Giáo trình giải tích 1 part 3 ppt

n ∈ N
n
√
x = x
1
n
x
n
n
R n [0, +∞)
✲
✻
y =
2n
√
x
✲
✻
y =
2n+1
√
x
m, n ∈ Z,n>0 x
m
n
=(
n
√
x)
m
n m

α x
α
= e
α ln x
(0, +∞) α>0 α<0
(xx

)
α
= x
α
x

α
a
x
= e
x ln a
a>0
R (0, +∞) a>1
0 <a<1
a
x+x

= a
x
a
x

✲

x
✻
y
r
1
y = a
x
(a>1)
✲
x
✻
y
r
1
y = a
x
(0 <a<1)
log
a
x =
ln x
ln a
(a>0,a=1
(0, +∞) R a>1
0 <a<1
log
a
x +log
a
x


=log
a
xx

log
a
x =log
a
b log
b
x
log
a
x
α
= α log
a
x
a
x
log
a
x y =log
a
x ⇔ a
y
= x
✲
x

✻
y
r
1
y =log
a
x (a>1)
✲
x
✻
y
r
1
y =log
a
x (0 <a<1)
x ∈ R
M (1, 0) M x 2π
x 2π
M sin x cos x
✲
✻
0−1
−1
1
M
❪
s
x
cos x

sin x
sin x R [−1, 1]
2π
cos x R [−1, 1]
2π
sin
2
x +cos
2
x =1
tan x =
sin x
cos x
x =
π
2
+ kπ, k ∈ Z R
π
cot x =
cos x
sin x
x = kπ, k ∈ Z R
π
✲
x
✻
y
y =sinx
02π
r

1
r
−1
✲
x
✻
y
y =cosx
02π
r
1
r
−1
arcsin x :[−1, 1] → [−
π
2
,
π
2
] sin : [−
π
2
,
π
2
] → [−1, 1]
arccos x :[−1, 1] → [0,π] cos : [0,π] → [−1, 1]
arctan x : R → (−
π
2

,
π
2
) tan : (−
π
2
,
π
2
) → R
x : R → (0,π) cot : (0,π) → R
✲
x
✻
y
y =tanx
0
π
2
−
π
2
✲
x
✻
y
y =arctanx
0
π
2

−
π
2
•
f(x)=2
x
+
3
√
x − ln(ln(ln(x
2
+ 1))) f(x)=
x
2
+ e
−x
tan 5x
√
x − 1+sin(πx)
a
x
, log
a
x, cos x, tan x, cot x, arcsin x =arctan

x
√
1 − x
2


,
x =
π
2
− arctan x, arccos x =

x
√
1 − x
2

f(x)=a
0
+ a
1
x + ···+ a
n
x
n
a
0
,a
1
, ··· ,a
n
∈ R
y =2x +1 y = x
2
+5x − 1
y = x

3
− 3x +1
f(x)=
P (x)
Q(x)
P, Q
y =
x − 1
x +1
y =
x
2
+1
x − 1
cosh x =
e
x
+ e
−x
2
, sinh x =
e
x
− e
−x
2
, tanh x =
sinh x
cosh x
, coth x =

cosh x
sinh x
cosh
2
x − sinh
2
x =1
sinh(x + y)=sinhx cosh y +sinhy cosh x
cosh(x + y)=coshx cosh y +sinhx sinh y
X ⊂ R a ∈ R
a a {x ∈ R : |x − a| <δ} =(a − δ, a + δ)
δ>0
a
X U a U ∩X \{a} = ∅
(x
n
) X \{a} a
(a, b) x ∈ [a, b] {
1
n
: n ∈ N}
0
f : X → R a X f
L ∈ R x a >0 δ>0
a  x ∈ X 0 < |x − a| <δ |f(x) − L| <
∀>0, ∃δ>0:x ∈ X, 0 < |x − a| <δ ⇒|f(x) − L| <
lim
x→a
f(x)=L f(x) → L x → a
>0 δ>0 f x ∈ (a−, a+)

(a, L) 2δ ×2
•
∀x
n
∈ X \{a}, lim
n→∞
x
n
= a ⇒ lim
n→∞
f(x
n
)=L
• lim
x→a
f(x)=L ⇔ lim
x→a
|f(x) − L| =0
•
• lim
x→a
f(x)
∀>0, ∃δ>0:x, x

