2015考研数学极限必做100题
2015考研数学极限必做100题
1 如果lim x→x 0
f (x )存在,则下列极限一定存在的为
(A) lim x→x 0
[f (x )]α (B )lim x→x 0
|f (x )| (C )lim x→x 0
ln f (x ) (D )lim x→x 0
arcsin f (x )
2 设f (x )在x =0处可导,f (0)=0,则lim x→0
x 2f (x )?2f(x 3)
x =
(A )?2f ′(0) (B ?f ′
(0) (C )f ′(0) (D )
3.设f (x ),g (x )连续x →0时,f (x )和g (x )为同阶无穷小则x →0时,∫f (x ?t )?t x
0为 ∫xg (xt )?t 1
0的
(A )低阶无穷小 (B )高阶无穷小 (C )等价无穷小 (D )同阶无穷小
4.设正数列{a n } 满足lim n→∞
∫x n ?x a
n 0 =2 则lim n→∞
a n =
(A )2 (B )1 (C )0 (D )1
2 5.x →1时函数x 2?1
x?1?
1x?1
的极限为
(A )2 (B )0 (C )∞ (D )不存在,但不为∞
6.设f (x ) 在x =0的左右极限均存在则下列不成立的为
(A )lim x→0+
f (x ) = lim x→0
?f (?x ) (B ) lim x→0
f (x 2) = lim x→0
+f (x ) (C )lim x→0
f (|x |) = lim x→0
+f (x ) (D )lim x→0
f (x 3) = lim x→0
+f (x )
6. 极限lim
x→∞?sin
1
x ?1(1+1x )α
?(1+1x
)
=A ≠0的充要条件为
(A )α>1 (B )α≠1 (C )α>0 (D )和α无关
7.
.已知lim x→∞[x 21+x
?ax ?b ]=0,其中a,b 为常数则
a,b 的值为
(A )a =l ,b =1 (B )a =?1 ,b =1 (C )a =1,b =?1 (D a =?1,b =?1
8. 当x →0 时下列四个无穷小量中比其他三个更高阶的无穷小为
(A )x 2 (B )1?cos x (C )√1?x 2?1 (D )x ?tan x 9.已知x n+1=√x n y n ,y n+1=12(x n +y n ) ,x 1=a >0,y 1=b >0 (a
{x n }和{y n }
(A) 均收敛同一值(B )均收敛但不为同一值 (C )均发散 (D )无法判定敛散性 10. 设α>0,β≠0,lim x→∞
[(x
2α
+x
α)1
α
?x 2]=β则α,β为
11. 若 lim x→x 0
[f (x )+g (x )]存在,lim x→x 0
[f (x )?g (x )]不存在,则正确的为
(A )lim x→x 0f (x )不一定存在 (B )lim x→x 0
g (x )不一定存在
(C )lim x→x 0
[f 2(x )?g 2(x )] 必不存在 (D )lim x→x 0
f (x )不存在
12. 下列函数中在
[1,+∞)无界的为
(A)f (x )=x 2
sin 1
x 2 (B )f (x )=sin x 2
+2√x
(C )f (x )=x cos √x +x 2??x (D )f (x )=arctan 1x
x
13. 设f (x )连续lim x→0
f (x )
1?cos x =2且x →0
时∫f (t )?t sin 2 x 0为x 的n 阶无穷小则n
=
(A )3 (B )4 (C )5 (D )6
14. 当x →0时下列四个无穷小中比其他三个高阶的为
(A )tan x ?sin x (B )(1?cos x )ln (1+x ) (C )(1+sin x
)x
?
