Giáo trình giải tích 1 part 5 ppsx

|R
n
| = |
e
θ
(n +1)!
|≤
3
(n +1)!
 =10
−3
n =6
 =10
−6
n =9
lim
x→+∞
(x − x
2
ln(1 +
1
x
))
ln(1 +
1
x
)) =
1
x
−
1

2x
2
+ o(
1
x
2
)
x − x
2
ln(1 +
1
x
)=
1
2
+ x
2
o(
1
x
2
) →
1
2
x → +∞
lim
x→0
e
x
2

−
√
1 − x
2
+ x
3
ln(1 + x
2
)
e
x
2
−
√
1 − x
2
+ x
3
=1+x
2
+o(x
3
)−(1+
1
2
(−x
2
+x
3
)+o(x

2
)=
3
2
x
2
+o(x
2
)
ln(1 + x
2
)=x
2
+ o(x
2
)
lim
x→0
e
x
2
−
√
1 − x
2
+ x
3
ln(1 + x
2
)

= lim
x→0
3
2
x
2
x
2
=
3
2
0
0
,
∞
∞
f,g I x
0
∈ I
g

(x) =0 ∀x ∈ I lim
x→x
0
f(x) = lim
x→x
0
g(x)=0
lim
x→x

0
f(x)
g(x)
= lim
x→x
0
f

(x)
g

(x)
g

(x) =0 ∀x ∈ I lim
x→x
0
f(x) = lim
x→x
0
g(x)=∞
lim
x→x
0
f(x)
g(x)
= lim
x→x
0
f


(x)
g

(x)
x
0
= ±∞ f,g
x
0
f(x
0
)=g(x
0
)=0
c x
0
,x
f(x) −f(x
0
)
g(x) − g(x
0
)
=
f

(c)
g


(c)
x → x
0
c → x
0
x
0
= ±∞ F (t)=f(
1
t
),G(t)=g(
1
t
)

p>0 lim
x→+∞
ln x
x
p
= lim
x→+∞
1/x
px
p−1
=0
p>0 p ≤ k
lim
x→+∞
x

p
e
x
= lim
x→+∞
px
p−1
e
x
= ···= lim
x→+∞
p(p − 1) ···(p − k +1)x
p−k
e
x
=0
0
0
∞
∞
0.∞ fg =
f
1/g
∞−∞ f − g =
1
1/f
−
1
1/g
=

1/g −1/f
1/fg
1
∞
, 0
0
, ∞
0
f
g
= e
g ln f
g ln f 0.∞
p>0 lim
x→0
+
x
p
ln x = lim
x→0a+
ln x
x
−p
= lim
x→0
+
1/x
−px
−p−1
= lim

x→0
+
x
p
−p
=0
lim
x→0

1
x
−
1
sin x

= lim
x→0
sin x − x
x sin x
= lim
x→0
cos x − 1
sin x + x cos x
= lim
x→0
−sin x
2cosx +sinx
=0
lim
x→0

+
x
x
= lim
x→0
+
e
x ln x
= e
lim
x→0
+
x ln x
= e
0
=1
lim
x→0
(1 + x
2
)
1
e
x
−x−1
1
∞
y =(1+x
2
)

1
e
x
−x−1
ln y =
ln(1 + x
2
)
e
x
− x − 1
0
0
lim
x→0
ln y = lim
x→0
2x
1+x
2
e
x
− 1
= lim
x→0
1
1+x
2
lim
x→0

2x
e
x
− 1
= lim
x→0
2
e
x
=2
lim
x→0
(1 + x
2
)
1
e
x
−x−1
= e
2
f(x)=sin
2
x sin
1
x
g(x)=e
x
−1 lim
x→0

f(x)
g(x)
lim
x→0
f

(x)
g

(x)
f I
f I f

≥ 0 f

≤ 0 I
f

> 0 f

< 0 I f I

e
x
> 1+x (x =0)
(x
p
+ y
p
)

