洛必达法则例题

习题3−2

1. 用洛必达法则求下列极限:

(1)x

x x )1ln(lim 0+→; (2)x

e e x

x x sin lim 0−→−; (3)a

x a x a x −−→sin sin lim ; (4)x

x x 5tan 3sin lim π→; (5)2

2)2(sin ln lim x x

x −→ππ; (6)n n m

m a x a

x a x −−→lim ;

(7)x x x 2tan ln 7tan ln lim 0+→; (8)x

x x 3tan tan lim

2π

→; (9)x

arc x x cot )11ln(lim ++∞→; (10)x

x x x cos sec )1ln(lim 20−+→; (11)x x x 2cot lim 0

→; (12)2

1

20lim x x e x →; (13)

−−−→1112lim 21x x x ; (14)x x x

a )1(lim +∞→; (15)x x x sin 0

lim +→; (16)x x x

tan 0)1(lim +→.

解 (1)111lim 111

lim )1ln(lim 000=+=+=+→→→x

x x x x x x . (2)2cos lim sin lim 00=+=−−→−→x

e e x e e x

x x x x x . (3)a x a

x a x a x a x cos 1cos lim sin sin lim ==−−→→. (4)5

35sec 53cos 3lim 5tan 3sin lim 2−==→→x x x x x x ππ. (5)812csc lim 41)2()2(2cot lim )2(sin ln lim 22

222−=−−−=−⋅−=−→→→x x x x x x x x πππππ. (6)n m n m n m a x n n m

m a x a n

m na mx nx mx a x a x −−−−−→→===−−1111lim lim . (7)177sec 22sec lim 277tan 2tan lim 2722sec 2tan 177sec 7tan 1lim 2tan ln 7tan ln lim 22002200=⋅⋅==⋅⋅⋅⋅=+→+→+→+→x x x x x x

x x x x x x x x . (8))sin (cos 23)3sin (3cos 2lim 31cos 3cos lim 3133sec sec lim 3tan tan lim 2

2222222x x x x x x x x x x x x x x −⋅−==⋅=→→→→ππππ 3sin 3sin 3lim cos 3cos lim

2

2=−−−=−=→→x x x x x x ππ

. (9)122lim 212lim 1lim 11

)1(111lim cot arc )11ln(lim 222

2==+=++=+−⋅+=++∞→+∞→+∞→+∞→+∞→x x x x x x x x x x x x x x x . (10)x x x

x x x x x x x x 22022020cos 1lim cos 1)1ln(cos lim cos sec )1ln(lim −=−+=−+→→→(注: cos x ⋅ln(1+x 2)~x 2) 1sin lim )sin (cos 22lim 00==−−=→→x

x x x x x x . (11)2122sec 1lim 2tan lim 2cot lim 2000

=⋅==→→→x x x x x x x x . (12)+∞====+∞→+∞→→→1lim lim 1lim lim 2

1

01202

2t t t t x x x

x e t e x e e x (注: 当x →0时, +∞→=21x

t ).

(13)2

121lim 11lim 1112lim 12121−=−=−−= −−−→→→x x x x x x x x . (14)因为)1ln(lim )1(lim x a

x x x x e x a +∞→∞→=+, 而 a a a x ax x

x a

x a x x a x a x x x x x x ==+=−−⋅+=+=+∞→∞→∞→∞→∞→1lim lim 1)(11lim 1)1ln(lim )1(ln(lim 22, 所以 a x a x x x x e e x a ==++∞→∞→)1ln(lim )1(lim . .

(15)因为x x x x x e x ln sin 0

sin 0lim lim +→+→=, 而 0cos sin lim cot csc 1

lim csc ln lim ln sin lim 20000=−=⋅−==+→+→+→+→x

x x x x x x x x x x x x x , 所以 1lim lim 0ln sin 0

sin 0===+→+→e e x x x x x x . (16)因为x x x x e x

ln tan tan 0)1(lim −+→=, 而 0sin lim csc 1

lim cot ln lim ln tan lim 202000=−=−==+→+→+→+→x

x x x x x x x x x x x , 所以 1lim )1(lim 0ln tan 0

tan 0===−+→+→e e x x x x x x . 2. 验证极限x x x 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 x x x x , 极限x

x x x sin lim +∞→是存在的. 但)cos 1(lim 1

cos 1lim )()sin (lim x x x x x x x x +=+=′′+∞→∞→∞→不存在, 不能用洛必达法则. 3. 验证极限x

x x x sin 1

sin lim 20→存在, 但不能用洛必达法则得出.