Giáo trình giải tích 1 part 9 pps

√
3,
√
2+
√
6,
3
√
5 −
4
√
3,
ax + b
cx + d
(a, b, c, d ∈
Q,ad− bc =0,x∈ Q)
sup A, inf A, max A, min A
A = {
1
n +1
: n ∈ N} A = {
1
2
n
+
(−1)
n
n +1
: n ∈ N}
A = {

1+(−1)
n
n +1
− n
2
: n ∈ N}
A, B ⊂ R A B ⊂ A sup A, sup B,inf A, inf B
A, B ⊂ R
sup(A ∪B) = max(sup A, sup B), inf(A ∪B) = min(inf A, inf B)
A ∩B
D = {
m
2
n
: m ∈ Z,n∈ N} R
D R F D D \ F
R
 ∈{
1
10
,
1
100
, ··· ,
1
10
n
} N






n
n +1
− 1




<, ∀n ≥ N

 N

lim
n→∞
n
n +1
=1
N
1
√
n +1
< 0, 03, ∀n ≥ N lim
n→∞
1
√
n +1
=0
a

n
=
1
2
n
a
n
=sin
nπ
2
a
n
=10
n
a
n
= n sin
π
n
a
n
=(−1)
n
tg(
π
2
−
1
n
) a

n
= −n
2
>0 N |a
n
| <
n ≥ N
a
n
=
(−1)
n
n
a
n
=sin
π
n
a
n
= q
n
|q| < 1
E>0 N |a
n
| >E
n ≥ N
a
n
=(−1)

n
n a
n
=lnlnn a
n
= q
n
|q| > 1
lim
n→∞
a
n
n!
= lim
n→∞
n
p
a
n
= a>1 lim
n→∞
n
√
n =
lim
n→∞
n
√
n!= lim
n→∞


1+
1
n

n
=
lim
n→+∞
n +(−1)
n
n −(−1)
n
lim
n→+∞
5n
2
+ n −7
7n
2
− 2n +6
lim
n→+∞
n
√
n
2
+ n +1
lim
n→+∞

5 −2
n
5+2
n+1
lim
n→+∞
1
n
cos
nπ
2
lim
n→+∞
1+2+···+ n
√
9n
4
+1
lim
n→+∞
(

n
2
+5−

n
2
+3) lim
n→+∞

√
n(
√
n +1−
√
n +2)
lim
n→+∞

1
1.2
+
1
2.3
+ ···+
1
n(n +1)

lim
n→+∞
(1 −
1
2
2
)(1 −
1
3
2
) ···(1 −
1

n
2
)
lim
n→+∞
1+a + ···+ a
n
1+b + ···+ b
n
(|a|, |b| < 1) lim
n→+∞
n
√
3+sinn
α ∈ R
α
π
∈ Z lim
n→+∞
sin nα lim
n→+∞
cos nα
lim
n→∞
a
n
= L =0 ((−1)
n
a
n

)
a
n
≤ M lim
n→∞
a
n
= L L ≤ M
lim
n→∞
a
n
= L lim
n→∞
|a
n
| = |L|
(|a
n
|) (a
n
) 0
a
n
=
1.3 (2n −1)
2.4.6 2n
a
n
<

1
√
2n +1
lim
n→∞
a
n
=0

1+
1
n

n
<e<

1+
1
n−1

n
lim
n→+∞
n(e
1
n
− 1)
(a
n
) (b

n
) 0 (a
n
b
n
) 0
lim
n→+∞
1
n
sin
nπ
2
= lim
n→+∞
1
n
lim
n→∞
sin
nπ
2
=0. lim
n→∞
sin
nπ
2
=0
a
0

=1,a
n
=
√
1+a
n−1
(a
n
)
ϕ = lim
n→∞
a
n
t
n
=

1+
1
n

n+1
1
n +1
< ln

1+
1
n


<
1
n
a
n
=1+
1
2
+ ···+
1
n
−ln n (a
n
)
γ = lim
n→∞
a
n
=0, 5772156649 ···
(a
n
) a
n+1
− a
n
≤
1
n
(a
n

