专转本高等数学复习二

limC

lim

(

limna1(ax

limnn1(a limqn0(|q|limax(a

limax0(a

limarctanxlimarctanx

limsinx

lim(11)x

x0limf(xf(x0例 同除法（”极限的计算

x2例 求x2

x

1解

x

1x2x2x

x21 例 求x

x3．43x2

1解

x

0xx43x2

1

3 例 求

x43x2 x4

x3解x10．此时可先求出x

4

0，再利用非零无穷小的倒数为无穷大这一性质，得

x43x23

x0b0axnaxn1 x lim n xbxmbxm1 x

(n21(n21 3n6(n(n21 3n6

(1(1131

4例 求

(2x3)20(3x2)30(5x1)50

(23)20(3

解lim(2x3)(3x

2 (5x3n1

(51x

7求n

xx0xx xx0x26x例 求x4

5xx26x (x4)(x x 解lim

x4

5x

x4(x4)(x

x4x 例 求

3x3x1解

x2

3x3x1(3x)(1(x1)(x1)(3x1x2(x2(x(x1)(x1)(3x11(x1(x1)(3x1

12x21例12x21

x) lim

x) x2x2x21

1111111111x1

3131x31lim

xx2x

x23x31x311x1 x

例 设f(x)

a为何值时limf(xa为何值时limf(xaxx

解limf(xlimex1，limf(xlim(axaa1limf(x

a1时limf(x例 证明

x2

x1|x证明因为

x2 (x1)(x

x2 (x1)(x

2

x

(x所以

x2

x1|x14求

2e41e

sinx|x

2e4

)211，lim

2exe

sinx

)011e

ex∴

2e41e

sin|x

)00例 求lim1cosx

(

sin()1．()

2sin2

sin2

sin2解lim1cosx

2

()2x ()

1

21

x

2 x 16求

sin

( ( x1tan1tanx1sinxx1cosx

解一：原式limx1cos

1tanx

1sinx2

2 解二：原式

1tanx11sinx

12

tanxsinx1limtanx2x例 求lim

nxcosn x0，原式x0时，原式

2nsinxcosxcosxcos

2nsin

2

4

2nsinx

sin

sinn2nsin

sin 1

lim(

]()elim1()]()e( (例 求lim(111)n

n

n2

n

n2n解 n

)n

)n1

lim

)n1

e

n

2x

x12

x1例 求 )x

)2

lim )2

e2

x

x

x

x 2x3例21 x2x122

limxx1记住常用的等价无穷小：若(x)0sin(x)~

tan(x)~

1cos(x)

22

arcsin(x)~arctan(x)~

[1(x)]1~

ln[1(x)]~

e(x)1~22求lim(tanxsinx)(1x1 (ex1)ln(13x

tanx(1cosx)(1xtanx(1cosx)(1xx21(tanx(tanxsinx)(1x

lim 22

(ex21)ln(13x2

(ex21)ln(13x2(xx)1

x0x2 lim(tanxsinx)(1x1)

2 (ex1)ln(13x21cos2xarctan

x2例24 求limsinx．x解limsinxlim1sinx03n2n3n2nn225求

n2n2

3n3n2n

31n 31n n根据有界变量乘无穷小仍是无穷小，可知原式例 求lim123 nnn

n 2 解lim123 nn n

n 2 n例27设l是正整数，求lim nnk1k(k1k(k

1(l

1k 1 ∴k(kl)l12lk )因此原式1(11) （1）

1

（lnnk1k(knn 1(11)3n

（l2nk1k(k 例 求lim(24282 2n2)2n2)解lim(24282 2n2) 例 求lim(1 3x11

13) x11

1 6x3x6x3x6

t 6解 t（t0则x6

x1t1，于是

6x6x3x

t1t2

t1t

limtanxxx0x2sin

limlnx例33求lim1 xx0x

e134求

sin22

cos2

x21sin2

2x4sin2xcosx解原式

sinx

x

x2sin2

x1sinlim lim1cos4x

lim4sin4x

135设a0b0limxaxbxx

limax

（用法则 x1

1(axlnabxln

1x

x lnb

0(xt)fx例 设函数f(x)连续x

00，求

xx0f(xx x0f(t)dt0tf x解原式 （分母作变量替换 ux xf0 f xf xfx （用法则，分子、分母各求导数x f(u)duxf0（用积分中值定理lim （0x之间）(0)

