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Giáo trình giải tích 1 part 2 ppt

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lim
n→∞

n(

n +2 −

n + 1) = lim
n→∞

n
(

n +2−

n +1)(

n +2+

n +1)

n +2+

n +1
= lim
n→∞

n
(n +2)− (n +1)

n +2+



n +1
= lim
n→∞

n

n(

1+
2
n
+

1+
1
n
)
= lim
n→∞
1

1+
2
n
+

1+
1
n

=
1
(lim

1+
2
n
+ lim

1+
1
n
)
=
1

1+

1
=
1
2
Q Q x
n
=(1+
1
n
)
n
R

(x
n
≤ x
n+1
, ∀n)&(∃M,x
n
<M,∀n) ⇒∃lim x
n
(x
n
≥ x
n+1
, ∀n)&(∃m, m < x
n
, ∀n) ⇒∃lim x
n
(x
n
)
(−x
n
) (x
n
)
a =sup{x
n
: n ∈ N}
lim x
n
= a >0

x
n
≤ a x
N
a −<x
N
n>N a − <x
n
≤ a<a+  |x
n
− a| <
lim x
n
= a 
(x
n
) lim x
n
=+∞
(x
n
) lim x
n
= −∞
I
n
=[a
n
,b
n

]
I
n
⊃ I
n+1
n ∈ N I
n

n∈N
I
n
= ∅
a
n
≤ a
n+1
≤ b
n+1
≤ b
n
(a
n
)
(b
n
)
a = lim a
n
lim b
n

= b a ≤ b
[a, b] ⊂ I
n
, ∀n 
a
0
≤ x
n
≤ b
0
, ∀n I
0
=[a
0
,b
0
]
x
n
I
1
n
1
x
n
1
∈ I
1
I
1

x
n
I
2
n
2
>n
1
x
n
2
∈ I
2
I
0
⊃ I
1
⊃···⊃I
k
I
k
b
0
−a
0
2
k
n
1
<n

2
< ···<n
k
x
n
k
∈ I
k
a ∈ I
k
, ∀k |x
n
k
−a|≤
b
0
−a
0
2
k
→ 0
k →∞ (x
n
k
)
k∈N
a 
(x
n
) (x

n
)
∀>0, ∃N : n, m > N ⇒|x
n
− x
m
| <
⇐ lim x
n
= a >0 N |x
n
− a| </2
∀n>N m, n > N |x
n
− x
m
|≤|x
n
− a| + |x
m
− a| </2+/2=
⇒ (x
n
)
(x
n
)  =1 N x
N
−1 <x
n

<x
N
+1, ∀n>N
M =max{|x
0
|, ··· , |x
N
|, |x
N
| +1} |x
n
|≤M,∀n
(x
n
k
)
k∈N
a
(x
n
) a |x
k
−a|≤|x
k
−x
n
k
|+|x
n
k

−a|
n
k
≥ k k →∞ n
k
→∞ |x
k
− x
n
k
|→0
|x
n
k
− a|→0 a lim
k→∞
x
k
= a 
|x
n
− x
n+p
|→0 , n →∞, p =0, 1, ···
lim
n→∞
1
n
p
=0 (p>0)

lim
n→∞
n

a =1 (a>0)
lim
n→∞
n

n =1
lim
n→∞
n

n!=+∞
lim
n→∞
n
p
a
n
=0 (a>1)
lim
n→∞
a
n
=0 |a| < 1 lim
n→∞
a
n

=+∞ a>1
a ≥ 1 x
n
=
n

a − 1 lim x
n
=0
x
n
≥ 0 a =(1+x
n
)
n
≥ 1+nx
n
0 ≤ x
n

a − 1
n
lim x
n
=0
0 <a<1
1
a
x
n

=
n

n − 1
n =(1+x
n
)
n

n(n − 1)
2
x
2
n
0 ≤ x
n


2

n − 1
lim x
n
=0 lim
n

n =1
n! >

n

3

n
n

n! >
n
3
a>1 a
1
p
=1+u (u>0)
(a
1
p
)
n
=(1+u)
n
>
n(n − 1)
2
u
2
lim
n
p
a
n
= lim



n
(a
1
p
)
n


p
=0
p =0 
s
n
=1+
1
1!
+
1
2!
+ ···+
1
n!
t
n
=

1+
1

n

n
lim
n→∞
s
n
= lim
n→∞
t
n
= e
(s
n
) s
n
=1+1+
1
1.2
+
1
1.2.3
+ ··· +
1
1.2 n
< 1+
1+
1
2
+

1
2
2
+ ···+
1
2
n−1
< 3 lim s
n
= e
t
n
=

1+
1
n

n
=
n

k=0
n!
k!(n − k)!
1
n
k
=
n


k=0
1
k!
n
n
n − 1
n

n − k +1
n
=
n

k=0
1
k!

1 −
1
n



1 −
k −1
n

t
n

<t
n+1
t
n
≤ s
n
< 3 lim t
n
= e

e = e

t
n
≤ s
n
e

≤ e
n ≥ m
t
n
=1+1+
1
2!

