f
k
) X
∀>0, ∃N : k, l > N ⇒ d(f

k
(x),f
l
(x)) ≤ , ∀x ∈ X.

R
[a, b]
1+
1
1!
x + ···+
1
k!
x
k
+ ··· e
x
x −
1
3!
x
3
+ ···+
1
(2k +1)!
x
2k+1
+ ··· sin x
1+
1

2!
x + ···+
1
2k!
x
2k
+ ··· cos x
f
k
(x)=x
k
,x ∈ [0, 1] (f
k
)
(f
k
) f
k
k (f
k
)
f f
x
0
>0 N k>N
d(f(x),f
k
(x)) <

3

d(f(x
0
),f
k
(x
0
)) <

3
k f
k
x
0
δ>0
d(f
k
(x),f
k
(x
0
)) <

3
d(x, x
0
) <δ.
d(f(x),f(x
0
)) ≤ d(f(x),f
k

(x)) + d(f
k
(x),f
k
(x
0
)) + d(f
k
(x
0
),f(x
0
)) <,
f x
0

R
n
X ⊂ R
n
BF(X,R
n
) f : X → R
n
f ∈ BF(X, R
m
) ⇔∃M>0:f(x)≤M,∀x ∈ X.
f =sup
x∈X
f(x)

BF(X,R
m
)
R
n
d(f,g)=f − g,f,g∈ BF(X, R
m
)
f,f
k
∈ BF(X, R
m
) (f
k
)
k∈N
f
f
k
→ f BF(X,R
m
) d(f
k
,f) → 0 k →∞
X C(X, R
m
)
g :[a, b] → R
a = a
0

<a
1
< ···<a
k
= b
g
g(x)=A
i
x + B
i
,x∈ [a
i−1
,a
i
],i=1, ···,k
g A
i
,B
i
g
f :[a, b] → R
f
[a, b]
f
f
f
g : K → R
K X
1
, ··· ,X

k
g
f :[a, b] → R
f
[a, b]
f max, min
f
f
R
n
f [a, b]
f
f
x = a + t(b − a) [a, b]=[0, 1]
f
B
k
(x)=B
k
f(x)=
k

p=0
C
p
k
f(
p
k
)x

p
(1 − x)
k−p
.
(x + y)
k
=
k

p=0
C
p
k
x
p
y
k−p
x x kx(x + y)
k−1
=
k

p=0
pC
p
k
x
p
y
k−p

x
2
k(k −1)x
2
(x + y)
k−2
=
k

p=0
p(p − 1)C
p
k
x
p
y
k−p
y =1−x r
p
(x)=C
p
k
x
p
(1 −x)
k−p
k

p=0
r

p
(x)=1,
k

p=0
pr
p
(x)=kx,
k

p=0
p(p − 1)r
p
(x)=k(k −1)x
2
.
k

p=0
(p − kx)
2
r
p
(x)=k
2
x
2

p=0
r

p
(x) − 2kx
k

p=0
pr
p
(x)+
k

p=0
p
2
r
p
(x)=kx
= k
2
x
2
− 2kx +(kx + k(k −1)x
2
)=kx(1 − x)
M =max
|x|≤1
|f(x)| >0 δ>0
|x − y| <δ |f(x) − f(y)| <
f(x) − B
k
(x)=f(x) −

k

p=0
C
p
k
f(
p
k
)x
p
(1 −x)
k−p
=
k

p=0
(f(x) − f(
p
k
))r
p
(x).

1
p : |
p
k
− x| <δ |f(x) −f(
p

k
)| < r
p
(x) ≥ 0
|

1
|≤
k

p=0
r
p
(x)=.