1
p
> (x
q
+ y
q
)
1
q
0 <x,y 0 <p<q)
f(x)=e
x
− x
f

(x)=e
x
−1 f

(x) < 0 x<0 f

(x) > 0 x>0 f
(−∞, 0) (0, +∞) f(x)=e
x
− x>f(0) = 1 x =0
g(t)=(x
t
+ y
t
)

1
t
(0, +∞)
g

(t) ln g(t)=
ln(x
t
+ y
t
)
t
g

(t)
g(t)
=
−x
t
ln(
x
t
+y
t
x
t
) − y
t
ln(
x

t
+y
t
y
t
)
t
2
(x
t
+ y
t
)
g

(t) < 0, ∀t>0 g (0, +∞)
f I
f

(x
0
)=0 f

(x) x x
0
f
x
0
f(x
0

) ≥ f(x) x x
0
f

(x
0
)=0 f

(x) x x
0
f
x
0
f(x
0
) ≤ f(x) x x
0

f n I x
0
f

(x
0
)=f

(x
0
)=···= f
(n−1)

(x
0
)=0 f
(n)
(x
0
) =0
n f
(n)
(x
0
) > 0 f x
0
n f
(n)
(x
0
) < 0 f x
0
n f x
0
f(x
0
+∆x)=f(x
0
)+
1
n!
f
(n)

(x
0
)∆x
n
+ o(∆x
n
)

e
x
> 1+x (x =0)
f(x)=e
x
+ e
−x
+2cosx
f

(x)=e
x
− e
−x
− 2sinx f

(0) = 0
f

(x)=e
x
+ e

−x
− 2cosx f

(0) = 0
f
(3)
(x)=e
x
− e
−x
+2sinx f
(3)
(0) = 0
f
(4)
(x)=e
x
+ e
−x
+2cosx f
(4)
(0) = 4 > 0
x =0
max, min x
√
1 − x
2
.
f(x)=x
√

1 − x
2
x ∈ [−1, 1]
f [−1, 1] max, min
x f

(x)=0
f(−1),f(1)
f

(x)=
1 − 2x
2
√
1 − x
2
=0 ⇔ x = ±
1
√
2
f(
1
√
2
)=
1
2
,f(−
1
√

2
)=−
1
2
,f(−1) = 0,f(+1) = 0
f
max
= f = f(
1
√
2
)=
1
2
,f
min
= f(−
1
√
2
)=−
1
2
S
r h
V = πr
2
h s = πr
2
+ πr

2
+2πrh
s h =
s − 2πr
2
2πr
V (r)=πr
2

s − 2πr
2
2πr

=
1
2
r(s −2πr
2
) r ∈ [0,

s
π
]
V

(r)=
1
2
(s − 6πr
2

) V

(r)=0 ⇔ r =

s
6π
V

r

s
6π
V (r) max
h =2

s
6π
=2r V
f I
f
x
1
,x
2
∈ I 0 <t<1
f(tx
1
+(1− t)x
2
) ≤ tf(x

1
)+(1− t)f(x
2
)
f
x
1
,x
2
∈ I 0 <t<1
f(tx
1
+(1−t)x
2
) ≥ tf(x
1
)+(1− t)f(x
2
)
f f
f f
✲
x
✻
y
❝
x
1
❝
❝

x
2
❝
s
tx
1
+(1−t)x
2
s
s
f(tx
1
+(1− t)x
2
) s
tf(x
1
)+(1− t)f(x
2
) s
✑
✑
✑
✑
✑
✑
✑
✑
✑
✑

✑
✑
✑
M(x
0
,f(x
0
)) f M
f
f I
x
1
, ··· ,x
n
∈ I, t
1
, ··· ,t
n
≥ 0,t
1
+ ···+ t
n
=1
f(t
1
x
1
+ ···+ t
n
x

n
) ≤ t
1
f(x
1
)+···+ t
n
f(x
n
)
x
1
, ··· ,x
n
∈ I, α
1
, ··· ,α
n
≥ 0
f(
α
1
x
1
+ ···+ α
n
x
n
α
1