1 −
1
n


+ ···+
1
n!

1 −
1
n



1 −
n − 1
n

≥ 1+1+
1
2!

1 −
1
n

+ ···+
1
m!

1 −
1
n




1 −
m − 1
n

m n →∞ e

≥ 1+1+
1
2!
+ ···+
1
m!
= s
m
m →∞ e

≥ e 
e e =2, 71828 ···
e =
m
n
∈ Q
0 <e−s
n
=
1
(n +1)!
+ ···<

1
n!n
0 <n!(e −s
n
) <
1
n
n!e, n!s
n

x
n
= a
0
+ a
1
x + ···+ a
n
x
n
|x| < 1 |a
k
| <M,∀k
x
n
=1+
1
2
+ ···+
1

n
|x
n+p
− x
n
| = |a
n+1
x
n+1
+ ···+ a
n+p
x
n+p
|≤|a
n+1
|x|
n+1
| + ···+ |a
n+p
||x
n+p
|
≤ M|x|
n+1
+ ···+ M|x|
n+p
≤ M|x|
n+1
(1 + ···+ |x|
p

)
≤ M|x|
n+1
1
1 −|x|
n →∞ |x
n+p
− x
n
|→0 p (x
n
)
n, m =2n |x
m
− x
n
| =
1
n +1
+ ···+
1
2n
>
1
2
(x
n
)

x ∈ R

a
0
=[x] ∈ Z,a
n
=[10
n
(x − a
0

a
1
10
−···−
a
n−1
10
n−1
)] ∈{0, 1, ···, 9},
x
n
= a
0
+
a
1
10
+ ···+
a
n
10

n
→ x, n →∞
x = a
0
,a
1
a
2
···a
n
···
R
a
0
=[x] a
0
≤ x<a
0
+1 0 ≤ x −a
0
< 1
a
1
= [10(x − a
0
)] ∈{0, 1, ··· , 9}
a
1
10
≤ x − a

0
<
a
1
+1
10
[0, 1] x −a
0
0 ≤ x − a
0

a
1
10
<
1
10
a
2
∈{0, 1, ··· , 9}
a
2
10
2
≤ x − a
0

a
1
10

<
a
2
+1
10
2
n 0 ≤ x − a
0

a
1
10
−···−
a
n
10
n
<
1
10
n
a
n+1
=[10
n+1
(x − a
0

a
1

10
−···−
a
n
10
n
)] a
n+1
∈{0, 1, ··· , 9}
0 ≤ x − a
0

a
1
10
−···−
a
n
10
n

a
n+1
10
n+1
<
1
10
n+1
x

n
0 ≤ x − x
n
<
1
10
n
lim x
n
= x 

1, 000 ···=0, 999 ··· 0, 5=0, 4999 ···

1
2
=0, 5 ,
1
3
=0, 333 ··· , 0, 123123123 ···= 123 ×
1
10
3
− 1
R
X, Y
X Y
n {1, 2, ··· ,n}
N
N → X X
2N, Z, Q

R
a, b ∈ R a = b [a, b]
[a, b]={x
n
: n ∈ N} [a, b]
I
1
x
1
∈ I
1
I
1
I
2
x
2
∈ I
2
I
1
⊃ I
2
⊃···⊃I
n
⊃··· x
n
∈ I
n
x ∈∩

n∈N
I
n
x ∈ [a, b]
x = x
n
, ∀n x ∈ [a, b] 
• N
X ⊂ N N X
0 → x
0
= min X, n → min(X \{x
0
, ··· ,x
n−1
})
• X f : X → Y Y
m : Y → X, m(y) = min f
−1
(y) m
Y → m(Y )
• N
2
f : N
2
→ N,f(m, n)=
(m + n)(m + n +1)
2
+ n )


N

N
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r❜❜❜❜❜❜
❜❜❜❜❜❜
❜❜❜❜❜❜
❜❜❜❜❜❜
❜❜❜❜❜❜
❜❜❜❜❜❜











❅■








❅■






❅■






❅■



❅■
• (X
n
)
n∈I
X = ∪
n∈I
X
n
N → I n → i
n
n N → X
n
,m→ f
n
(m)
N
2
→ X, (m, n) → f
i
n
(m)
• 0 1
X
N → X, n → x

n
x
0
= x
0,0
x
0,1
x
0,2
··· ···
x
1
= x
1,0
x
1,1
x
1,2
··· ···
x
2
= x
2,0
x
2,1
x
2,2
··· ···
x
n

= x
n,0
x
n,1
x
n,2
··· x
n,n
···
y =(y
n
) y
n
=1
x
n,n
=0 y
n
=0 x
n,n
=1 y X 0, 1
X y = x
n
, ∀n
n!
n!=

n
e


n

2πne
θ
n
12n
, 0 <θ
n
< 1

f : X → Y, x → y = f(x)
X, Y R
x ∈ X
y = f(x) ∈ Y
X
f
f(X)={y ∈ R : ∃x ∈ X, y = f(x)}
f
y x
y =2πx, y = mx, y = mx
2
{x ∈ R : f (x) }
f(x)=

x − 1
x − 2
{x ∈ R : x − 1 ≥ 0,x− 2 =0} =[1, 2) ∪(2, +∞)
f(x)=[x]=n n ≤ x<n+1
f(x)= x =






