极限必做150题一刘刈4.docx
- 文档编号:6745870
- 上传时间:2023-01-09
- 格式:DOCX
- 页数:21
- 大小:64.43KB
极限必做150题一刘刈4.docx
《极限必做150题一刘刈4.docx》由会员分享,可在线阅读,更多相关《极限必做150题一刘刈4.docx(21页珍藏版)》请在冰豆网上搜索。
极限必做150题一刘刈4
极限必做150解答
刘刈
17级
刈
言:
我的个人解答,经过老司机(柯)检查,如果还有错误,欢迎大佬联系我及时改正,我的QQ:
198924030
111
1.
lim(
x→0xsinx
-)
tanx
=limtanx-sinx=limtanx(1-cosx)=1
x→0x3x→0x32
2.limln(a+x)+ln(a-x)-2lna
x→0
x2
1-1-
=lima+xa-x
=lim
2x=-1
x→02x
x→02x(a+x)(a-x)a2
3.lim
x→0
1-cosx
=lim=
x→012
2
4.lim
x→a
=lim
x-a+lim
=lim
x-a
+1=1
x→a
x2-a2
x→ax→a
5.
lim
x→0
=lim
x→0
-
lim=1
x→0
6.limtanmx(m、n为正整数)
x→0sinnx
=limmx=m
x→0nxn
7.lim
x→0
ln(1+x+x2)+ln(1-x+x2)secx-cosx
=lim
x→0
ln⎡⎣(x2+1)2-x2⎤⎦
secx(1-cos2x)
=lim
x→0
x4+x2
x21
1ex+e2x+..+enx
8.limln()
x→0xn
=1⎛ex-1
e2x-1
enx-1⎫12nn+1
limç
++..+
⎪=++..+=
x→0x⎝
nnn
⎭nnn2
9.limsin
n→∞
n2+a2π)
=n22
na2
lim(-1)sin(n+aπ-nπ)=lim(-1)sinπ=0
n→∞
n→∞
n2+a2
⎛3n2-2⎫n(n+1)
10.limç2
n→∞⎝3n
+4⎪
⎭
6
n(n+1)
lim-6n(n+1)
=lim⎛1-
n→∞
⎫
3n2+1⎪
=en→∞
3n2+1
=e-2
⎝⎭
2n+1n
11.lim
n→∞⎝2n-1⎭
=lim⎛1+
2⎫n
⎪
lim2n
=en→∞2n-1=e
n→∞⎝
2n-1⎭
⎛⎫n
12.limç⎪
n→∞⎝2⎭
⎛na-1nb-1⎫
limn+
lnalnbab
n→∞ç22⎪
limn(+)+
=e⎝⎭=en→∞nn=e22=
⎡(2+1)(2-1)⎤
13.limn2⎢en+en-2e2⎥
n→∞⎣⎦
令1=tn
=lim
t→0
e(2+t)+e(2-t)-2e2
t2
=lim
t→0
e(2+t)-e(2-t)
2t
=lim
t→0
e(2+t)+e(2-t)=2
2
⎡1⎤
14.limn⎢an-1⎥(a为整数)
n→∞⎣⎦
=limnlna=lna
n→∞n
n
15.lim
n→∞ç
n+1⎪
⎝⎭
⎛⎫-2n
limnç
-1⎪
limn
lim
n→∞ç
e
n+1
⎪=en→∞ç
n+1
⎪=en→∞
=e-1
16.limn2⎡ln(a+1)+ln(a-1)-2lna⎤
n→∞⎢⎣nn⎥⎦
令1=t,同第二题
n
⎛ab⎫
17.limnçen-en⎪
n→∞⎝⎭
a
=
⎛b⎫
limn(en-1)-limnçen-1⎪=a-b
n→∞
n→∞⎝⎭
18.lim⎛1
n→∞⎝n
limn
1⎫n
+en⎪
⎭
1
=en→∞+n(en-1)=e2n
19.
[]
limnln(n+1)-lnn
n→∞
=limnln⎛n+1⎫=limnln(1+1)=1
n→∞
çn⎪
n→∞n
⎝⎭
20.
lim
x2-1
x→-1lnx
=lim(x+1)(x-1)=lim(x+1)(x-1)=2
x→-1
ln(-x)
x→-1
-(x+1)
21.
[]
limln(1+x)-ln(x-1)x
x→+∞
=limln(1+x)x=limln(1+2
)x=lim
2x=2
x→+∞
x-1
x→+∞
x-1
x→+∞x-1
22.limlncosx
x→0x2
=limln(1+cosx-1)=limcosx-1=-1
x→0x2x→0x22
[]
23.lim(x+2)ln(x+2)-2(x+1)ln(x+1)+xlnxx
x→+∞
=lim⎡xln(x+2)-xln(x+1)+xlnx-xln(x+1)+2ln(x+2)⎤x
x→+∞⎢⎣
x+1⎥⎦
=lim⎡xlnx+2+xln
x⎤x+2=limx2ln(1-
1)+2=2-1=1
x→+∞⎢⎣
x+1
x+1⎥⎦
x→+∞
(x+1)2
24.
lim(
1
+xx
x→0
lim+1
ex→0xe
25.lim(cos
x→0
limcosx-1
1
x)x
-1x
lim2-1
=ex→0x
=ex→0x
=e2
x→0
⎡π⎤cotx
26.lim⎢⎣tan(4-x)⎥⎦
sin(π-x)-cos(π-x)
=
lim
x→0
=e
4
cos(π
4
4
-x)tanx
=
2lim-cos(
ex→0
π-x)-sin(π44
-x)
e-2
27.
lim(sinx+cosx)xx→0
limsinx+cosx-1limcosx-1+1
=ex→0x
=ex→0x=e
2
28.lim(sinx)tanxx→π
2
limsinx-1
x→πcos2x
e2
limcosxx→π-2cosxe2
1
=e2
⎛2x2-x+1⎫x
29.