1 (D )∫arcsin t ?t x 2
15. 设[x ]表示不超过x 的最大整数,则y =x ?[x ]是
(A )无界函数 (B )单调函数 (C )偶函数 (D )周期函数
16. 极限lim x→∞
[x 2
(x?a )(x+b )]
x
=
(A )1 (B )? (C) ?a?b (D )?b?a
17. 函数
f (x )=
x 2?x x ?1
√1+
1x 的无穷间断点的个数为
(A ) 0 (B ) 1 (C ) 2 (D ) 3
18. 如果lim x→0
[1
x ?(1
x ?a)?x ]=1,则a=
(A ) 0 (B ) 1 (C ) 2 (D ) 3 19. 函数f (x )=x?x 3
sin πx 的可去间断点的个数为
(A ) 1 (B ) 2 (C ) 3 (D )无穷多个
20. 当x →0+时,与√x 等价的无穷小量是 (A ) 1-?√x (B ) ln
1?x
(C ) √1+√x ?1 (D ) 1?cos √x 21.设函数f (x )=
1
?x x?1?1
,则
(A ) x =0,x =1都是f (x )的第一类间断点 (B )x =0,x =1都是f (x )的第二类间断点
(C )x =0是f (x )的第一类间断点,x =1是f (x )的第二类间断点 (D )x =0是f (x )的第二类间断点,x =1是f (x )的第一类间断点 22 lim n→∞
ln √(1+1n )2(1+2n )2…(1+n n )2
n
等于
(A )∫ln 2 x ?x 2
1 (B) 2∫ln x ?x 21 (C) 2∫ ln (1+x )?x 21 (D) ∫ln
2 (1+x )?x 2
1 23.若lim
x→0
sin 6x+xf (x )
x =0,则lim
x→0
6+f (x )x 为
(A )0 (B )6 (C )36 (D )∞
24.对任意给定的ε∈(0,1),总存在正整数N ,当n ≥N 时,恒有“|x n ?a |≤2ε” 是数列收敛于a 的
(A )充分必要条件 (B )充分非必要条件
(C )必要非充分条件 (D )非充要条件 25.设函数f (x )=lim
n→∞1+x
1+x 2n
,讨论函数f (x )的间断点,其结论为
(A ) 不存在间断点 (B )存在间断点x =0
(C )存在间断点x =1 (D )存在间断点x =?1
26. .lim
n→∞[tan (π
4
+2
n
)]
n
=
27.x[sin ln (1+3
x
)?sin ln (1+1
x
)] =
28.已知lim
x→∞3xf(x)=lim
x→∞
[4f(x)+5]则lim
x→∞
xf(x)=
29.在[0,1]上函数f(x)=nx(1?x)n的最大值记为M(n)则lim
n→∞
M(n) =
30. 设k、L、δ>0则lim
x→0
[δk?x+(1?δ)L?x]?1x =
31.lim
x→+∞
arcsin (√x2+x?x) =
32. lim
x→0
∫(3sin t+t2cos 1
t
)?t
x
(1+cos x)∫ln (1+t)?t
x
=
33.lim
x→+∞(1+2x+3x)1
x+sin x =
34. α~β(x→a)则lim
x→a (β
α
)
β2
β2?α2 =
.lim x→0
∫tsin (x2?t2)?t
x
(1?cos x)ln (1+2x2)
=
35.lim
x→0+
(?x?1?x)1ln x =
36.f(x)有连续的导数f(0)=0,f′(0)=6,则lim
x→0
∫f(t)?t
x3
[∫f(t)?t
x
]
3
=
37.f(x)的周期T=3且f′(?1)=1,则lim
?→0
?
f(2?3?)?f(2)
=
38.lim
n→∞2n n!
n n
=
39.设f(x)在x=1连续且lim
x→1f(x)+x x?3
x?1
=?3,则f
′(1)=
40.极限p=∫lim
n→∞√2n+x2n n
2
?2
?x =
41.lim
x→0[1+tan x
1+sin x
]
1
x3 =
42.lim
x→+∞
(ln x)1x?1 =
43.x→0时f(x)=?x?1+ax
1+bx
为x的3阶无穷小则a=,b =
44. 极限lim
x→?√4x2+x?1+x+1
√2
=
45.lim
n→∞(1?1
2
)(1?1
3
)?(1?1
n
) =
46.lim
x→+∞(√x6+x5
6?√x6?x5
6) =
47. f′′(x)存在f(0)=f′(0)=0,f′′(x)>0,u(x)为曲线f(x)在(x,f(x))处切
线在x轴的截距则lim
x→0
x
u(x)
=
48. a>0,bc≠0,lim
x→+∞[x a ln (1+b
x
)?x] =c (c≠0)则a= b= c=
49.lim
n→∞
sin (√n2+1π) =
50.已知x→0时x?(a+bcos x)sin x为x的5阶无穷小则a = ,b=
lim x→0[ (1+x)
1
x
?
]
1
x
=
35.lim
x→+∞∫|sin t|?t
x
x
=
36.f(x)可导对于?x∈(?∞,+∞)有|f(x)|≤x2则f′(0)=
37.lim
n→∞∫x n
1+x
?x
1
=
38.如果lim
x→∞(1+x
x
)
ax
=∫t?t?t
a
?∞
则a=
39.设x→1+时2?2x?1ln x与(x?1)n为同阶无穷小则n=
40 .lim
x→+∞
?x
(1+1
x
)
x2
=
41.lim
x→0ln (sin2 x+?x)?x
ln (x+?)?2x
=
42. |x|<1时lim
n→∞
(1+x)(1+x2)?(1+x2n)=
43. 设极限lim
x→+∞
[(x5+7x4+2)a?x]=b(b≠0)则a= b =
44. lim
x→∞[x?x2ln (1+1
x
)] =
45. w =
lim x→0
[
1
(√)?1ln (1+x
)
]
= 46. 设y =y (x )由y 2+xy +x 2?x =0确定满足y (1)=?1的连续函数
则lim x→1
(x?1)2
y (x )+1 =
47 .设a 1,a 2…a m 为正数(m ≥
2)则lim n→∞
(a 1
n
+
a 2
n +?+
a m n )1
n =
48. f (x )连续x →0时F (x )=∫(x 2+1?cos t )f (t )?t x
0为x 3的等价无穷小 则f (0)= 49. f (x )连续 f (0)=0,
f ′(
0)≠0则lim
x→0∫
f(x 2?t)?t
x 2
x
3∫f (xt )?t 10
=
50. f (x )=∫
sin (xt )t
?t x
x 2则lim
x→0f (x )
x 2
=
51. 极限lim x→∞
x 2
[ a
1x+1
?a 1
x
] =
52. 已知f (x )在x =a 可导f (x )>0 ,n ∈N ,f (a )=1,f ′(a )=2
则极限lim n→∞
[
f(a+1n
)f (a ) ]n
=
53. lim x→0
(cot 2 x ?1
x 2)=
54. lim
x→1lncos
(x?1)1?sin π
2x
=
55. 如果lim x→?∞
(√x 2+x +1+ax +b)=0 则a = b =
56. lim x→0
(arcsin x x )1
1?cos x
=
57. 已知曲线y =f (x )在点(0,0)处切线经过点(1,2)则极限 lim x→0
[cos x +
∫f (t )?t x 0]1
x 2
=
58. 已知f (x )在x =0邻域内可导且lim x→0
[sin x x 2
+
f (x )x
]=2 则f (0)=
f ′(0)= lim x→0x
f (x )+? =
59. lim
x→0
√1+tan x?√1+sin x xln
(x+1)?x 2 =
60 lim
x→1
ln x ln(1?x)=
61. lim
n→∞[1
2
+3
22
+5
23
+? +2n?1
2n
] =
62. lim
x→0[a
x
?(1
x2
?a2)ln(1+ax)] = (a≠0)
63 .lim
x→0?
