浙江大学2009数学分析
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b
1 n
4.
f (x) 3 [a, b] þëY,… min f (x) = 1. y²: lim
x∈[a,b] n→∞ a
dx = 1. (f (x))n
y². 1 = min f (x) = f (ξ ),
x∈[a,b]
Ø”
ξ ∈ (a, b)(ξ = a ½ ξ = b aq?Ø).
0
e−tx [f (x) − C ]dx .
(b) d lim f (x) = C •
x→∞
ε ∀ ε > 0, ∃ A > 0, s.t. x > A ⇒ |f (x) − C | < . 2 (c) q f 3 [0, A] þŒÈ, k., • M , u´éþã?¿ ε ∃δ= > 0, s.t. 2A(M + C )
1
ln x dx 1 + x2
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52 =
0
http://www.sciencenet.cn/u/zjzhang/48
1 1
ln x dx − 1 + x2
0
ln x dx 1 + x2
= 0.
(d)
D
(x + y )sgn(x − y )dxdy ,Ù¥D = [0, 1] × [0, 1]. )‰.
ε ε ≤ + = ε. 2 2
6. y² f (x) = ˜—ëY.
|sin x| 3 (0, 1) † (−1, 0) þþ˜—ëY, x
3 (−1, 0) ∪ (0, 1) þØ
y². (a) f (x) = ½Â g1 (x) =
|sin x| 3 (0, 1) † (−1, 0) þþ˜—ëY. x
π 2
r3 √ dr = 1 − r2
sin θdθ =
0 0
3
π 2
sin θ(1 − cos2 θ)dθ =
2 3
= πabc
4 4 1 + 2+ 2 2 3a 3b c
.
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52
http://www.sciencenet.cn/u/zjzhang/50
1 n
dx (f (x))n
dx (1 + ε)n
(2δ ) n = 1+ε 1 → , 1+ε dε ?¿5•
n→∞ b
1 n
n → ∞.
lim
a
dx = 1. (f (x))n
5P. •[y²I‡^ 5.
þe4•. sŒ±š~N´/ Ñ.
é?¿ a > 0, f (x) 3 [0, a] þiùŒÈ,… lim f (x) = C . y²:
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52
http://www.sciencenet.cn/u/zjzhang/47
úôŒÆ 2009 cêÆ©Ûë•)‰
Š Á ö ‡ Üy< ©‰Ñ úôŒÆ 2009 cêÆ©Û•ïÁK ˜‡ë•)‰. úôŒÆ êÆ©Û •ïÁK
'…c
x 2 y 2 z 2 c2 + 4 + 4 · a4 b c z c 1−
x2 a2
=8
x2 + y 2 ≤1 a 2 b2 x,y ≥0
− 1
y2 b2
x2 y 2 1 + 4 + 2 4 a b c
1−
dxdy
= 8abc
x2 +y 2 ≤1 x,y ≥0
1 − x2 − y 2
π 2
x2 y 2 1 + 2 + 2 (1 − x2 − y 2 ) dxdy 2 a b c 1 r(1 − r2 ) drdθ c2
b
cos f (x)dx ≤
a
2 . f (b)
y².
b f (b)
cos f (x)dx
a
=
f (a)
cos y ·
1 f (f −1 (y ))
dy
CþO† y = f (x)
f (b)
=
1 f (f −1 (ξ ))
f (a)
cos ydy
í2 È©1˜¥Š½n 1 ≤ |sin f (b) − sin f (a)| f (b) 2 ≤ . f (b)
, +
z2 c4
dS =8 f (x, y, z )
Σ
x2 + y 2 + z 2 =1 a2 b2 b2 x,y,z ≥0
x2 y 2 z 2 + 4 + 4 dS a4 b c x2 y 2 z 2 + 4 + 4 dxdy a4 b c x2 y 2 − 2 a2 b
=8
x2 + y 2 ≤1 a 2 b2 x,y ≥0
1 n 1 dx ≤ (b − a) n → 1, n (f (x))
(a) ˜•¡,
a b
n → ∞.
(b) ,˜•¡,é?¿ ½
ε > 0,
∃ δ < ε, s.t. |x − ξ | < δ ⇒ 1 ≤ f (x) < 1 + ε,
a
wenku.baidu.com
b
1 n
ξ +δ ≥
ξ −δ
1
x2
e 2 cos x − 1 e 2 (x cos x − sin x) = lim = lim 4 x→0 x→ 0 5x 20x3 1 x sin x 1 = lim = . 60 x→0 x2 60
+∞
(c)
0
ln x dx. 1 + x2 )‰.
1 ∞
ª =
0
ln x dx + 1 + x2
a
cos2
1 dx(ab = 0). x + b2 sin2 x
x
e 2 cos tdt − x (b) lim
0 x→0
t2
(ex − 1)2 (1 − cos2 x) arctan x
x
.
