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1、习题 32 1. 用洛必达法则求下列极限: (1) x x x )1ln( lim 0 + ; (2) x ee xx x sin lim 0 ; (3) ax ax ax sinsin lim; (4) x x x 5tan 3sin lim ; (5) 2 2 )2( sinln lim x x x ; (6) nn mm ax ax ax lim; (7) x x x 2tanln 7tanln lim 0+ ; (8) x x x 3tan tan lim 2 ; (9) xarc x x cot ) 1 1ln( lim + + ; (10) xx x x cossec )1ln(
2、lim 2 0 + ; (11)xx x 2cotlim 0 ; (12) 2 1 2 0 lim x x ex ; (13) 1 1 1 2 lim 2 1 xx x ; (14) x x x a) 1 (lim+ ; (15) x x xsin 0 lim + ; (16) x x x tan 0 ) 1 (lim + . 解 (1)1 1 1 lim 1 1 1 lim )1ln( lim 000 = + = + = + x x x x xxx . (2)2 cos lim sin lim 00 = + = x ee x ee xx x xx x . (3)a x ax ax axax
3、cos 1 cos lim sinsin lim= . (4) 5 3 5sec5 3cos3 lim 5tan 3sin lim 2 = x x x x xx . (5) 8 1 2 csc lim 4 1 )2()2(2 cot lim )2( sinln lim 2 22 2 2 = = = x x x x x xxx . (6) nm n m n m ax nn mm ax a n m na mx nx mx ax ax = 1 1 1 1 limlim. (7)1 77sec 22sec lim 2 7 7tan 2tan lim 2 7 22sec 2tan 1 77sec 7ta
4、n 1 lim 2tanln 7tanln lim 2 2 00 2 2 00 = = = + x x x x x x x x x x xxxx . (8) )sin(cos2 3)3sin(3cos2 lim 3 1 cos 3cos lim 3 1 33sec sec lim 3tan tan lim 2 2 2 2 2 2 22 xx xx x x x x x x xxxx = = 3 sin 3sin3 lim cos 3cos lim 22 = = x x x x xx . (9)1 2 2 lim 21 2 lim 1 lim 1 1 ) 1 ( 1 1 1 lim cotarc
5、) 1 1ln( lim 2 2 2 2 = + = + + = + + = + +xxxxx x x xx x x x x x x . (10) x x x xx xx x xxx 2 2 0 2 2 0 2 0 cos1 lim cos1 )1ln(cos lim cossec )1ln( lim = + = + (注: cosxln(1+x2)x2) 1 sin lim )sin(cos2 2 lim 00 = = x x xx x xx . (11) 2 1 22sec 1 lim 2tan lim2cotlim 2 000 = = xx x xx xxx . (12)+= + 1 l
6、imlim 1 limlim 2 1 0 1 2 0 2 2 t t t t x x x x e t e x e ex(注: 当x0时, += 2 1 x t). (13) 2 1 2 1 lim 1 1 lim 1 1 1 2 lim 1 2 1 2 1 = = = xx x xx xxx . (14)因为 )1ln( lim)1 (lim x a x x x x e x a + =+, 而 a a ax ax x x a x a x x a x a x xxxxx = + = + = + =+ 1 limlim 1 )( 1 1 lim 1 )1ln( lim)1(ln(lim 2 2 ,
7、 所以 a x a x x x x ee x a =+ + )1ln( lim)1 (lim. . (15)因为 xx x x x ex lnsin 0 sin 0 limlim + =, 而 0 cos sin lim cotcsc 1 lim csc ln limlnsinlim 2 0000 = = + xx x xx x x x xx xxxx , 所以 1limlim 0lnsin 0 sin 0 = + eex xx x x x . (16)因为 xxx x e x lntantan 0 ) 1 (lim + =, 而 0 sin lim csc 1 lim cot ln liml
8、ntanlim 2 0 2 000 = = + x x x x x x xx xxxx , 所以 1lim) 1 (lim 0lntan 0 tan 0 = + ee x xx x x x . 2. 验证极限 x xx x sin lim + 存在, 但不能用洛必达法则得出. 解 1) sin 1 (lim sin lim=+= + x x x xx xx , 极限 x xx x sin lim + 是存在的. 但)cos1 (lim 1 cos1 lim )( )sin( limx x x xx xxx += + = + 不存在, 不能用洛必达法则. 3. 验证极限 x x x x sin
9、1 sin lim 2 0 存在, 但不能用洛必达法则得出. 解 001 1 sin sin lim sin 1 sin lim 0 2 0 = x x x x x x x xx , 极限 x x x x sin 1 sin lim 2 0 是存在的. 但 x xx x x x x xx cos 1 cos 1 sin2 lim )(sin ) 1 sin( lim 0 2 0 = 不存在, 不能用洛必达法则. 4. 讨论函数 + = 0 0 )1 ( )( 2 1 1 1 xe x e x xf x x 在点x=0处的连续性. 解 2 1 )0( = ef, )0(lim)(lim 2 1 2 1 00 feexf xx = , 因为 1)1ln( 1 1 0 1 1 00 lim )1 ( lim)(lim + + = + = x xx x x x xx e e x xf, 而 2 1 )1 (2 1 lim 2 1 1 1 lim )1ln( lim 1)1ln( 1 1 lim 00 2 00 = + = + = + =+ + xx x x xx x xx xxxx , 所以 )0(lim )1 ( lim)(lim 2 1 1)1ln( 1 1 0 1 1 00 fee e x xf x xx x x x xx = + = + + . 因此f(x)在点x=0处连续.
