洛必达法则习题

习题 32 1.−用 洛达法则求必列下限极:( 1 l)m

xi→ 0

l

n1(+ x); x

2) (iml

e

x −e−x x →0; sinx isn x− sin a ; 3() limx→a x a

x−→ π

(4)li m5()lim

x→

s

n 3x ;i tan 5xl n in s (xπ − 2x ) 2x

m am−

π

;

2

;

xn −a ln tnan7 x 7) li( m ;x→ +0 ln ant 2

x(6 lim

)

→x

a(8

)li mx

→

πantx ;ta n 3

2x

1

nl1(+ ) x ;()9lim x → ∞ +ar coc tx

(1

0) li

mln

1( +x 2 ); x0→sec x − cosx

(1 1)l imx c t 2o x ;

x 0→1

12) (lm i

xx→0

2

2

e

x

;

1

2 (1) lim 23 − x; 1→x − 1 x 1 a−( 14) im (l1 )+ ; x →x x∞(15 l)imx s ni ;xx → +

01

(1) 6iml( ) t n ax. x →+0

x

1

n(1l+ ) 1 x + x =lmi 1 =1 解 .(1) iml =ilmx →0 x → x 0→ 10 x+ 1

x(

) l2m

i

xe − e− x xe+ e x−= lm i=2. x →0s nix →0 xco xss n i −x is a ncs o (3x)li =m im =lc osa . x →a x a 1 x→−ax

→π

() 4iml

si n3x 3cso3x 3= l m i− .=2 t a n5 x→xπ5 s ce5 x

5

5( lim

x)→

l

nsnix π( − x)22

π

=li m

→

π

2

x

2cot

x 1 csc 2− 1 =x li−m = −. 2π −( x2)⋅ (−2 )4 x→ − 2π8

2

(6

)lm

xia

→

m −x a x ma−

nn

=lim

x→

amx

m 1 −xn

n1

−=

m

xm −1 a

n −1

n=

mmn− a. n

1

⋅ ecs 72x 7 ⋅nl tna7 x 7 tan 2x 7se c22 x ⋅2 t a n7x ( )7 lmi = il = mil m l=i =m1.x → + 0l natn 2 xx→+ 01 2 x→+ t0na7 x 2x +0 sec→ 2 7 ⋅ x72 ⋅s e c x2 ⋅ 2 ant2 x ()8 lmi

→x

π

t2a xns e c2x 1 cs o2 x31 2 oc 3sx− (sn i3)x⋅ 3 li= m=l im = ilmta n3x →x πsec 3x2 3 ⋅ x→ π3c os x2 3x→ 2 πco xs−(s n ix)2 2 2

=

− lm

ix

→

2π

co sx3− 3sin 3 x= − il =m3 . cπs x − osi x nx→2

1 ⋅1(− ) 21 1x 1+l n(1 + )+ x2 2x 12 = xim xl ()9 lim= l m i= iml= lm = i .12 x→ + ar∞cc o t xx →∞ x + +→∞ x x +x +→∞ 1 + 2xx → +∞ 122 + x1 (01)lim l(n1+ x 2 ) ocsx ln( + x 21) x =2li m =li m 注(: csx⋅on(l+12)~x2)xx →0 sec x −c s ox x→ 0 →0 x1 − cs 2o 1x− c o 2 xs2 xx = lmi l=mi =1. x → 0 −2co sx ( − si nx ) x0 →sin

xx→ x 00

→

1() lim1x co t 2 x= l i

m

1

x 1 1 =il = .m2 x →0 atn 2x ecs 2 ⋅x 2

212

(

1) 2im xl 2e x

→x

02

e

x e te t =1 ilm= lm = liim= ∞ +注: (当 →x0 , t =时2 → ∞ +.)x → 0 1 t→+ ∞t t → ∞+ 1 x x2

1

 1 x− − 11 2 1() lim3 2 −= lmi = −. = im 2 xl 1→ x − 1 x → 1x→ 1 −x 21 2x x− 1x l

(n+1) ax , (1)因为4li m 1(+ ) x = lm i x →∞ex ∞→x a

而

)

2 aa lnx(1+ ) 1+ ax a xa= l im x il x(ml(1n+ )= lmi =l im= il m=a ,x →∞x → ∞ x →∞ →x x∞→ ∞ 1 x1 x+ a1 −2x x

lx(n1 )+ ax = ea .lim(1 + ) x= im le →x x ∞∞→ xa

1

⋅ −(

a

所

以

.15()为因l i xm in xs =l i m ein s lx n ,xx

+→ 0 → +0

x而

所以

1 l nx sn 2ix xlmi sinx nl x =il m=l m =i l−i m=0, →x+ 0x →+ 0ccs x →x +0 −cs xc⋅ c otx x → 0+ x ocs

x x +→0

l

i mx

si n x =il mesin x ln x = e0 = 1.

x→+

0

1 1(6因为)lim ( ta) n x = −e atn lx n x , x→+0 x 而

1ln x s i 2 xnx lim t n ax nlx l=mi= lim = − lm =i ,0x → +0 →x 0 c+o t xx→ 0 − csc 2 x+ x →+ x

0

以所

lim

(1 )tnax = ilme ta− n xlnx = 0e =. 1x→0 + x→+x0

2.验证 限极l m

解i

x∞

→ + xsn i x在,存 不能用洛但必法达得则出.x

x+ sin xisnx x sin+ = limx( 1+ ) = 1 , 极限lim 存在是的. x → ∞x → ∞ x xx( x+ in sx)′ 1 +cosx 但 lm i li=m= lim (1+ c s o) 不存x在 ,能不洛必用法则.达x ∞ →x ∞→x →∞ ( ) ′x1 x

∞→

l

i

x m2s n

i3

验.极证限 lm

x →i0

sin x

1

x 在,存但不 能用洛必法达得出.

则1

x1 2s inx 1 x 是存 的在 .x 解li m=lim ⋅ six n = ⋅1 0 = ,0极限 li mx→0 s i nxx →0 ins xx →0 xsi x n 2 sxni

1 11( x 2 s n )i 2 ′x sni −c osx l=i xmx 不存在 ,不 能洛用达法则. 必 但limx →0 (s in x ) x →′0 osc

x 1 1 (1 + x ) xx] [ . 4论讨数函 f( )x=  1 e−  e 2

x

>0 ≤0x−

1

在点2 =x 处的连0性.

续

解f( 0 = e

)

−1

2,li f (mx )= lim e

→ −0 xx→ −

0e=

11

−

1

2

= (f0) ,

1 [ln 1( x+ −)]1( 1 +x ) x ]x = im el x x ,因为 im fl ( ) x l=mi [ →x 0 + x→ 0−x →−0 e1

而

1

−1 − 111 ln 1 +(x ) x 1 −1 + xl i m[ n(l1+ x )− 1] =l mi= lim = l i =m−, 2 x → + 0 x x → x+ 0x→ + 0 x→ 0 +x22 ( 1+ ) 2x

所x以

x

→+

0

ilmf (x =)l i m[

x →−0

(1 +e

1

x )x

1

x → 0

− ] x l=im e

x1

1 ln[(1+x ) 1]− x

=

e

1 2−

=

f ( ) 0

因此.f(x )点在 x0 =处连续

.