∈ X, 0 < |x − a| <δ,0 < |x

− a| <δ ⇒|f(x) − f (x

)| <
lim

x→0
x
2
=0 >0 δ =
√

x |x − 0| <δ |x
2
− 0| <δ
2
= 
lim
x→0
sin
1
x
x
n
=
1
2nπ
x

n
=
1
π
2
+2nπ
0 sin

1
x
n
=sin2nπ =0 sin
1
x

n
=sin(
π
2
+2nπ)=1
n →∞
a
f(x)=x sin
1
x
lim
x→0
f(x)=0 |f(x) − 0| = |x sin
1
x
|≤|x|→0 x → 0
a lim
x→a
f(x) = f(a)
f(x)=[1−|x|] lim
x→0
f(x)=0= f(0) = 1
f,g, ϕ : X → R a X

lim
x→a
f(x)=L lim
x→a
g(x)=M
lim
x→a
(f ± g)(x)=L ± M lim
x→a
fg(x)=LM lim
x→a
f
g
(x)=
L
M
( M =0)
f(x) ≤ g(x) x
a L ≤ M
f(x) ≤ ϕ(x) ≤ g(x) x
a L = M lim
x→a
ϕ(x)=L
lim
x→a
f(x)=L, lim
y→L
g(y)=A δ>0
0 < |x − a| <δ f(x) = L lim
x→a

g ◦f(x) = lim
y→L
g(y)=A

f(x)=g(x)=

0 x =0
1 x =0
lim
x→0
g(f (x)) = lim
y→0
g(y)
f a
lim
x→a
f(x)=f (a)
lim
x→a
e
x
= e
a
|x|≤1





1+

x
n

n
− 1




≤|x|(e − 1)
n → +∞ |e
x
− 1|≤|x|(e − 1) lim
x→0
|e
x
− 1| =0 lim
x→0
e
x
=1
u = x − a
lim
x→a
e
x
= lim
x→a
e
x−a

e
a
= lim
u→0
e
u
e
a
= e
a
lim
x→a
sin x =sina 0 ≤|sin t|≤|t|
|sin x − sin a| = |2cos
x + a
2
sin
x − a
2
|≤2|
x − a
2
|→0 x → a
ln x arctan x

lim
x→a
f(x) > 0 =1 lim
x→a
g(x) =0

lim
x→a
f(x)
g(x)
= lim
x→a
f(x)
lim
x→a
g(x)
f : X → R
L
f(x) x a
∀>0, ∃δ>0:|f(x) − L| <,∀x ∈ X, 0 <x− a<δ ( 0 <a−x<δ)
L = f(a+) = lim
x→a
+
f(x) = lim
x→a+0
f(x)( L = f(a−) = lim
x→a
−
f(x) = lim
x→a−0
f(x))
(a, a + δ) ∩ X = ∅ (a − δ, a) ∩ X = ∅
δ>0
lim
x→0
+

(x)=1 lim
x→0
−
(x)=−1
lim
x→n
+
[x]=n lim
x→n
−
[x]=n − 1 n ∈ Z
lim
x→1
+
√
x − 1=0 lim
x→1
−
√
x − 1
x<1
lim
x→a
f(x) lim
x→a
+
f(x) = lim
x→a
−
f(x)

f (a, b) lim
x→x
+
0
f(x) lim
x→x
−
0
f(x)
x
0
∈ (a, b)
a = ±∞ L = ±∞
+∞ (R, +∞) −∞
(−∞, −R) R>0
lim
x→a
f(x)=+∞⇔∀E>0, ∃δ>0:f(x) >E,∀x ∈ X, 0 < |x −a| <δ
lim
x→a
f(x)=−∞ ⇔ ∀E>0, ∃δ>0:f (x) < −E,∀x ∈ X, 0 < |x − a| <δ
lim
x→+∞
f(x)=L ⇔∀>0, ∃R>0:|f(x) −L| <,∀x ∈ X, x > R
lim
x→−∞
f(x)=L ⇔∀>0, ∃R>0:|f(x) −L| <,∀x ∈ X, x < −R
a X
lim
x→+∞

f(x)=+∞, lim
x→+∞
f(x)=−∞, lim
x→−∞
f(x)=+∞, lim
x→−∞
f(x)=−∞
lim
x→a
+
f(x)=∞, lim
x→a
−
f(x)=∞
p>0 lim
x→+∞
x
p
=+∞ lim
x→0
+
1
x
p
=+∞
a>1 lim
x→+∞
a
x
=+∞ lim

x→−∞
a
x
=0
a>1 lim
x→+∞
log
a
x =+∞ lim
x→0
+
log
a
x = −∞
lim
x→
π
2
+
tan x = −∞ lim
x→
π
2
−
tan x =+∞
0
0
,
∞
∞