)
a
1
>a
2
> 0 a
n+1
=
a
n
+ a
n−1
2
, (n ≥ 2) a
1
,a
3
,a
5
,
a
2
,a
4
,a
6
, lim
n→∞
a
n

= L L
s
n
= a
0
+ a
1
x + ···+ a
n
x
n
|x| < 1 |a
k
| <M,∀k
H
n
=1+
1
2
+
1
3
+ ···+
1
n
0 <r<1 |a
n+1
− a
n
|≤Cr

n
, ∀n (a
n
)
a
0
=1,a
n
=1+
1
a
n−1
3
2
≤ a
n
≤ 2 n ≥ 1 (a
n
)
ϕ = lim
n→∞
a
n
a
n
= e
sin 5n
lim sup
n→∞
a

n
lim inf
n→∞
a
n
a
n
=(−1)
n
(2 +
3
n
) a
n
=1+
n
n +1
cos
nπ
2
(a
n
) lim
n→∞
a
n
= a
s
n
=

a
1
+ ···+ a
n
n
,p
n
=
n
√
a
1
···a
n
a
(a
n
) lim
n→∞
a
n+1
a
n
= a lim
n→∞
n
√
a
n
= a

a
n
=
n
n
n!
lim
n→∞
n
n
√
n!
= e
f : I → R I ⊂ R (x
n
)
x
0
∈ I ,x
n+1
= f(x
n
)(n =0, 1, 2, ···)
f (x
n
)
···
(x
n
,f(x

n
)), (x
n+1
,x
n+1
), (x
n+1
,f(x
n+1
) n =0, 1, 2, ···
(x
n
) x
0
f
f(x)=
√
1+x f(x)=1+
1
x
f(x)=x
2
− x +1
>0 δ>0  a |f(x) −L| < |x −a| <δ
f(x)=
1
x
,a=1,L=1 f(x)=x
2
,a=2,L=4

lim
x→0
sin
1
x
(x
n
) (x

n
)
0 (sin
1
x
n
) (sin
1
x

n
)
lim
x→0
sin x
x
= lim
x→+∞

1+
1

x

x
= lim
x→0
ln(1 + x)
x
=
lim
x→0
a
x
− 1
x
= lim
x→0
(1 + x)
p
− 1
x
=
lim
x→0
x
2
− 1
2x
2
− x −1
lim

x→∞
x
2
− 1
2x
2
− x −1
lim
x→3
√
x +13−2
√
x +1
x
2
− 9
lim
x→+∞
(
3
√
x +1−
3
√
x) lim
x→1
m
√
x −1
n

√
x −1
lim
x→0
sin 5x
tan 8x
lim
x→0
(1 + x
2
)
2
x
2
lim
x→+∞

x
2
− 2x −1
x
2
− 4x +2

x
lim
x→0

sin x
x


3sinx
x−sin x
x → 0
(1 + x)
p
=1+px + o(x) sin x = x + o(x) cos x =1−
x
2
2
+ o(x)
e
x
=1+x + o(x) ln(1 + x)=x + o(x)
∼
a
x
(a>1),x
p
, ln x x → +∞
0
x [x]
√
x
f a f(a) > 0 h>0 f(x) > 0
x, a −h<x<a+ h
f g |f|, max(f,g). min(f,g)
f(x)=
x + x
2

x
2
− 1
,x= ±1 f(±1) = 0
f(x)=
sin x
x
,x=0 f(0) = α
f(x)=x
1
x−1
,x=1 f(1) = α
f(x)= x g(x)=x(1 −x
2
)
f(g(x)) f ◦ g
f [a, b] g [b, c]
h h(x)=f(x),x∈ [a, b] h(x)=g(x),x∈ (b, c]
h f(b)=g(b)
f(x)=arctg(
1
x
2
− 1
) f(x)=e
x+
1
x
f(x)= (sin
π

x
)
f :[0, 1] → [0, 1] x =
p
q
f(x)=
1
q
x f(x)=0 f
>0
p
q
1
q
>
f : R → R f(tx)=tf(x) t, x ∈ R f
f : R → R f(x + y)=f(x)+f(y)
f : R → R f(x + y)=f(x)f(y)
f : R
+
→ R f(xy)=f(x)+f(y)
f : R
+
→ R f(xy)=f(x)f(y)
f R f max, min
f [0, 1) max min
f : R → R lim
x→±∞
f(x)=+∞
min{f(x):x ∈ R}

