最后一次串讲（修改于1月4号）

Tiao Lu
Email: tlu@math.pku.edu.cn
School of Mathematical Sciences, Peking University
2006-12-27
1 /18

I
n → +∞, x → x0, x → +∞,
1
n = 1
103
limx→0+ xx
1
= 1 ⇒ limn→+∞ n
2 /18

I
n → +∞, x → x0, x → +∞,
1
n = 1
103
limx→0+ xx
1
= 1 ⇒ limn→+∞ n
2 /18

I
n → +∞, x → x0, x → +∞,
1
n = 1
103
limx→0+ xx
1
= 1 ⇒ limn→+∞ n
2 /18

I
n → +∞, x → x0, x → +∞,
1
n = 1
103
limx→0+ xx
1
= 1 ⇒ limn→+∞ n
2 /18

I
n → +∞, x → x0, x → +∞,
1
n = 1
103
limx→0+ xx
1
= 1 ⇒ limn→+∞ n
2 /18

II
sin x
limx→0 x
limx→∞
= 1
1 x
1 + = e
4224 lim
sinx ∼ x, tanx ∼ x,
x
tanx − sinx
limx → 0
x3 tan x−sinx
x3 ,:
= lim
x→0
x − x
= 0.
x3 3 /18

II
sin x
limx→0 x
limx→∞
= 1
1 x
1 + = e
4224 lim
sinx ∼ x, tanx ∼ x,
x
tanx − sinx
limx → 0
x3 tan x−sinx
x3 ,:
= lim
x→0
x − x
= 0.
x3 3 /18

II
sin x
limx→0 x
limx→∞
= 1
1 x
1 + = e
4224 lim
sinx ∼ x, tanx ∼ x,
x
tanx − sinx
limx → 0
x3 tan x−sinx
x3 ,:
= lim
x→0
x − x
= 0.
x3 3 /18

II
sin x
limx→0 x
limx→∞
= 1
1 x
1 + = e
4224 lim
sinx ∼ x, tanx ∼ x,
x
tanx − sinx
limx → 0
x3 tan x−sinx
x3 ,:
= lim
x→0
x − x
= 0.
x3 3 /18

II
sin x
limx→0 x
limx→∞
= 1
1 x
1 + = e
4224 lim
sinx ∼ x, tanx ∼ x,
x
tanx − sinx
limx → 0
x3 tan x−sinx
x3 ,:
= lim
x→0
x − x
= 0.
x3 3 /18

II
sin x
limx→0 x
limx→∞
= 1
1 x
1 + = e
4224 lim
sinx ∼ x, tanx ∼ x,
x
tanx − sinx
limx → 0
x3 tan x−sinx
x3 ,:
= lim
x→0
x − x
= 0.
x3 3 /18

Problem
limx→1x 1
x−1
t = x − 1
limx→1x 1
x−1 = limt→0(1 + t) 1
t = e
4 /18

Problem
limx→1x 1
x−1
t = x − 1
limx→1x 1
x−1 = limt→0(1 + t) 1
t = e
4 /18

Problem
limx→0 1 1
x − ex −1
∞ − ∞
∞
∞
∞
∞
1 1
lim −
x→0 x ex e
= lim
− 1 x→0
x − 1 − x
x(ex − 1)
lim
x→0
= lim
x→0
e x − 1
e x − 1 + xe x
ex ex + ex = lim
+ xex x→0
1
2 + x
= 1
2
5 /18

Problem
limx→0 1 1
x − ex −1
∞ − ∞
∞
∞
∞
∞
1 1
lim −
x→0 x ex e
= lim
− 1 x→0
x − 1 − x
x(ex − 1)
lim
x→0
= lim
x→0
e x − 1
e x − 1 + xe x
ex ex + ex = lim
+ xex x→0
1
2 + x
= 1
2
5 /18

Problem
limx→0 1 1
x − ex −1
∞ − ∞
∞
∞
∞
∞
1 1
lim −
x→0 x ex e
= lim
− 1 x→0
x − 1 − x
x(ex − 1)
lim
x→0
= lim
x→0
e x − 1
e x − 1 + xe x
ex ex + ex = lim
+ xex x→0
1
2 + x
= 1
2
5 /18