, 0.∞, ∞−∞, 0
0
, 1
∞
, ∞
0
.
lim
x→+∞


x
2
+7−

x
2
− 1

∞−∞
lim
x→+∞


x
2
+7−

x
2

− 1

= lim
x→+∞
(
√
x
2
+7−
√
x
2
− 1)(
√
x
2
+7+
√
x
2
− 1)
√
x
2
+7−
√
x
2
− 1
= lim

x→+∞
(x
2
+7)−(x
2
− 1)
√
x
2
+7+
√
x
2
− 1
= lim
x→+∞
8
√
x
2
+7+
√
x
2
− 1
= lim
x→+∞
8
+∞
=0

lim
x→1
3
√
x − 1
√
x − 1
0
0
lim
x→1
3
√
x − 1
√
x − 1
= lim
x→1
3
√
x − 1
√
x − 1
(
3
√
x
2
+
3

√
x +1)
(
3
√
x
2
+
3
√
x +1)
(
√
x +1)
(
√
x +1)
= lim
x→1
x − 1
x − 1
(
√
x +1)
(
3
√
x
2
+

3
√
x +1)
= lim
x→1
√
x +1
3
√
x
2
+
3
√
x +1)
=
√
1+1
3
√
1
2
+
3
√
1+1)
=
2
3
lim

x→0
sin x
x
=1
lim
x→∞
(1 +
1
x
)
x
= lim
x→0
(1 + x)
1
x
= e
lim
x→0
ln(x +1)
x
=1
lim
x→0
a
x
− 1
x
=lna
lim

x→0
(1 + x)
p
− 1
x
= p
0 < |x| <
π
2
|sin x| < |x| < |tan x|
1 <




x
sin x




<
1
|cos x|
(x
n
) +∞ n
k
=[x
k

]
1
n
k
+1
≤
1
x
k
≤
1
n
k

1+
1
n
k
+1

n
k
≤

1+
1
x
k

x

k
≤

1+
1
n
k

n
k
+1
lim
k→∞

1+
1
k

k
= e lim
x→+∞
(1 +
1
x
)
x
= e
lim
x→−∞
(1+

1
x
)
x
= lim
y→+∞
(1−
1
y
)
−y
= lim
y→+∞
(
y
y −1
)
y
= lim
y→+∞
(1+
1
y −1
)
y−1
(1+
1
y −1
)=e
lim

x→0
(1 + x)
1
x
= lim
y→∞
(1 +
1
y
)
y
= e
lim
x→0
ln(x +1)
x
= lim
x→0
ln(x +1)
1
x
= ln(lim
x→0
(x +1)
1
x
)=lne =1
u = a
x
− 1 x =log

a
(u +1)=
ln(u +1)
ln a
lim
x→0
a
x
− 1
x
= lim
u→0
u ln a
ln(u +1)
=lna
lim
x→0
(1 + x)
p
− 1
x
= lim
x→0
e
p ln(1+x)
− 1
x
= lim
x→0
e

p ln(1+x)
− 1
p ln(1 + x)
p ln(1 + x)
x
= lim
u→0
e
u
− 1
u
lim
x→0
p ln(1 + x)
x
=lne.p = p
lim
x→0
1 − cos x
x
2
= lim
x→0
2sin
2
x
2
x
2
= lim

x→0
1
2

sin
x
2
x
2

2
= lim
u→0
1
2

sin u
u

2
=
1
2
lim
x→∞

x − 2
x − 3

3x

= lim
x→∞

1+
1
x − 3

(x−3)
3x
x−3
= lim
u→∞

1+
1
u

u lim
x→∞
3x
x−3
= e
3
lim
x→0
tan x
x
, lim
x→0
sin kx

x
, lim
x→0
sin kx
sin lx
, lim
x→0
(1+5x)
1
x
a ∈ R a = ±∞
a
f(x) ∼ g(x) x → a lim
x→a
f(x)
g(x)
=1 f(x) g(x)
f(x)=o(g(x)) x → a lim
x→a
f(x)
g(x)
=0 f(x) g(x)
f(x)=O(g(x)) x → a ∃C>0:|f(x)|≤C|g(x)| x a
f(x)=o(1) x → a ⇔ lim
x→a
f(x)=0
f(x)=O(1) x → a ⇔ f(x) a
f(x)=g(x)+o(g(x)) x → a ⇔ f(x) ∼ g(x) x → a
o(g(x)) O(g(x))
f(x) ∈ o(g(x)) f(x)=o(g(x))