P (x)=a
0
+ a
1
x + ···+ a
j
x
j
− a
j+1
x
j+1
−···−a
n
x
n
,
a
k
≥ 0, ∀k a
0
+ ···+ a
j
> 0,a
j+1
+ ···+ a
n
> 0 P (x)
P (x)
x

j
(0, +∞)
tan x = x
f I x
1
, ··· ,x
n
∈ I
c ∈ I f(c)=
1
n
(f(x
1
)+···+ f(x
n
))
f :[a, b] −→ [a, b] f
x
0
∈ [a, b] f(x
0
)=x
0
f :[1, 2] −→ [0, 3] f(1) = 0,f(2) = 3 f
f :[a, b] −→ R f(a)f(b) < 0
f(x)=0
√
2  x
2
−2=0

>0 δ>0 |sin x −sinx

| < |x − x| <δ
sin R
f : X → R f
X
∃L>0: |f(x) −f(x

)|≤L|x − x

|, ∀x, x

∈ X
f X
f(x)=x
3
0 ≤ x ≤ 1 0 ≤ x<∞
f(x)=x +sinx −∞ ≤ x<+∞ f(x)=
1
1+x
2
0 ≤ x<∞
f(x)=sin
π
x
0 <x<∞
f x
0
f
x

0
f(x
0
+ h)=a + bh + o(h), khi h → 0
a = f(x
0
) b = lim
h→0
f(x
0
+ h) −f(x
0
)
h
h → 0
(x + h)
2
= x
2
+2x.h + o(h) sin(x + h)=sinx +cosx.h + o(h)
f

(x)
f(x)=sinx
2
f(x)=cos
3
(2x) f(x) = ln(sin(x
2
+1))

f(x)=

x

x +
√
x f(x)=x
x
f(x)=(a
x
)
a
f(x)=(x
a
)
x
f(x)=
x
√
x f(x)=x(1 + x
2
)
tan x
f

+
(x),f

−
(x)

f(x)=|x
2
− 1| f(x)=
3
√
x
2
f(x)=x
n
sin
1
x
x =0 f (0) = 0 n ∈ N
f(x)=x
2
sin
1
x
,f(0) = 0 f

a f(x)=ax
2
g(x)=lnx
y = x
2
x = y
2
f (a, b) c f

(c) > 0 f

c
f(x)=x x f(x)=sinx x f

(0) = 1
f
f c ∈ (a, b) f

(c) > 0 x, c < x < b
f(x) >f(c)
1+x<e
x
(x =0)
x −
x
2
2
< ln(1 + x) <x (x>0) x −
x
3
6
< sin x<x (x>0)
(x
p
+ y
p
)
1/p
< (x
q
+ y

q
)
1/q
(0 <x,y;0<q<p)
ϕ :(a, b) → R M>0 |ϕ(x)| <M,∀x ∈
(a, b)
f(x)=x + ϕ(x),x∈ (a, b) f 
k ∈ R x
3
−3x +k =0
[0, 1]
a
0
n +1
+
a
1
n
+ ···+
a
n−1
2
+ a
n
=0
a
0
x
n
+ a

1
x
n−1
+ ···+ a
n
=0 [0, 1]
f(x)=
a
0
x
n+1
n +1
+ ···+ a
n
x
f [a, b]
f(x)=0 [a, b] c ∈ (a, b) f

(c)=0
f(x)=0 n +1
f [a, b]
f

f

(a) f

(b)
g(x)=f(x) − γx γ f


(a) f

(b)
g max min c ∈ (a, b)
f(x)=0 −1 ≤ x<0 f (x)=1 0 ≤ x ≤ 1
[−1, 1]
c
f(b) −f(a)=f