−1 x<0
0 x =0
+1 x>0
D χ
D
(x)=

1 x ∈ D
0 x ∈ D
[1, 5], [−π], [e], [sinx], (−2), (2
64
), (−[0, 3])
f = {(x, y):x ∈ X, y = f(x)} R × R = R
2
R
2
(0, 0) O R × 0
Ox 0 ×R Oy (x, y) ∈ R
2
Ox (x, 0)
Oy (0,y)
f
f
s
O


x

y
s
(x, f (x))
(x
0
,f(x
0
)), (x
1
,f(x
1
)), ··· , (x
n
,f(x
n
))
f
[x] (x)
x
x
0
x
1
··· x
n
y y
0
y

1
··· y
n
f,g : X → R
f ± g, fg,
f
g
g(x) =0, ∀x ∈ X
(f ±g)(x)=f(x) ±g(x) ,fg(x)=f(x)g(x),
f
g
(x)=
f(x)
g(x)
,x∈ X
f : X → Y g : Y → Z g ◦f : X → Z
g ◦f(x)=g(f(x))
f : X → Y f
−1
: Y → X
f
−1
(y)=x ⇔ y = f(x)
f(x)=x − [x]
f(x)=[x] g(x)= (x) f ◦ g g ◦ f
f X
∀x
1
,x
2

∈ X, x
1
<x
2
⇒ f (x
1
) ≤ f(x
2
)( f(x
1
) <f(x
2
))
f
X
∀x
1
,x
2
∈ X, x
1
<x
2
⇒ f (x
1
) ≥ f(x
2
)( f(x
1
) >f(x

2
))
f(x)=x
n
n ∈ N [0, +∞)
f(x)=[x] g(x)= (x) R
n f(x)=x
n
R
X x ∈ X −x ∈ X
f
X f(−x)=f(x), ∀x ∈ X
f
X f(−x)=−f(x), ∀x ∈ X
x
2
, cos x x
3
, sin x R
f
f(x)=
1
2
(f(x)+f(−x)) +
1
2
(f(x) −f(−x))
Oy
O (x, y = f(x))


x

y
0
s
(x, y)
s
(y, x)
s
(−x, y)
s
(−x, −y)
s
(x + T,y)












y = x
f X
T>0
f(x + T )=f(x), ∀x ∈ X

T
f
x ∈ X x + T ∈ X x + nT ∈ X n ∈ N
f(x + nT )=f(x)
T
k ∈ Z \{0} sin kx cos kx

k
f(x)=x − [x] 1
Q χ
Q

x
α
e
x
ln x sin x arctan x
x
exp(x)=e
x
= lim
n→+∞

1+
x
n

n
R (0, +∞)
e

0
=1,e
x+x

= e
x
e
x






1+
t
n

n
− 1




≤|t|(e − 1) |t|≤1(∗)

1+
t
n


n
=1+t
n

k=1
C
k
n
t
k−1
n
k
|t|≤1





1+
t
n

n
− 1




≤|t|
n


k=1
C
k
n
|t|
k−1
n
k
≤|t|
n

k=1
C
k
n
|t|
k−1
n
k
≤|t|
n

k=1
C
k
n
1
n
k

= |t|

1+
1
n

n
− 1

≤|t|(e −1)
x ∈ R x
n
=

1+
x
n

n
(x
n
)
x>0 e (x
n
)
N ∈ N x ≤ N
x
n



1+
N
n

n


1+
1
n

N.n


1+
1
n

n.N
≤ 3
N
exp(x) = lim
n→+∞
x
n
x ≥ 0
x<0 −x>0

1+
x

n

n
=
(1 −
x
2
n
2
)
n
(1 −
x
n
)
n
(∗) t =
x
2
n
lim
n→∞
(1 −
x
2
n
2
)
n
=1

lim
n→+∞

1+
x
n

n
=
1
exp(−x)
exp(x) x ∈ R
e
0
=1

1+
x
n

n

1+
x

n

n

1+

x+x

n

n
=

1+
xx

n
2
(1 +
x+x

n
)

n
n →∞ (∗) t =
xx

n(1 +
x+x

n
)
e
x
e

x

e
x+x

=1
e
x
> 0 e
t
> 1 t>0 x<x

e
x
− e
x

= e
x
(1 − e
x

−x
) < 0
e
x

ln x e
x
(0, +∞) R

ln e =1, ln x +lnx

=lnxx

x
α
α ∈ R
n ∈ N x
n
= x ···x n
R n n (−∞, 0)
[0, +∞)
n ∈ N x
−n
=
1
x
n
R \ 0 n n
(−∞, 0) (0, +∞)


y = x
2n


y = x
2n+1



y =
1
x
2n+1


y =
1
x
2n

×