Problem
limx→0 1 1
x − ex −1
∞ − ∞
∞
∞
∞
∞
1 1
lim −
x→0 x ex e
= lim
− 1 x→0
x − 1 − x
x(ex − 1)
lim
x→0
= lim
x→0
e x − 1
e x − 1 + xe x
ex ex + ex = lim
+ xex x→0
1
2 + x
= 1
2
5 /18

Problem
limx→0 1 1
x − ex −1
∞ − ∞
∞
∞
∞
∞
1 1
lim −
x→0 x ex e
= lim
− 1 x→0
x − 1 − x
x(ex − 1)
lim
x→0
= lim
x→0
e x − 1
e x − 1 + xe x
ex ex + ex = lim
+ xex x→0
1
2 + x
= 1
2
5 /18

Problem
limx→0 1 1
x − ex −1
∞ − ∞
∞
∞
∞
∞
1 1
lim −
x→0 x ex e
= lim
− 1 x→0
x − 1 − x
x(ex − 1)
lim
x→0
= lim
x→0
e x − 1
e x − 1 + xe x
ex ex + ex = lim
+ xex x→0
1
2 + x
= 1
2
5 /18

1 f (x) =
2 1 x = 1
x−1 +1
(A)(B)
(C)(D)
6 /18

1 f (x) =
2 1 x = 1
x−1 +1
(A)(B)
(C)(D)
6 /18

f ′ (−x)f ′ (x)−x.
f (−x)8811
7 /18

f ′ (−x)f ′ (x)−x.
f (−x)8811
7 /18

f ′ (−x)f ′ (x)−x.
f (−x)8811
7 /18

f ′ (−x)f ′ (x)−x.
f (−x)8811
7 /18

f ′ (−x)f ′ (x)−x.
f (−x)8811
7 /18

f ′ (−x)f ′ (x)−x.
f (−x)8811
7 /18

f ′ (−x)f ′ (x)−x.
f (−x)8811
7 /18

y = f (x)f (x) = f (0) + x + α(x),limx→0 α(x)
x = 0,
f ′ (0) =
f ′ f (x) − f (0) x + α(x) x
(0) = lim = lim = lim
x→00 x − 0 x→0 x x→0 x +lim
α(x)
= 1
x→0 x
8 /18

y = f (x)f (x) = f (0) + x + α(x),limx→0 α(x)
x = 0,
f ′ (0) =
f ′ f (x) − f (0) x + α(x) x
(0) = lim = lim = lim
x→00 x − 0 x→0 x x→0 x +lim
α(x)
= 1
x→0 x
8 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

” ∞
∞ ”, ”0
0 ”, ”∞ − ∞”,
”1∞ ”
,
f ′′ (x) < 0,
f ′′ (x) > 0,
9 /18

Problem
: ln(1 + x) < x
√ 1+x x ∈ (0,1)
f (x) = ln(1 + x) − x
√ 1+x f (0) = 0
f ′ (x) = − 2 + x − 2√x + 1
2(x + 1) 3
2
2 + x − 2 √ x + 1 > 0 if x ∈ (0,1)
.f ′ (x) < 0x ∈ (0,1)
ξ ∈ (0,x)x ∈ (0,1)
f (x) = f ′ (ξ)x
f (x) < 0.
10 /18

Problem
: ln(1 + x) < x
√ 1+x x ∈ (0,1)
f (x) = ln(1 + x) − x
√ 1+x f (0) = 0
f ′ (x) = − 2 + x − 2√x + 1
2(x + 1) 3
2
2 + x − 2 √ x + 1 > 0 if x ∈ (0,1)
.f ′ (x) < 0x ∈ (0,1)
ξ ∈ (0,x)x ∈ (0,1)
f (x) = f ′ (ξ)x
f (x) < 0.
10 /18

Problem
: ln(1 + x) < x
√ 1+x x ∈ (0,1)
f (x) = ln(1 + x) − x
√ 1+x f (0) = 0
f ′ (x) = − 2 + x − 2√x + 1
2(x + 1) 3
2
2 + x − 2 √ x + 1 > 0 if x ∈ (0,1)
.f ′ (x) < 0x ∈ (0,1)
ξ ∈ (0,x)x ∈ (0,1)
f (x) = f ′ (ξ)x
f (x) < 0.
10 /18

