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 ∴+ <X1< . 2 n+nT n+n丌 1 < < lin Im x n+丌 n->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