)‰. e 2 cos tdt − x ª = lim
0 x→0
x2 t2
x5
x2 x2 x · · 2 · 2 x (e − 1) 1 − cos x arctan x
1. OŽ (10 × 4 = 40 ) (a) a2 )‰. sec2 x cos2 x + sin2 x dx = dx ª = a2 + b2 tan2 x a2 cos2 x + b2 sin2 x b da tan x 1 a 1 a arctan tan x + C. = 2· · 2 = b a b ab b 1 + tan x
L: (x0 , y0 , z0 )
ƒ²¡• x0 y0 z0 (x − x0 ) + 2 (y − y0 ) + 2 (z − z0 ) = 0, 2 a b c
= y0 z0 x0 x + 2 y + 2 z = 1. 2 a b c u´ f (x, y, z ) =
x2 a4
1 +
y2 b4
1
= 8abc
0 1 0
√
r 1 − r2
r2 cos2 θ r2 sin2 θ + a2 b2
+
= 8abc
0
= 2abc
1 0
r3 π r3 π 1 π √ · + √ · + 2 r(1 − r2 ) · dr 2 2 2 2 2 a 1−r 4 b 1−r 4 c 1 1 2π 1 π + 2 · + 2· 2 a b 3 c 2
http://www.sciencenet.cn/u/zjzhang/52
(b) f (x) 3 (−1, 0) ∪ (0, 1) þؘ—ëY. ^‡y{. e f (x) 3 (−1, 0) ∪ (0, 1) þ˜—ëY, K ∀ ε > 0, ∃ δ > 0, s.t. AO/,
δ <x<0 −2
∃ δ > 0, s.t. 0 < |x − x0 | < δ ⇒
1 f (x) 3 < < , 4 x − x0 4
−δ < x − x0 < 0 ⇒ f (x) < 1 (x − x0 ) < 0, 4 0 < x − x0 < δ = f 3 x0 3.
1 ⇒ f (x) > 4 (x − x0 ) > 0.
x→∞ +∞ t→0+
lim t
0
e−tx f (x)dx = C.
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52
∞
http://www.sciencenet.cn/u/zjzhang/51
y². (a) d t
0
e−tx dx = 1
+∞
e
+∞
t
0
e
−tx
f (x)dx − C = t
1 x 1 y
ª =
0 1
dx
0 x
(x + y )dy −
0 1
dy
0 x
(x + y )dx
=
0
dx
0
(x + y )dy −
0
dx
0
(x + y )dy
= 0.
2. XJ f (x) 3 x0 4 Š. y². dK¿,
,+•SŒ
, … lim
x→ x0
1 f (x) = . y ² f ( x) 3 : x0 ? x − x0 2
|sin x| , x
x ∈ (0, 1], x = 0.
1,
g2 (x) =
|sin x| , x
x ∈ [−1, 0), x = 0.
−1,
=• g1 (x), g2 (x) ©O3 [0, 1],[−1, 0] þ˜—ëY, —ëY.
§‚ •› f (x) •˜
[pìŒÆêÆ,“ 2010 c11 ò17 Ï 47–52
Σ
x2 y 2 z 2 + 2 + 2 =1 a2 b c xdx ydy zdz + 2 + 2 = 0, a2 b c
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52 ý¥¡þ: (x0 , y0 , z0 ) ?
http://www.sciencenet.cn/u/zjzhang/49 {•þ• x0 y0 z0 , , , a2 b2 c2
A
ε > 0,
0<t<δ⇒ t
0
ε e−tx [f (x) − C ]dx < δ · A · (M + C ) < . 2
o( k
+∞
t
0
e−tx f (x)dx − C
+∞
= t
0 A
e−tx [f (x) − C ]dx
+∞
≤ t
0
e−tx [f (x) − C ]dx + t
A
e−tx [f (x) − C ]dx
†+•þ4~, m+•þ4O, u´ f 3 x0 ? : ý¥¡ Σ :
4 Š.
f (x, y, z ) L«l : p(x, y, z ) ? )‰. é ¦‡©k
x2 y 2 z 2 + 2 + 2 = 1(a > 0, b > 0, c > 0) þ a2 b c dS ƒ²¡ ål. ¦1˜.-¡È© . f (x, y, z )
x, x ∈ (−1, 0) ∪ (0, 1) |x − x | < δ
|f (x) − f (x )| < ε.
0<x < - δ → 0+ , k
δ 2
⇒ |f (x) − f (x )| < ε.
2 = |−1 − 1| < ε, ù´‡gñ!
