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# 一、极限四则运算法则 PowerPoint PPT Presentation

§2 － 4 极限运算法则. 一、极限四则运算法则. 定理 1. 若 lim f ( x )= A , lim g ( x )= B 存在, 则. (1) lim[ f ( x )  g ( x )] =. lim f ( x )  lim g ( x ) =. A  B. (2) lim[ f ( x ) g ( x )] =. lim f ( x ) · lim g ( x ) =. A · B. (3). 证 : (2) 因 lim f ( x )= A , lim g ( x )= B , 均存在 ,.

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§2－4 极限运算法则

(1) lim[f (x) g(x)] =

limf (x)limg(x) =

AB

(2) lim[f (x)g(x)] =

limf (x) ·limg(x) =

A ·B

(3)

[A+(x)]·[B+(x)]

f (x) ·g(x)=

= AB+[A(x)+ B(x)+(x)(x)]

(1) lim[Cf (x)] = C limf (x)

(2) lim[f (x)]n = [limf (x)]n

= limf (x)–limg(x)

0.

= 2–6 = –4

= 2 ·23 + 22 – 4 =16,

(2) 也可直接利用例3后介绍的结论, 有

§2－5 夹逼定理 两个重要极限

u

u

u0

= 1 ·1=1

= 3·1= 3

= 1

= lne

= 1

§2－6 无穷小量的比较

(i)

(ii)

(iii)

= 1

= 0

= 0 ·1= 0

m

o

t

§2－7 函数的连续性

m0

t0

y

y

o

o

x

x

y = (x)

y = f (x)

B

f (x0)

y

y

A

y

y

y

y

x0

x0

x

x

x

x

x x0

x x0

(x)在x0的极限不存在, 而

C(a, b)表示在(a, b)内连续的函数全体所成集合.

u可正, 可负, 还可为0.

xU(x0),

(令x = xx0)

y

o

x

y=(x)

C

y=CD的长

A

y

B=(x0)

D

x>0

x0

x0+ x

y

o

x

y=f (x)

y= –(MN的长)

C

y=CD的长

y

M

f (x0)

D

N

x>0

x<0

x0+x

x0

x0+x

(1) af (x)+bg(x)在x0处连续, 其中a, b为常数.

(2) f (x) ·g(x)在x0连续.

(3) 当g(x0)0时,

u0=(x0),

>0, 因y=f (u)在u0连续,

>0, 当|x–x0|<时, 有|u–u0|= |(x) – (u0)|< .

| f [(x)] – f [(x0)]| = | f (u) –f (u0)|< 

= f [(x0)]相当于

limf [(x)] = f [lim(x)]

(2) 初等函数在其定义域内连续.

= 21 = 2