xf()xf f f(0)f 37求lim(secxtanx2138求limx(ex139lim(sinx)tanx 40lim(arctanx)x（ex

lnsin

解lim(sinx)tanxlimetanxln(sinx)limecotxlimecsc2xlimesinxcosx

1

41lim(cotx)lnx（e142求limsin

1xx3sinxx2cos43求lim

x0

cosxln 00

”型，但分子，分母分别求导数后的极限不存在，因此不能 法则3sinxxcos1原式lim x2x01cosx2用定理求极

n2n2nkn例 求nk1

kn121222n2 1222 n3nk1n2nn3n

1n(n1)(2n而lim12

lim n3

n(n2

1n(n1)(2nlim12

lim n3n n3n 由定理可知nk1

2n 46求lim(1n

354

2n)xn

132

2n6 ，yn

23

2n则0xn

，于是0x2x

2nnnn由定理可知：limx20，于是原极限为nnnn(1xn例当|x|1lim(1x)(1x2(1xn

k2

k1k 例

2)(12

2

2)，12

47f(x2

f(x03x)f(x0 [ 解原式=

(0

[(

) f(x3x)f(x =3

2

f(x02x)f(x0

n k limnfn0f(x)dxnk n48求nk1

2k分析如果还想用定理中的方法来考n22n2 n2 k1n2k

n222而 1， 22nn2

nn2由此可见，无法再用定理，因此我们改用定积分定义来考nn解lim lim1 nn22nk1n22

nnk1

(kn1 arctanx1101 49设0x13xn1

x(3x，证明limx

0，3

0

0

x1(3x1)x1(3x1x1(3x1（几何平均值≤算术平均值n10

333

{xn}xn(3xnn1xnxn(3xn

xnxn(3xn(32xn3xn

xn

xn∴xn1xn，则{xn}

limxll(3l(3nxn1

xn(3xn两边取极限，得ll23ll2l0（舍去）得l32

limxnn nx2ax例 已知x1

3x

ab解依题意有

x2ax2

(x1)(x

(1c5，所以c6x1

3x

x1(x1)(xx2axbx1)(x6a7，b6x2例 已知

abx

x2解因为

(x21)(axb)(x (1a)x2(ab)x

x

0ax2ax例 设x1

3a和解：由题设可知lim(x2axb)0，1ab0，再由法则

x2ax

2x

2

3，a4,bx1sin(x2 x12xcos(x2 例 已知当x0时

1~sin2xa1111ax2解因为1

sin2

lim

1aa2253x0x2ln(1x2是sinnx的高阶无穷小，而sinnx又是1cosxnnx2ln(1x2 x2 解因为0

sinn

lim

，所以4n0n4又0

sinn

1lim

n20n2nn3x01cos

x012

21x例 已知当x0时 1x111 x x(x x x

11[x2(1x21

11x4(1x21解由1

lim lim2 lim2 ，

m1，n1

xc例55设f(x)在(,)内可导，且limf(x)e，

limf(xf(x1，求c

xxc

例 设f(x)

a，x

f(xx0a解af(0limf(xlimsinx1．

ax

1ln(1

例 设f(x)

sin2x2

ab c1xx

x 1x limf(x)limaxsin1ln(12x)1，

x0

sin2x2

2x1x1 limf(x)limc1xxclim1 2x

ce4limf(xlimf(xf(0

1x

x0

1x

sin2x

x 例 设f(x)2(1x2ln

0剟

1f(x

x limf(xlimsin2x2limf(xlim2(1x22f(02f(xx0

limf(x