大连理工大学2020年数学分析考研真题
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x→+∞
8. Á¦
lim
x→1
∞
xn 3n
cos(nπx3).
n=0
9. OŽ
1x
lim 2x 1 + − e .
x→+∞
x
+∞
10. ä
(−1)[x]dx
1
. OŽK.
ñÑ5.
1. OŽ -È©
x−y
e x+y dxdy.
D
Ù¥ D = {(x, y) | x ≥ 0, y ≥ 0, x + y ≥ 1}.
2. a ´knê, …š ê, ¦ f (x) = cos ax 3 [−π, π) Fp“?ê.
n. {‰K.
1. f (x, u) 3 [0, +∞) × [a, b] þëY, é ∀u ∈ [a, b),
+∞
y²:
f (x, u)dx 3 [a, b) þؘ—Âñ.
0
+∞
f (x, u)dx Âñ,
19. ŒënóŒÆ 2020 cêÆ©Û考研真题
˜. ( 6 × 10 = 60 ©) y²K.
1. e h(x) 3 [0, 1] þëY, … h(0) = 0, y²:
3 [0, 1] þ˜—Âñ.
x
gn(x) = h(sn)ds
0
2. ®• g ∈ C[0, +∞), lim g(x) = c ∈ R, ÁFra Baidu bibliotek:
5. f (x) ëYŒ‡, J • f (x) Jacobi Ý , Áy: f (x) ´à¼ê ¿©7‡^‡´é ∀a, Ñk
f (x) ≥ f (a) + Jf (a)f (a).
6.
L:
x2 y2 +
= 1, Á¦:
25 16
y3dx − xy2dy L (x2 + y2)2 .
7. e f : Rn → R, lim f (x) = A •3…•k•ê, ¦y: f (x) 3 Rn þ˜—ëY.
x→+∞
+∞ g(x)
π
lim n
n→∞ 0
n2
+ x2 dx
=
c. 2
3. r > 1, {an} ´ 4. ®•
‘?ê, ÷v n ln an > r, Áy: an+1
∞
an Âñ.
n=1
ex + f (x, y) = u2;
ey − f (x, y) = v2.
… f (0, 0) = 0, Áy: 3 (1, 1) •S (u, v) •˜(½ (x, y).
0
+∞
f (x, b)dx uÑ,
0
8. Á¦
lim
x→1
∞
xn 3n
cos(nπx3).
n=0
9. OŽ
1x
lim 2x 1 + − e .
x→+∞
x
+∞
10. ä
(−1)[x]dx
1
. OŽK.
ñÑ5.
1. OŽ -È©
x−y
e x+y dxdy.
D
Ù¥ D = {(x, y) | x ≥ 0, y ≥ 0, x + y ≥ 1}.
2. a ´knê, …š ê, ¦ f (x) = cos ax 3 [−π, π) Fp“?ê.
n. {‰K.
1. f (x, u) 3 [0, +∞) × [a, b] þëY, é ∀u ∈ [a, b),
+∞
y²:
f (x, u)dx 3 [a, b) þؘ—Âñ.
0
+∞
f (x, u)dx Âñ,
19. ŒënóŒÆ 2020 cêÆ©Û考研真题
˜. ( 6 × 10 = 60 ©) y²K.
1. e h(x) 3 [0, 1] þëY, … h(0) = 0, y²:
3 [0, 1] þ˜—Âñ.
x
gn(x) = h(sn)ds
0
2. ®• g ∈ C[0, +∞), lim g(x) = c ∈ R, ÁFra Baidu bibliotek:
5. f (x) ëYŒ‡, J • f (x) Jacobi Ý , Áy: f (x) ´à¼ê ¿©7‡^‡´é ∀a, Ñk
f (x) ≥ f (a) + Jf (a)f (a).
6.
L:
x2 y2 +
= 1, Á¦:
25 16
y3dx − xy2dy L (x2 + y2)2 .
7. e f : Rn → R, lim f (x) = A •3…•k•ê, ¦y: f (x) 3 Rn þ˜—ëY.
x→+∞
+∞ g(x)
π
lim n
n→∞ 0
n2
+ x2 dx
=
c. 2
3. r > 1, {an} ´ 4. ®•
‘?ê, ÷v n ln an > r, Áy: an+1
∞
an Âñ.
n=1
ex + f (x, y) = u2;
ey − f (x, y) = v2.
… f (0, 0) = 0, Áy: 3 (1, 1) •S (u, v) •˜(½ (x, y).
0
+∞
f (x, b)dx uÑ,
0