+ ···+ α
n
) ≤
α
1
f(x
1
)+···+ α
n
f(x
n
)
α
1
+ ···+ α
n
,
f 1 2 I
f f

f f

≥ 0 f

≤ 0
f f
f
⇔ ⇔
f I x
1

,x
2
∈ I x
1
<x
2
x = tx
1
+(1−t)x
2
∈
(x
1
,x
2
)
f(x)= f(tx
1
+(1−t)x
2
) ≤ tf(x
1
)+(1− t)f(x
2
)
⇔ (x
2
− x
1
)f(x) ≤ (x

2
− x)f(x
1
)+(x − x
1
)f(x
2
)
⇔ (x
2
− x)f(x)+(x − x
1
)f(x) ≤ (x
2
− x)f(x
1
)+(x − x
1
)f(x
2
)
⇔ (x
2
− x)(f(x) −f(x
1
)) ≤ (x − x
1
)(f(x
2
) − f(x))

⇔
f(x) −f(x
1
)
x − x
1
≤
f(x
2
) − f(x)
x
2
− x
⇒ f I x → x
1
x → x
2
f

(x
1
) ≤ f

(x
2
) f

⇐ f

x

1
,x
2
∈ I x
1
<x
2
x ∈ (x
1
,x
2
)
f(x) −f(x
1
)
x − x
1
= f

(c
1
) c
1
(x
1
,x)
f(x
2
) − f(x)
x

2
− x
= f

(c
2
) c
2
(x, x
2
)
f

f(x) −f(x
1
)
x − x
1
≤
f(x
2
) − f(x)
x
2
− x
f I 
f I f

(x
0

)=0 f

(x)
x x
0
M(x
0
,f(x
0
)) f
f(x)=lnx I =(0, ∞) x
1
, ···x
n
> 0
ln

x
1
+ ···+ x
n
n

≥
ln x
1
+ ···+lnx
n
n
x

1
+ ···+ x
n
n
≥
n
√
x
1
···x
n
x
k
x
−1
k
n
√
x
1
···x
n
≥
n
1
x
1
+ ···+
1
x

n
, (x
1
, ··· ,x
n
> 0)
f(x)=e
x
e
t
1
x
1
+t
2
x
2
≤ t
1
e
x
1
+ t
2
e
x
2
,x
1
,x

2
∈ R,t
1
,t
2
> 0,t
1
+ t
2
=1
a = e
t
1
x
1
,b= e
t
2
x
2
p = t
−1
1
,q = t
−1
2
ab ≤
a
p
p

+
b
q
q
, (a, b > 0,p,q > 0,
1
p
+
1
q
=1)
p, q > 0
1
p
+
1
q
=1





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
y =
x
3
x
2
− 1
R \{±1}
y

=
x
2
(x
2
− 3)
(x
2
− 1)
2
,y

=0 ⇔ x =0,x= ±
√

3
y

=
2x(x
2
+3)
(x
2
− 1)
3
,y

=0 ⇔ x =0
x = ±1 lim
x→±1
y = ∞
y = x y = x +
x
x
2
− 1
lim
x→±∞
(y −x)=0
x
−∞ −
√
3 −10 1
√

3+∞
y

+0−||−0 −||−0+
y
−∞

−3
√
3
2

−∞
||
+∞
 0 
−∞
||
+∞

3
√
3
2
 +∞
✲
x
✻
y
s

s
√
3
3
√
3
2
s
−
√
3
−
3
√
3
2
✑
✑
✑
✑
✑
✑
✑
✑
✑
✑
✑
✑
✑
✑

✑
✑
✑
✑
✑
✑
✑
✑
✑
✑
✑
✑
y = x
1−1
y = x +
√
x
2
− 1
•

x = x(t)
y = y(t)
,t∈ (α, β)
C = {(x, y):x = x(t),y = y(t),α<t<β}
x = a cos t, y = a sin t 0 ≤ t ≤ 2π x
2
+ y
2
= a