(c)(b −a)
f(x)=
x
x −1
(0 ≤ x ≤ 2) f(x)=
x
x −1
(2 ≤ x ≤ 4)
f(x)=Ax + B (a ≤ x ≤ b) f(x)=1− x
2/3
(−1 ≤ x ≤ 1)
f [3, 5] (3, 5) f(3) = 6,f(5) = 10
c ∈ (3, 5) f
c
c
f

(c)
g

(c)

=
f(b) −f(a)
g(b) −g(a)
f(x)=x, g(x)=x
2
(0 ≤ x ≤ 1)
f(x)=sinx, g(x)=cosx (−
π
2
≤ x ≤ 0)
|sin a −sinb|≤|a −b| |arctan a −arctan b|≤|a − b|

1+
1
x

x
<e<

1+
1
x

x+1
(x>0)
|f(x)−f(a)|≤ sup
c∈[a,x]
|f

(c)||x−a| f(x) f (a)

 δ >0 δ |x −a| <δ
|f(x) −f(a)| <
f(x)=x
2
f(x)=
1
x
f g n h = fg
h

(c)=f

(c)g(c)+2f

(c)g

(c)+f(c)g

(c)
h
(n)
(c)=
n

k=0
n!
k!(n −k)!
f
(k)
(c)g

(n−k)
(c)
f
(100)
f(x)=x
3
sin x f(x)=x
2
e
−
x
a
f(x)=
1+x
√
1 −x
n a
x
, sin(ax + b), log
a
x, (1 + x)
p
f
(n)
(x) f(x)=
1
x
2
− 3x +2
f(x)

x
0
f(x)=x
n
x =0 f (x)=(1+x)
n
4 P(2) = −1,P

(2) = 0,P

(2) = −12,P

(2) = 24
a>0,h>0,n∈ N θ ∈ [0, 1]
1
a + h
=
1
a
−
h
a
2
+
h
2
a
3
+ ···+
(−1)

n−1
h
n−1
a
n
+
(−1)
n
h
n
(a + θh)
n+1
.
0
4
f(x)=ln(2cosx +sinx) f(x)=e
√
1+x
f(x)=(1+x)
1
x
f(x)=ln(1+x) g(x) = arctan x
f
(n)
(0) g
(n)
(0) f g x
0
=0
ln 2 = 1 −

1
2
+
1
3
+ ···+(−1)
n−1
1
n
+ R
n
π
4
=1−
1
3
+
1
5
+ ···+(−1)
n
1
2n +1
+ R
n
N n>N |R
n
| < 10
−3
3

√
29, sin 46
o
, ln(1, 05)
< 10
−3
lim
x→0
+
tan x −x
x −sin x
lim
x→+∞
ln x
x
0,0001
lim
x→0
(
1
x
−
1
sin x
) lim
x→+∞
x
1
x
lim

x→+∞

a
1
x
+ b
1
x
2

x
(a, b > 0) lim
x→0
sin(x −sinx)
√
1+x
3
− 1
lim
x→1
1 −x +lnx
1 −
√
2x −x
2
lim
x→0
x
2
sin

1
x
sin x
lim
x→∞
x −sin x
x +sinx
f n
f a ∆
h
f(a)=f(a + h) −f(a)
f a ∆
k
h
f(a)=∆
h
(∆
k−1
h
f(a)) k =2, 3, ··· ,n
∆
2
h
f(a)
f

(a) ∼
∆
2
h

f(a)
h
2
=
f(a +2h) −2f(a + h)+f(a)
h
2
, h → 0
∆
k
h
f(a)
f
(n)
(a) ∼
∆
n
h
f(a)
h
n
h → 0
max min
f(x)=|x
2
− 3x +2|,x∈ [−10, 10] f (x)=
√
5 −4x, x ∈ [−1, 1]
f(x)=x
n

(1 −x)
m
,x∈ [0, 1]
a, b > 0 m, n ∈ N
a
m
b
n
a + b
a
m
+ b
n
ab
x
2
a
2
+
y
2
b
2
=1
a p(x)=x
2
+ a 0 [−1, 1]
a sup
x∈[−1,1]
|p(x)|

f(x) g(x) [a, b] sup
x∈[a,b]
|f(x) −g(x)|
f(x)=x
n
,g(x)=1,x∈ [−1, 1]
f(x)=1+x + ···+ x
n
,g(x)=
1
1 −x
,x∈ [−r, r](0<r<1
f(x)=arcsinx + arccos x f(x)=2arctanx +arctan