⎭
limç2
x→∞⎝2x
+x-1⎪
⎛2x2-x+1⎫
⎛-2x+2⎫
limxçç2-1⎪⎪
limxç2⎪
=ex→∞⎝2x+x-1
⎭=ex→∞⎝2x+x-1⎭=e-1
2x+13x
30.lim
x→∞⎝2x-1⎭
lim3x⎛2⎫
=lim(1+
2)ex→∞ç2x+1⎪=e6
x→∞
⎝⎭
2x+1
1
31.lim(1-2x)x=e-2
x→0
32.limcosx(π)
x→+∞
⎡⎛π⎫⎤
π2π2
limx⎢cosç-1⎪⎥
limx(-)-
=ex→+∞⎣
⎝x⎭⎦=ex→+∞
2x=e2
1
33.lim⎛cosx⎫x-a
⎝⎭
limcosx-cosalim-sinx
=ex→acosa(x-a)=ex→acosa=e-tana
34.
limln(x0+x)+ln(x0-x)-2lnx0同第二题-1
x→0
+x22
35.limln(1+eax)ln(1+b)
x→+∞
=limln(1+eax)
x
bbln(1+eax)
lim
=lim
abeax
=ab
x→+∞
xx→+∞x
x→+∞1+eax
36.limln(secx+tanx)
x→0
sinx
=limln[(1+sinx)cosx]=limln(1+sinx)+limlncosx=1
x→0x
x→0x
x→0x
11
37.limx2(ax-ax+1)
x→+∞
11-1
=limx2ax+1(axx+1-1)=limx2(
1-1
2
)lna=limlna=lna
x→+∞
x→+∞
xx+1
x→+∞x(x+1)
1
⎛1+xax⎫x2
38.lim
x→0⎝1+xb⎭
1
⎛xax-xbx⎫x2
=limç⎪
=explim
ax-bx
=explim
ax-1
-
bx-1=(lna-lnb)a
x→0⎝
1+xbx⎭
x→0x(1+xbx)
x→0
xxb
39.lim
x→0
e5x-1
5
x
40.lim
x→0
ex+e-x-2
x2
=lim
ex-e-x
=lim
ex-1
-
lim
e-x-1=1+1=
x→02x
x→02x
x→02x22
41.lim
x→0
etanx-e3x
sinx
=lim
x→0
etanx-1
x
-
lim
x→0
e3x-1
x
=1-3=-2
42.lim
x→0
a3x-1
x
=lim3xlna=3lna
x→0
43.lim
x→a
x
ax-aa
x-a
=lim
x→a
aa(ax-a)
x-a
=lim
x→a
aa(x-a)lnax-a
=aa
lna
44.
limlnx-lnx0=1
x→x0
x-x0x0
45.lim
x→1
xn-1
x-1
令x-1=t
=lim
t→0
(1+t)n-1
t
=limnt=n
t→0t
1
⎛ax+bx⎫x
46.limç⎪
x→0⎝2⎭
令1=t,如12题
x
1
47.lim(ax+ebx)xx→0
=explim
x→0
ax+ebx-1
x
=explim
x→0
ebx-1
x
+
a=e
a+b
48.证明不等式:
ln⎛1+1⎫<1
其中n为正整数
çn⎪n
⎝⎭
解:
令f(x)=ln(1+x)-x
f'(x)=
1
1+x
-1=
-x
1+x
≤0,当x∈[0,+∞)
所以f(x)在[0,+∞)递减所以f(x) 即ln(1+x)-x<0⇒ln(1+x) nn 证毕 49.设α(x)=x3-3x+2,β(x)=c(x-1)n,确定c及n,使当x→1时,α(x)~β(x) 解: lim α(x) =1⇒lim x3-3x+23x2-3 1lim=1 x→1β(x) x→1 c(x-1)n x→1cn(x-1)n-1 ⇒lim3(x+1)(x-1)=1⇒lim3(x+1)=1 x→1 cn(x-1)n-1 x→1cn(x-1)n-2 所以n-2=0,6 cn =1⇒n=2,c=3 50.设f(x)= -2+ x,g(x)= A,确定K及A,使当x→+∞,f(x)~g(x)xk 解: limf(x)=1⇒lim=1 x→+∞g(x) -2 x→+∞ += Ax-k 1-1 = limf(x)=1⇒lim 1 ~-x2 =1 x→∞ x→+∞g(x) ⇒lim x→+∞ x→+∞ 1 -x2xk Ax-k =1 所以k+1=1,k=-1,-1 =1,A=-4 224A
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- 极限 150 题一刘刈