o(g(x)) − o(g(x)) = O(g(x)) −O(g(x)) =
x → a
f(x) ∼ f
1
(x) g(x) ∼ g
1
(x) f(x)g(x) ∼ f
1
(x)g
1
(x),
f(x)
g(x)
∼
f
1
(x)
g
1
(x)
f(x) ∼ f
1
(x) g(x) ∼ g
1
(x) f(x)+g(x) ∼ f
1
(x)+g
1
(x)
x → a

f(x)=o(ϕ(x)) g(x)=o(ϕ(x)) f(x) ± g(x)=o(ϕ(x))
f(x)=O(ϕ(x)) g(x)=O(ϕ(x)) f(x)+g(x)=O(ϕ(x))
f(x)=o(ϕ(x)) g f(x)g(x)=o(ϕ(x))
f(x)=O(ϕ(x)) g f(x)g(x)=O(ϕ(x))
(x − a)
n
,e
x
, ln x
x → 0
(1 + x)
α
=1+αx + o(x) (1 + x)
α
∼ 1+αx
e
x
=1+x + o(x) e
x
∼ 1+x
ln(1 + x)=x + o(x)ln(1+x) ∼ x
sin x = x + o(x)sinx ∼ x
cos x =1−
x
2
2
+ o(x)cosx ∼ 1 −
x
2
2

x → +∞
(a
0
+ a
1
x + ···+ a
n
x
n
)
1
m
∼ a
1
m
n
x
n
m
(a
n
=0)
a
0
+ a
1
x + ···+ a
n
x
n

b
0
+ b
1
x + ···+ b
m
x
m
∼
a
n
x
n
b
m
x
m
(a
n
,b
m
=0)
log
a
x = o(x
n
)(a>1,n>0)
x
n
= o(a

x
)(a>1,n>0)
lim
x→0
√
1+x − 1
sin 2x
√
1+x − 1 ∼
x
2
sin 2x ∼ 2x x → 0
lim
x→0
√
1+x − 1
sin 2x
= lim
x→0
x/2
2x
=
1
4
lim
x→0
ln(1 + sin x)
x +tan
3
x

ln(1 + sin x) ∼ ln(1 + x) x +tan
3
x ∼
x + x
3
∼ x x → 0
lim
x→0
ln(1 + sin x)
x +tan
3
x
= lim
x→0
ln(1 + x)
x
=1
n ∈ N n
n! ∼

n
e

n
√
2πn, n!=O(n
n
),a
n
= o(n!).

f X a f
a lim
x→a
f(x)=f(a)
f a
• f a lim
x→a
f(x)=L L = f(a)
• ∀>0, ∃δ>0:∀x ∈ X, |x −a| <δ ⇒|f(x) −f(a)| <
• (x
n
) X lim
n→∞
x
n
= a lim
n→∞
f(x
n
)=f(a)
C(X) X
f
a lim
x→a
+
f(x)=f(a)
f
a lim
x→a
−

f(x)=f(a)
f a f a
a
a

f(x)=
1
x
0
f(x)= x 0 lim
x→0
−
f(x)=
−1 = f(0) = 0
f(x)=
sin x
x
x =0 f(0) = L lim
x→0
f(x) = lim
x→0
sin x
x
=1 f
0 1=f(0) = L
f(x)=sin
1
x
x =0 f(0) = L L f
0 lim

x→0
sin
1
x
D(x)=

0 x
1 x
a f(a)=0
R (x
n
) a
f(x
n
)=1 f(a)=0 a
f(x)=x sin
1
x
x =0 f(0) = 0
g(x)=[x]
f
a lim
x→a
−
f(x)=f (a
−
) lim
x→a
+
f(x)=

f(a
+
) |f(a
+
) − f(a
−
)| =0
a a
I
✲
x
✻
y
s
✒
✛
✻
✙
[a, b]
0 a
f a g f(a) g ◦f a
f a f(a) >L f(x) >L a δ>0
f(x) >L x |x −a| <δ
 =
f(a) − L
2
δ>0 |x − a| <δ
f(a)−<f(x) <f(a)+ f(x) >f(a)− =
f(a)+L
2

>
L + L
2
= L 
f g a |f|, max(f,g), min(f,g)
a |f| f x →|x|
max(f,g)=
1
2
(f +g+|f −g|), min(f,g)=
1
2
(f +g−|f −g|)
f [a, b]
f(a) f(b) c ∈ (a, b) f (c)=0
γ f(a),f(b) c ∈ (a, b) f(c)=γ