NAN DA XUE JING PIN KE CHENG 定理1.设Q(x),a'(x),B(x),B(x)是某极限过程中的 无穷小量.f(x)是另一变量,且,a(x)~a'(x), B(x)~B(x),则 a'(x) (1)lim B(x)6(x) (2) lima(xf(x)=lima(x)f(x) (3)lima()f( lim a(x)f(x) f"(x) (x) 只须右端极限存在或为无穷大 OD 高等數粤
定理1. 设(x), (x), (x), (x)是某极限过程中的 无穷小量. f (x)是另一变量, 且, (x) ~ (x), (x) ~ (x), 则 , ( ) ( ) lim ( ) ( ) (1) lim x x x x = (2) lim(x) f (x) = lim(x) f (x), . ( ) ( ) ( ) lim ( ) ( ) ( ) (3) lim x x f x x x f x = 只须右端极限存在或为无穷大
NAN DA XUE JING PIN KE CHENG 证:(1)因为a(x)~a'(x),B(x)~B(x) 所以lim a() B(x) liml a(x)a(x) b(r) a(x) B(x) B(x) lim a(r) (x) 类似可证(2),(3) OD 高等數粤
证: (1) 因为(x) ~ (x), (x) ~ (x), 所以 ( ) ( ) lim x x = ( ) ( ) ( ) ( ) ( ) ( ) lim x x x x x x ( ) ( ) lim x x = 类似可证(2), (3)
NAN DA XUE JING PIN KE CHENG 例!.求1m(2x x→)0Sln5x 解:由于当x->0,tgx~x,从而tg2x~2x 当x>0,sinx~x,从而sin5x~5x 故,lim 522x lim g 2x2 x-0sin 5x x,0 5x 5 OD 高等數粤
例1. . sin 5 tg2 lim 0 x x x→ 求 解: 由于当x→0, tgx ~ x, 从而tg2x ~ 2x. 当x→0, sinx ~ x, 从而sin5x ~ 5x. 故, x x x sin 5 tg2 lim →0 x x x 5 2 lim →0 = 5 2 =
NAN DA XUE JING PIN KE CHENG e 例2.lim (a≠b) x→>0 sIn ax- sin bx bx e a-b)x 解:lim =im x>osin ax-sin bx x=0 atb 2 cOs Xsin bx e (a-b)x Im 0 +6 b 2 COS SIn 11(a-b)x x→>0a b 2 OD 高等數粤
例2. , ( ) sin sin lim 0 a b ax bx e e ax bx x − − → 解: x a b x a b e e bx a b x x 2 sin 2 2cos ( 1) lim ( ) 0 + − − = − → x a b e x a b e a b x x bx x 2 sin ( 1) lim 2 2cos lim ( ) 0 0 − − + = − → → x a b a b x x 2 ( ) lim 2 1 0 − − = → = 1 ax bx e e ax bx x sin sin lim 0 − − →
NAN DA XUE JING PIN KE CHENG 例3.求lmx2m、3 x→)0 解:imx2ln(1+3)=1im/3 m x->0x 3 x 或,limx2ln(1+-3)=lim x>0x3 n(1+-3) =1im3ln(1+3)3=1im3 lim In(1+ x→>∞x x→)0X =0·1=0 OD 高等數粤
例3. ). 3 lim ln(1 3 2 x x x + → 求 解: ) 3 lim ln(1 3 2 x x x + → 3 2 3 lim x x x = → x x 3 lim → = = 0 或, ) 3 lim ln(1 3 2 x x x + → ) 3 ln(1 3 3 lim 3 3 x x x x = + → 3 3 3 ) 3 ln(1 3 lim x x x x = + → 3 3 3 ) 3 lim ln(1 3 lim x x x x x = + → → = 0 ·1= 0