浙江大学数学分析历年考研试题

|bn|
=
∞;
1 n−1
2
|bn| i=1 bi+1 − bi
k..
n1
y²: e lim an+1 − an •3, K lim an ••3.
n→∞ bn+1 − bn
n→∞ bn
18
14. úôŒÆ 2011 cïÄ)\Æ•ÁÁKêÆ©Û
Βιβλιοθήκη Baidu
1. (z K 10 ©, 60 ©) OŽK
tan x − sin x
x
.
+∞ ln x
(3)
0
1 + x2 dx.
(4) (x + y)sgn(x − y)dxdy. Ù¥ D = [0, 1] × [0, 1].
D
2. ( 15 ©) XJ f (x) 3 x0
,
•SŒ
,
… lim
x→x0
f (x) x − x0
=
1 .
2
y²: f (x)
3: x0
?
4
Š.
3. ( 15 ©) f (x, y, z) Ll : O(0, 0, 0) ²¡ ål. ¦1˜a-¡È©:
5. ( 20 ©) é?¿ a > 0, f (x) 3 [0, a] þiùŒÈ, … lim f (x) = C. y²:
x→+∞
+∞
lim t
e−txf (x)dx = C.
t→0+ 0
6. ( 20 ©) y² f (x) = | sin x| 3 (0, 1) † (−1, 0) þþ˜—ëY, 3 (−1, 0) ∪ (0, 1) ¥Ø˜—ëY. (5: ¡ x
Σ
ù8‡²¡¤Œá•NL¡, {•••á•NL¡ ý.
1 ln x
(5)
dx.
0 1−x
1 arctan x
(6)
√
dx.
0 x 1 − x2
2. ( 15 ©) ¼ê
xy
f (x, y) =
, x2 + y2
0,
(x, y) = (0, 0); (x, y) = (0, 0).
Áy: f (x, y) 3²¡ R2 þëY, ê fx(x, y), fy(x, y) k., f (x, y) 3 : (0, 0) ?ØŒ‡.
b
cos f (x)dx
a
2 .
f (b)
17
13. úôŒÆ 2010 cïÄ)\Æ•ÁÁKêÆ©Û
1. (z K 10 ©, 60 ©) OŽe
(n+1)2
1
(1) lim
√.
n→∞ k=n2
k
4•ÚÈ©
(2)
y sin(xy)dxdy.
[0,π]×[0,1]
ex sin x − x(1 + x)
ý
¡Σ:
x2 a2
+
y2 b2
+
z2 c2
=
1 (a, b, c
>
0)
þ: P (x, y, z)
?
ƒ
ds .
f (x, y, z)
Σ
4. ( 20 ©) f (x) 3 [a, b] þëY, … min f (x) = 1. y²:
x∈[a,b]
lim
n→∞
1
b dx
n
a (f (x))n = 1.
(3) lim
x→0
sin3 x
.
(4) OŽ zdxdy. Ù¥ Σ ´n / {(x, y, z) : x, y, z
Σ
2π √
(5)
1 + sin xdx.
0
1 ln(1 + x)
(6)
0
1 + x2 dx.
2. ( 15 ©) an = sin an−1, n
2, … a1 > 0, OŽ:
3. ( 15 ©) f (x) ´ [a, a + 1] ( a •~ê) þ ëY Š¼ê, P M = max f (x). y²:
x∈[a,a+1]
a+1
An = n
f (x) ndx
a
'u n üN4O, …
a+1
lim n
f (x) ndx = M.
n→∞
a
4. ( 15 ©)
shx
·
shy
+∞ sin xy
dy
0
y
5. ( 20 ©)
f (x) ëY, y² Poisson úª:
1
f (ax + by + cz)dS = 2π f
−1 x2 +y 2 +z 2 =1
a2 + b2 + c2t dt.
6. ( 20 ©) {an}n 1, {bn}n 1 •¢êê , ÷v
1
lim
n→∞
=
1,
Ù¥
shx
=
ex
−
e−x .
OŽ
2
+∞
y(x)dx.
0
5. ( 15 ©) ?Ø?ê 3 (−∞, +∞) þ Âñ5Ú˜—Âñ5.
∞ (−1)n (1 + x2)n
n=1
6. ( 15 ©) a1, b1 •?¿À½ ¢ê, an, bn ½Â•:
1
an = max{bn−1, x}dx,
01
y = f (x) 38Ü D (D ⊂ R) þ˜—ëY´•: é ∀ ε > 0, •3 δ > 0, ¦ é x , x ∈ D, |x − x | < δ ž, k f (x ) − f (x ) < ε)
7. ( 20 ©) f (x) 3 [a, b] þŒ , ¼ê f (x) 3 [a, b] þüNeü, … f (b) > 0. y²:
úôŒÆ 2009 − 2020 cýK®o
12. úôŒÆ 2009 cïÄ)\Æ•ÁÁKêÆ©Û
1. (z K 10 ©, 40 ©) OŽ 1
(1) a2 cos2 x + b2 sin2 x dx (ab = 0).
(2)
lim
x→0
(ex
−
e x t2 2
0
cos tdt − x
1)2(1 − cos2 x) arctan
0, x + y + z = 1}, Ù{••† (1, 1, 1) ••ƒÓ.
n
lim
n→∞
3 an.
3. ( 15 ©)
¼ê
f (x)
3
(−∞, +∞)
þëY,
n
•Ûê.
y²:
e
lim
x→+∞
f (x) xn
=
lim
x→−∞
f (x) xn
=
1,
K•§
f (x) + xn = 0
k¢Š.
4. ( 20 ©) y² 3 [δ, +∞) þ˜—Âñ (Ù¥ δ > 0).
bn = min{an−1, x}dx,
0
√
√
Áy:
lim
n→∞
an
=
2
−
2,
lim
n→∞
bn
=
2 − 1.
n = 2, 3, · · · n = 2, 3, · · ·
19
7. ( 15 ©) a1 ∈ (0, 1), … an+1 = an(1 − an), n 1, y²: lim nan = 1.
(1) lim
x→0
sin x3
.
(2)
[x + y]dxdy, Ù¥ [α] L« α êÜ©.
[0,2]×[0,2]
(3) F (x) =
x2
sin xy dy,
x
>
0,
¦
F
(x).
x
y
(4) OŽ y(x − z)dydz + x2dzdx + (y2 + xz)dxdy. Ù¥ Σ ´ x = 0, y = 0, z = 0, x = a, y = a, z = a