2
C
y x
dy
dx
(x(t)) =
y

(t)
x

(t)
x = a lim
t→t
0
x(t)=a lim
t→t
0
y(t)=∞
y = b lim
t→t
0
x(t)=∞ lim
t→t
0
y(t)=b
y = kx + b lim
t→t
0
x(t) = lim

t→t
0
y(t)=∞ lim
t→t
0
y(t)
x(t)
= k
lim
t→t
0
(y(t) − kx(t)) = b
x = a cos
3
t, y = a sin
3
t (a>0)
R 2π t ∈ [0, 2π]
x

(t)=−3a cos
2
t sin t, x

(t)=0 ⇔ t =0,π/2,π,3π/2, 2π
y

(t)=3a sin
2
t cos t, y


(t)=0 ⇔ t =0,π/2,π,3π/2, 2π
−a ≤ x(t),y(t) ≤ a
t
0 π/2 π 3π/22π
x

0 − 0 − 0 0 0
x
a
 0 
−a
 0 
a
y

0 0 − 0 − 0 0
y 0 
a
 0 
−a
 0
✲
x
✻
y
0
a
−a
a

−a
•
y

x

(x(t)) = −tan t t =0,π,2π
t = π/2, 3π/2
• t x
2/3
+ y
2/3
= a
2/3
x
3
+ y
3
− 3axy =0 (a>0)
y = tx x
3
+ t
3
x
3
− 3atx
2
=0








x =
3at
1+t
3
y =
3at
2
1+t
3
(t = −1)
x

=3a
1 − 2t
3
(1 + t
3
)
2
,x

=0 ⇔ t =
1
3
√

2
y

=3a
2 − t
3
(1 + t
3
)
2
,y

=0 ⇔ t =0,
3
√
2
t →−1 x →∞,y →∞
y
x
→ k = −1 y −kx →−a
y = −x −a
t
−∞ −101/
3
√
2
3
√
2+∞
x


|| 0 −−
x 0
+∞
||
−∞
 0  a
3
√
4  a
3
√
2  0
y

−||−0 0 −
y 0
−∞
||
+∞
 0  a
3
√
2  a
3
√
4  0
✲
x
✻

y
t
s
❅
❅
❅
❅
a
3
√
4
a
3
√
4
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅

y = −x −a
dy
dx
(x(t)) =
t(2 − t
3
)
1 − 2t
3
t =0,
3
√
2 t =1/
3
√
2
•
O Ox
(r, ϕ) ∈ R
+
× [0, 2π) M |
→
OM | = r (
→
Ox,
→
OM)=ϕ
(r, ϕ) M r
ϕ
✟

✟
✟
✟
✟
✟
✟
✟
✟
✟
✟✯
M
0
O
s ✲
x
✻
y
x
0
y
0
r
ϕ
(x, y) (r, ϕ) M

x = r cos ϕ
y = r sin ϕ




r =

x
2
+ y
2
cos ϕ =
x
r
, sin ϕ =
y
r
(1, 0) (1, 1) (0, 1)
(−1, 1) (−1, 0)
ϕ ∈ R
r = r(ϕ),ϕ∈ Φ
C = {M(x, y):x = r(ϕ)cosϕ, y = r(ϕ)sinϕ, ϕ ∈ Φ}
r = a (a>0) O a
C ϕ
r
ϕ
r = ae
−kϕ
(a>0,k >0)
ϕ ∈ R
r

= −ake
−kϕ
< 0 r ϕ

✲
x
✻
r = a sin 3ϕ
R 2π/3
ϕ ∈ [0,
π
3
]
r

=3a cos 3ϕ, r

=0 ⇔ ϕ =
π
6
ϕ
0 π/6 π/3
r

+0−
r
0

a

0
✲
x0
a

f [a, b] F (x)
f [a, x] [x, x +∆x]
min
[x,x+∆x]
f.∆x ≤ F (x +∆x) −F(x) ≤ max
[x,x+∆x]
f.∆x
∆x → 0 F

(x)=f(x)
✲
x
✻
y
abxx+∆x
min
[x,x+∆x]
f
max
[x,x+∆x]
f
f F
x
f :(a, b) → R F
f
F

(x)=f(x) dF(x)=f

(x)dx , ∀x ∈ (a, b)

F G f (a, b) F −G =
f
f

f(x)dx