2x
1 −x
2

f(x)=
x
3
+4
x
2
f(x)=
3
√
1 −x
3
f(x)=xe

−x
f(x)=ln

x
x −1

x
3
+ px + q =0
p, q
(p, q)
e
a+b
2
≤
1
2
(e
a
+ e
b
) ln
a + b
2
≥ ln
√
ab a, b > 0

x + y
2


n
≤
1
2
(x
n
+ y
n
) x, y > 0,n > 1
x ln x + y ln y ≥ (x + y)ln
x + y
2
x, y > 0
p, q > 0,
1
p
+
1
q
=1
ab ≤
a
p
p
+
b
q
q
, (a, b > 0, )

n

k=1
a
k
b
k
≤

n

k=1
|a
k
|
p

1
p

n

k=1
|b
k
|
q

1
q

p




n

k=1
|a
k
+ b
k
|
p
≤
p




n

k=1
|a
k
|
p
+





n

k=1
|b
k
|
p
|a
k
+ b
k
|
p
≤|a
k
||a
k
+ b
k
|
p−1
+ |b
k
||a
k
+ b
k
|

p−1
f :[a, b] → R
f(a) < 0 <f(b), f

(x) > 0,f

(x) > 0, ∀x
f(x)=0 x
0
= b
x
n+1
= f (x
n
,f(x
n
))
x
n+1
= x
n
−
f(x
n
)
f

(x
n
)

(x
n
) f(x)=0
√
2 10
−6
f(x)=x
2
− 2,x∈ [1, 2]
f
◦

x

4+x
2
dx

xe
−x
2
dx

ln xdx
x
√
1+lnx

sin x cos
3

xdx
1+cos
2
x

dx
(1 + x)
√
x


4 −x
2
dx


a
2
+ x
2
dx
◦

x
2
e
−x
dx

x

2
ln xdx

ln x
x
3
dx

e
x
sin xdx

arcsin x
x
2
dx
◦

dx
x
4
− x
2
− 2

dx
(x
2
− 1)(x
2

+1)

x +1
(x
2
+1)
2
dx

x
2
dx
x
6
− 1

dx
x(x
2
+1)
2

dx
x
4
+1

x
2
dx

(1 −x)
100
◦

dx
x(1 + 2
√
x +
3
√
x)

x

x −2
x +1
dx

1 −
√
x +1
1+
3
√
x +1
dx

dx
(x +1)
√

x
2
+ x +1

dx
x +
√
x
2
+2x

√
−x
+
4x +10dx
◦

dx
2sinx −cos x +5

dx
1+ cos x
>0

sin
4
xdx

cos
5

xdx

cos 3x sin 5xdx

sin
4
x cos
5
xdx

sin
2
x cos
4
xdx
n ∈ N
I
n
(a)=

dx
(a
2
+ x
2
)
n
I
n
=


sin
n
xdx J
n
=

cos
n
xdx
I
n
=

x
n
e
−x
dx
e
−x
2
,
sin x
x
,
1
ln x
,
1


(1 −x
2
)(1 −k
2
x
2
)
(0 <k<1)
e
x
√
x
, lnx cos x,
e
x
x
, sin x
2
,
1

1 −k
2
sin ϕ
f P
f(x)=x, x∈ [0, 1] P = {0,
1
3
,

2
3
, 1}
f(x)=x, x∈ [0, 1] P = {0,
1
n
,
2
n
, ··· ,
n
n
}
f(x)=x
2
,x∈ [0, 1] P = {0,
1
n
,
2
n
, ··· ,
n
n
}
f(x)=sinx, x ∈ [0,
π
2
] P = {0,
π

2n
,
2π
2n
, ··· ,
nπ
2n
}
f(x)=
1
x
,x∈ [a, b] P = {a, aq, ···,aq
n
= b} (0 <a<b)