Problem
: sinx > 2
π
πxx ∈ (0, 2 )
f (x) = sinx − 2
π x
f (0) = f ( π
2
f ′′
) = 0 .f (π
4
) =
√ 2
2
− 1
2
> 0
(x) = − sinx < 0 ξ ∈ (0, π
2 )
f (ξ) = 0
11 /18

Problem
: sinx > 2
π
πxx ∈ (0, 2 )
f (x) = sinx − 2
π x
f (0) = f ( π
2
f ′′
) = 0 .f (π
4
) =
√ 2
2
− 1
2
> 0
(x) = − sinx < 0 ξ ∈ (0, π
2 )
f (ξ) = 0
11 /18

I
cf (x)dx = c f (x)dx, c
(f (x) ± g(x)) dx = f (x)dx ± g(x)dx
b
a
b
f (x)dx = F(x)|ba
= F(b) − F(a),F (x) = f (x)dx
a cf (x)dx = c b
a f (x)dx, c
b
a (f (x) ± g(x)) dx = b
a f (x)dx ± b
a g(x)dx
a
a f (x)dx = 0
b
a f (x)dx = − a
b f (x)dx
b
a f (x)dx = c
a f (x)dx + b
c f (x)dx
b
a cdx = c(b − a)
f (x) ≥ 0 (a ≤ x ≤ b) ⇒ b
a f (x)dx ≥ 0
f (x) ≥ g(x) (a ≤ x ≤ b) ⇒ b
a f (x)dx ≥ b
a g(x)dx
12 /18

e.g.
b
a f (x)dx = b
a f (t)dt
u-
b
a f (x)dx = F(x)|b a
13 /18

D, 1dxdy =D
14 /18

Problem
a
0
√ a 2 − x 2 dx
a x a 2 − x2dx = a2 − x2 +
2
2
2 sin−1 u
+ C
a
a √
0 a2 − x2dx = a
√
2 a2 − a2 a + 2
2 sin−1 √
a 0
a − 2 a2 − 02 a + 2
2 sin−1
0 πa
a = 2
4
15 /18

Problem
1
1+ 3√ 1+x dx
u = 3√ 1 + x x = u 3 − 1 dx = 3u 2 du
1
1 + 3√
3u
dx =
1 + x 2
1 + u du
3(v + 1) 2
= dv
v
v = u + 1
16 /18

Problem
1
1+ 3√ 1+x dx
u = 3√ 1 + x x = u 3 − 1 dx = 3u 2 du
1
1 + 3√
3u
dx =
1 + x 2
1 + u du
3(v + 1) 2
= dv
v
v = u + 1
16 /18

Problem
a, bf (x) = x3 + ax2 + bxx = 1−2
a =?, b =?, b
a ex2f
(x)dx =?
−2
a = −2, b = −1
u = x2 du = 2xdx
e x2
x 3 dx =
e u u 1
2 du
17 /18

Problem
a, bf (x) = x3 + ax2 + bxx = 1−2
a =?, b =?, b
a ex2f
(x)dx =?
−2
a = −2, b = −1
u = x2 du = 2xdx
e x2
x 3 dx =
e u u 1
2 du
17 /18

Problem
a, bf (x) = x3 + ax2 + bxx = 1−2
a =?, b =?, b
a ex2f
(x)dx =?
−2
a = −2, b = −1
u = x2 du = 2xdx
e x2
x 3 dx =
e u u 1
2 du
17 /18

ln(x + 1) <
f (x) = ln(x + 1) − x(2x+1)
(x+1) 2
f ′ (x) = − x(1−x)
(x+1) 3
x(2x + 1)
(x + 1) 2
(x ∈ (0,1))
18 /18

ln(x + 1) <
f (x) = ln(x + 1) − x(2x+1)
(x+1) 2
f ′ (x) = − x(1−x)
(x+1) 3
x(2x + 1)
(x + 1) 2
(x ∈ (0,1))
18 /18