7.
f (x) 3 [a, b]þ Œ
,
¼ê f (x) 3 [a, b] þüNeü, … f (b) > 0. y²:
1 n
4.
f (x) 3 [a, b] þëY,… min f (x) = 1. y²: lim
x∈[a,b] n→∞ a
dx = 1. (f (x))n
y². 1 = min f (x) = f (ξ ),
x∈[a,b]
Ø”
ξ ∈ (a, b)(ξ = a ½ ξ = b aq?Ø).
0
e−tx [f (x) − C ]dx .
(b) d lim f (x) = C •
x→∞
ε ∀ ε > 0, ∃ A > 0, s.t. x > A ⇒ |f (x) − C | < . 2 (c) q f 3 [0, A] þŒÈ, k., • M , u´éþã?¿ ε ∃δ= > 0, s.t. 2A(M + C )
1
ln x dx 1 + x2
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52 =
0
http://www.sciencenet.cn/u/zjzhang/48
1 1
ln x dx − 1 + x2
0
ln x dx 1 + x2
= 0.
(d)
D
(x + y )sgn(x − y )dxdy ,Ù¥D = [0, 1] × [0, 1]. )‰.
ε ε ≤ + = ε. 2 2
6. y² f (x) = ˜—ëY.
|sin x| 3 (0, 1) † (−1, 0) þþ˜—ëY, x
3 (−1, 0) ∪ (0, 1) þØ
y². (a) f (x) = ½Â g1 (x) =
|sin x| 3 (0, 1) † (−1, 0) þþ˜—ëY. x
π 2
r3 √ dr = 1 − r2
sin θdθ =
0 0
3
π 2
sin θ(1 − cos2 θ)dθ =
2 3
= πabc
4 4 1 + 2+ 2 2 3a 3b c
.
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52
http://www.sciencenet.cn/u/zjzhang/50
1 n
dx (f (x))n
dx (1 + ε)n
(2δ ) n = 1+ε 1 → , 1+ε dε ?¿5•
n→∞ b
1 n
n → ∞.
lim
a
dx = 1. (f (x))n
5P. •[y²I‡^ 5.
þe4•. sŒ±š~N´/ Ñ.
é?¿ a > 0, f (x) 3 [0, a] þiùŒÈ,… lim f (x) = C . y²:
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52
http://www.sciencenet.cn/u/zjzhang/47
úôŒÆ 2009 cêÆ©Ûë•)‰
Š Á ö ‡ Üy< ©‰Ñ úôŒÆ 2009 cêÆ©Û•ïÁK ˜‡ë•)‰. úôŒÆ êÆ©Û •ïÁK
'…c
x 2 y 2 z 2 c2 + 4 + 4 · a4 b c z c 1−
x2 a2
=8
x2 + y 2 ≤1 a 2 b2 x,y ≥0
− 1
y2 b2
x2 y 2 1 + 4 + 2 4 a b c
1−
dxdy
= 8abc
x2 +y 2 ≤1 x,y ≥0
1 − x2 − y 2
π 2
x2 y 2 1 + 2 + 2 (1 − x2 − y 2 ) dxdy 2 a b c 1 r(1 − r2 ) drdθ c2
b
cos f (x)dx ≤
a
2 . f (b)
y².
b f (b)
cos f (x)dx
a
=
f (a)
cos y ·
1 f (f −1 (y ))
dy
CþO† y = f (x)
f (b)
=
1 f (f −1 (ξ ))
f (a)
cos ydy
í2 È©1˜¥Š½n 1 ≤ |sin f (b) − sin f (a)| f (b) 2 ≤ . f (b)
, +
z2 c4
dS =8 f (x, y, z )
Σ
x2 + y 2 + z 2 =1 a2 b2 b2 x,y,z ≥0
x2 y 2 z 2 + 4 + 4 dS a4 b c x2 y 2 z 2 + 4 + 4 dxdy a4 b c x2 y 2 − 2 a2 b
=8
x2 + y 2 ≤1 a 2 b2 x,y ≥0
1 n 1 dx ≤ (b − a) n → 1, n (f (x))
(a) ˜•¡,
a b
n → ∞.
(b) ,˜•¡,é?¿ ½
ε > 0,
∃ δ < ε, s.t. |x − ξ | < δ ⇒ 1 ≤ f (x) < 1 + ε,
a
wenku.baidu.com
b
1 n
ξ +δ ≥
ξ −δ
1
x2
e 2 cos x − 1 e 2 (x cos x − sin x) = lim = lim 4 x→0 x→ 0 5x 20x3 1 x sin x 1 = lim = . 60 x→0 x2 60
+∞
(c)
0
ln x dx. 1 + x2 )‰.
1 ∞
ª =
0
ln x dx + 1 + x2
a
cos2
1 dx(ab = 0). x + b2 sin2 x
x
e 2 cos tdt − x (b) lim
0 x→0
t2
(ex − 1)2 (1 − cos2 x) arctan x
x
.
