《函数与极限》课程PPT教学课件（讲稿）第一章（1.6）极限存在准则 两个重要极限（1/2）

6(1)极限存在准则

October, 2004 1.6 (1) 极限存在准则

极限存在准则 准则I(数列的夹逼准则) Squeeze Theorem 设有三个数列:{xn}{yn}{ 若它们满足条件: (1)yn≤x≤zn(n=1,2,3,…) (2)lim yn =A limin=A n→0 n→00 则limx n->0々个 October 2004

October, 2004 一、极限存在准则 准则 I （数列的夹逼准则） 设有三个数列： { }n x { }n y { }n z 若它们满足条件： （1） n n n y x z   ( 1,2,3,...) n = （2） lim n n y A → = lim n n z A → = 则 lim n n x A → = Squeeze Theorem

(1)yn≤xn≤zn(n=1,2,3,…) (2) lim ym=A limin= A n→0 之2= 则lim 示意图 A October 2004

October, 2004 示意图 （1） n n n y x z   ( 1,2,3,...) n = （2） lim n n y A → = lim n n z A → = 则 lim n n x A → = n n n A y x z

(1)yn≤x≤zn(n=1,2,3,) (2)lim yn=A liman=A n→0 n→0 则 limx=A n→0 证明VE>0 lim yn=A= 3N A-EM1) lim z=A→彐N2A-0 (ctober 2004

October, 2004 （ 1 ） n n n y x z   ( 1,2,3,...) n = （2 ） lim n n y A → = lim n n z A → = 则 lim n n x A → = 证明    0 lim n n y A → =  N1 A y A n −   +  1 ( ) n N lim n n z A → = N2 A z A n −   +   2 ( ) n N

lim yn=A= A-EN n→0 A-8N2) n→00 VE>0彐N=max{N1,N2 n>N=max{N1,N2}→ E<yn Sx<zn< A n→00 A-8 A A+8

October, 2004 lim n n y A → =  A y A n −   +   1 ( ) n N lim n n z A → =  A z A n −   +   2 ( ) n N    0  = N N N max{ , } 1 2 max{ , } 1 2 n N N N  =  A yn −   n z A  + n   x lim n n x A →   = y x z n n n A− A A+

准则I(函数的夹逼准则) Suppose in a neighbourhood of a, we have (1)g(x)≤f(x)≤h(x) (2) lim g(x)=lim h(x)=A X→a x→a Then we must have lim f(x)=A x→a This theorem is called the squeeze Theorem October 2004

October, 2004 准则 I’ （函数的夹逼准则） Suppose in a neighbourhood of a, we have (1) ( ) ( ) ( ) g x f x h x   (2) lim ( ) lim ( ) x a x a g x h x A → → = = Then we must have lim ( ) x a f x A → = This theorem is called the Squeeze Theorem

Geometrical interpretation of the Squeeze Theorem lim f(x)=A x→a h(x) g(x) f(x) October 2004

October, 2004 a f x( ) A g x( ) h x( ) Geometrical interpretation of the Squeeze Theorem lim ( ) x a f x A → =

例证明: lim cos x=1 x->0 解im ∴sinx≤ x->0 x 等价于lim(1- COSx)=0 x-0 Cosx=2n≤2.(n)1 x020由夹im( (1-cosx)=0 逼准 x->0 则 lim cos x=1 x->0 October 2004

October, 2004 例 证明 ： 0 limcos 1 x x → = 解 0 limcos 1 x x → = 等价于 0 lim(1 cos ) 0 x x → − = 1 cos − x 2 2sin 2x = sin x x  2 2 ( ) 2x   1 2 2 = x 2 0 1 limx 2 x → = 0 由夹 逼准则 0 lim(1 cos ) 0 x x → − = 0 limcos 1 x x →  =

例求lim( 十, n→>on2+丌n2+2 n+n丌 p.56,题4(2)《学习指导》p.25,例1.22 解 , n2+丌n2+2丌 n+1元 1 ∴+ n+丌 n→>0 十,, n→>On2+丌n2+2兀 2 n+nT October 2004

October, 2004 例 求 2 2 2 lim( ... ) n 2 n n n →  n n n n    + + + + + + 解 p.56, 题4(2) 2 2 2 ... 2 n n n n x n n n n    = + + + + + + 2 n n xn  2 n n  +... + 2 n n n +  2 n n n + + +... n x  1 n n   + lim n n → n + = 1 lim 1 n n x →  = 2 2 2 lim( ... ) 1 n 2 n n n →  n n n n    + + + = + + + 《学习指导》p.25, 例1.22

利用夹逼准则,可以证明下列有用的极限: m√n n→ ya=1(a>0) n→>00 lim=0 (a>1) n→0 October 2004

October, 2004 利用夹逼准则，可以证明下列有用的极限： lim 1 n n n → = lim 1 n n a → = ( 0) a  lim 0 n n n → a = ( 1) a 