高数课件-1.极限运算法则

—二、极限的四则运算法三、复合函数的极限运算法一、无穷小运算法注意无穷多个无穷小的和未必是无穷小. 当n，1是无穷小，但n111nn1n

limn(n1)lim1

1)

2n2

n 定理 ①、ysinx②、有ysinx例 求limsinx limsinxlim1sinx xxx

说明：y0ysin说明：y0ysinx .二、极限的四定理3若limf(x)A,limg(x) lim[f(x)g(x)]limf(x)limg(x)A证因limf(x)A,limg(x) f(x)A,g(x)B（其中,为无穷小） f(x)g(x)(A)(B(AB)(.思考若limf(x存在，limg(xlimf(xg(x)]不存在limf(xg(x)]g(x)[f(x)g(x)]f(x)由定理1可知limg(x)存在

若limf(xlimg(x)limf(xg(x)] 若limf(x)A,limg(x) lim[f(x)g(x)]limf(x)limg(x)

lim[Cf(x)]Climf(x)

（C为常数

limf(x)]nlimf(x)]n （n为正整数axnn说明axnn例 设n次多项式

limPn(x)Pn(x0证limPn(x)lim

a1limx alim

Pn(x0

n 若limf(x)A,limg(x) 且B≠0，则limf(x)limf(x)Ag(x) limg(x) 证因limf(x)A,limg(x) f(x)A,g(x)B,（其中为无穷小 f(x)AAA

g(x) B B(B

因此γ为无穷小，故f(x)A g(x) 由极限与无穷小关系定理，得limf(x)Alimf(x)g(x) limg(x)定理 若limf(x)A,limg(x)AB

f(x)g(x),提示令(x)f(x)g(x), nn 若limxnA,lim B, lim(xnynAlimxnyn③、当 0且B0时,lim yB yBn例 求lim(2xlim(2x1)lim2xlim12limx1211 例

x2

x3 5x解lim(x25x3)limx25limx (limx)2523221033

x3

lim(x3 limx3 x2x25x

lim(x25x

81 F(xP(x),P(xlimP(x)P(x0), limQ(x)Q(x0 limF(x)3①、若Q(x0) ②、P(x00Q(x00,

limF(x)

F(x) ③、若P(x0)0,Q(x0) 例 求

2x x1x25x分析分析x1lim(2x3)0,lim(x25x4)limx5x41514 2x 21

2x x1x25x例 求

x24x3

x2分析x3lim(x24x3)

lim(x29)

x24x

lim(x3)(x解

x2

(x3)(xlimx12x3x 例7

4x23x95x22xlim(4x23x9)lim(5x22x1)解x222

3x

4319 x5x22x

521 limF(x的情形，按“抓大头”法axmaxm limF(x)lim

bxnbxn1a0, 当nm 0

当n当n其中：a00b00mn为非负整数例8

3x22x .x.

x解x3,2lim2

2x

3121 x2x3x2

215 例

2x3x253x22x解7lim3x22x

∴

2x3x25x2x3x2 x3x22x二、特殊函数①、分子或分母x22x15x22x152x12x1(2x1(2x15)(2x15) x25

x22xx22x15x2x2

5)

x2

2x1 x22x

x251lim(x2) 5 2 ②、分段函数 x x例 设f(x)x23x

求limf(x),limf(x),limf(x).

x3

x

解limf(x)lim(x1) limf

lim

x23xx3

所以limf(x) 又limf(x)

x23x xlimf(x) lim(x1) 三、复合函数的极限运算法 设函数yfg(x)是由函数yf(u)0ug(x)yfg(x)0limg(x)u0,limf(u)A

且存在

0,当xU(x0,0)时，有g(x)u0, limfg(x)limf(u) 说明若定理中limg(x) lim limf(u) xx2xx2 令u