1
x+1
?
1
x?1
arctan1
x
=
64.设f(x)在[a,b]连续则lim
n→+∞∫x n f(x)?x
1
=
65. w=lim
x→0arcsin x?sin x
arctan x?tan x
=
66 . lim
x→0(x+3)x?3x
x2
=
67 .lim
x→+∞1
x
∫(1+t2)?t2 ?x2?t
x
=
68. lim
x→0?2?(x+1)
2
x
x
=
69. lim
x→0
2
√1+xsin x?cos x
=
70. lim
n→∞[(1+12
n2
)(1+22
n2
)+?+(1+n2
n2
)]
1
n =
71. 设x n=1
n2+1+2
n2+22
+…+n
n2+n2
则lim
n→+∞
x n=
72 .P=lim
x→0
[ln (1+?
2
x)
ln (1+?
1
x)
+a[x]
]
存在求p及a的值.
73.lim
x→+∞∫(1+t2)?t2?t
x
x?x2
=
74. lim
x→0[ 1
ln (1+x2)
?1
sin2 x
] =
75. lim
x→+∞
(x+?x)1x =
76. lim
x→1
x?x x
1?x+ln x
=
77. lim
n→∞1.3.5.7…(2n?1)
2.4.6.8…(2n)
=
78. lim
n→∞1
n
√n(n?1)?(2n?1)
n =
79. 极限lim
x→0(1?√cos x)(1?√cos x
3)…(1?√cos x
n)
(1?cos x)n?1
=
80. 设f (x )一阶连续可导且f (0)=0,f
′(0)=1则下列极限lim x→0
[1+f (x )]
1arcsin x
=
81. 函数f (x )满足f (0)=0 ,f ′(0)>0则极限lim x→0
+
x f (x )
= 82. lim x→+∞
[x +√1+x 2]2
x
=
83. lim x→+∞[ π
2?arctan x ]
1ln x
=
84. lim
x→0
1?cosx √cos 2x √cos3x
3
x 2
=
85. 函数f (x )=
x ln
|1?x |的第一类间断点的个数为
86. lim x→0
(cot x )2sin x = 87.lim
x→+∞
?x ?2
π
xarctan x
x+? =
88.lim n→∞(
1√2+1√n +22+?+1√22) = 89. lim x→+∞ x 2[lnarctan (x +1)?lnarctanx ] =
90. lim x→+∞
x 32
(√x +2?2√x +1+√x) = 91 设x ≠0时 lim n→∞
cos x
2cos x
4…cos x
2n =
92极限w =lim x→+∞1+2|x |
1+x
arctan x =
93. lim
x→0tan x+(1?cos x )
ln (1?2x )+(1??x 2
)
=
94 f (x )=arcsin x 在[0,b ]上用拉格朗日中值定理且中值为ε则lim b→0
ε
b =
95 已知曲线y =f (x )与y =sin x 在(0,0)处相切则lim n→∞
[ 1+f (2
n ) ]n
=
96 lim n→∞(
1
n +n+1
+
2n +n+2
+?+
n n +n+n
) =
97 lim x→+∞
(
a 1
x +b 1
x +c 1
x
3
)x
= 98 极限lim x→0
(1+x )1
x ??
x
=
99.设f (x ) 在x =1处可导且在(1,f (1))处的切线方程为y =x ?1,
求极限P = lim
x→0∫?t f(1+?x2??t)?t x2
x2ln cos x
100.如果lim
x→+∞
(x n+7x4+1)m?x=b(n>4 ,b≠0)求m,n及b的值