)‰. e 2 cos tdt − x ª = lim
0 x→0
x2 t2
x5
x2 x2 x · · 2 · 2 x (e − 1) 1 − cos x arctan x
1. OŽ (10 × 4 = 40 ) (a) a2 )‰. sec2 x cos2 x + sin2 x dx = dx ª = a2 + b2 tan2 x a2 cos2 x + b2 sin2 x b da tan x 1 a 1 a arctan tan x + C. = 2· · 2 = b a b ab b 1 + tan x
L: (x0 , y0 , z0 )
ƒ²¡• x0 y0 z0 (x − x0 ) + 2 (y − y0 ) + 2 (z − z0 ) = 0, 2 a b c
= y0 z0 x0 x + 2 y + 2 z = 1. 2 a b c u´ f (x, y, z ) =
x2 a4
1 +
y2 b4
1
= 8abc
0 1 0
√
r 1 − r2
r2 cos2 θ r2 sin2 θ + a2 b2
+
= 8abc
0
= 2abc
1 0
r3 π r3 π 1 π √ · + √ · + 2 r(1 − r2 ) · dr 2 2 2 2 2 a 1−r 4 b 1−r 4 c 1 1 2π 1 π + 2 · + 2· 2 a b 3 c 2
http://www.sciencenet.cn/u/zjzhang/52
(b) f (x) 3 (−1, 0) ∪ (0, 1) þؘ—ëY. ^‡y{. e f (x) 3 (−1, 0) ∪ (0, 1) þ˜—ëY, K ∀ ε > 0, ∃ δ > 0, s.t. AO/,
δ <x<0 −2
∃ δ > 0, s.t. 0 < |x − x0 | < δ ⇒
1 f (x) 3 < < , 4 x − x0 4
−δ < x − x0 < 0 ⇒ f (x) < 1 (x − x0 ) < 0, 4 0 < x − x0 < δ = f 3 x0 3.
1 ⇒ f (x) > 4 (x − x0 ) > 0.
x→∞ +∞ t→0+
lim t
0
e−tx f (x)dx = C.
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52
∞
http://www.sciencenet.cn/u/zjzhang/51
y². (a) d t
0
e−tx dx = 1
+∞
e
+∞
t
0
e
−tx
f (x)dx − C = t
1 x 1 y
ª =
0 1
dx
0 x
(x + y )dy −
0 1
dy
0 x
(x + y )dx
=
0
dx
0
(x + y )dy −
0
dx
0
(x + y )dy
= 0.
2. XJ f (x) 3 x0 4 Š. y². dK¿,
,+•SŒ
, … lim
x→ x0
1 f (x) = . y ² f ( x) 3 : x0 ? x − x0 2
|sin x| , x
x ∈ (0, 1], x = 0.
1,
g2 (x) =
|sin x| , x
x ∈ [−1, 0), x = 0.
−1,
=• g1 (x), g2 (x) ©O3 [0, 1],[−1, 0] þ˜—ëY, —ëY.
§‚ •› f (x) •˜
[pìŒÆêÆ,“ 2010 c11 ò17 Ï 47–52
Σ
x2 y 2 z 2 + 2 + 2 =1 a2 b c xdx ydy zdz + 2 + 2 = 0, a2 b c
[pìŒÆêÆ,“ 2010 c 1 1 ò 1 7 Ï 47–52 ý¥¡þ: (x0 , y0 , z0 ) ?
http://www.sciencenet.cn/u/zjzhang/49 {•þ• x0 y0 z0 , , , a2 b2 c2
A
ε > 0,
0<t<δ⇒ t
0
ε e−tx [f (x) − C ]dx < δ · A · (M + C ) < . 2
o( k
+∞
t
0
e−tx f (x)dx − C
+∞
= t
0 A
e−tx [f (x) − C ]dx
+∞
≤ t
0
e−tx [f (x) − C ]dx + t
A
e−tx [f (x) − C ]dx
†+•þ4~, m+•þ4O, u´ f 3 x0 ? : ý¥¡ Σ :
4 Š.
f (x, y, z ) L«l : p(x, y, z ) ? )‰. é ¦‡©k
x2 y 2 z 2 + 2 + 2 = 1(a > 0, b > 0, c > 0) þ a2 b c dS ƒ²¡ ål. ¦1˜.-¡È© . f (x, y, z )
x, x ∈ (−1, 0) ∪ (0, 1) |x − x | < δ
|f (x) − f (x )| < ε.
0<x < - δ → 0+ , k
δ 2
⇒ |f (x) − f (x )| < ε.
2 = |−1 − 1| < ε, ù´‡gñ!
7.
f (x) 3 [a, b]þ Œ
,
¼ê f (x) 3 [a, b] þüNeü, … f (b) > 0. y²: