高考導(dǎo)數(shù) 洛必達法則

第部：勒開

xxnee3!!(1)!

(0

ln(1)

nn2!3!!

n

R

()(1)!1

sinxx

35x2kk5!(21)!(2

cos

cosx

xx2kx2k(2)!

cos

第三部：新課標高命題趨及方法00第四部：洛必達法及其解洛達則f()x)limf()limg(x)U

)f

g

lim

fg

Alim

f()fg(xg例已

f(x)

alnxbyf()fx

xy

lnxf(x)x

求

方一分類論假反法ln(x)xx

k(fx))xx12

2

)

()2lnx

(k1)(((xh'(x

x2

k

h'(

(x2x22

x

'()

h(1)

(0,1)

(x)

x)

x(1,

()

11

2

x

x

x

lnklnk))f()xx

0

x(1,

1)(2x0

h)

x(1,

h()

11

2

)

k'()(1)(1,(x)

11

2

)

k

(注0k

x

x

x

lnklnx1lnxkf(x)xxxx

k

xlnxxlnx2xlnln(x)xx11

x

x

g'(

)2(x=x)2)2(1)22

hx)ln

1x2

1(12)h'(x)+(1+2)2xx2)2

(x)

hx(0,1)(

hx)

x(0,1)g'()0g'()g(在(0,1)(1,

lim()lim(

xln2lnx2lnx1x1x

g()xgx)g(x)

kxxf(x

lnk(xx注

k

()

xln

=1g(x)

例

fx)

x

fx)

f()

應(yīng)洛達則導(dǎo)(x)e

ax

2

x

R

x

x

a

x

x2

gx)

exx(0g'(xx23

h(x2)

x

(0h'(x)e

x

(0h''(x)xe

x

h'(ex

h'(x)'(0)

h(x2)

hx)

x(0

'()

()x3

g(

x2

lim(xx0x

x

eelimx2xx0210g()x(0g()22

f()

自

sinx

x

2

)

解應(yīng)洛達則導(dǎo)

x

2

)

a

xx3

xsinf(x)x3

'(x

3sincosx4

g(x)xcosx

g'(x)2cosxsinx

g''()xcosxcos(xtanx

gx)sinxg''(x(0,

2

)

g)

g'()

)'(x).g()22

g()f)

g(x)xf(x)(0,x4

lim(xlimx

xsinxsinlimlimxxx06x066

0)

11(sinx3x(0,)66

002010海寧文21）

f(x)x(

x2

f(x)

f(x)

f(x

解應(yīng)用必法和數(shù)

x

f(x

x(x

x

a

x

x(

xax2

e

a

exx(xe()(0,gx

h()exxh'()x0h()x

(0,

h(x(0)

g'(x)

hx)x

g()

e

(0,

(x

e

x

lim0gx)g(x)a.f()

2010全大理22）

f()

f(x)

xf()ax

解應(yīng)用必法和數(shù)xfx).

1f(x)aaxf(x

1ax

aR1

xax

1xeax

g)

xx

g'(x

e

x

xex=e(xex)2(xex2

x

)

h()

x

h'(x)x

h''(x)

x

+

h'(x)

h'(0)

hx

x≥kZf(xx≥kZf(x

(x)

h(0)

h(x

g'(x)=

(xe

exx)

h(

g(x

(0

g(x)x0

xxx0e

x

xexe

x

e1limx02xx

x

()()2

a

11a(]22

（2008例

f(x)

x2cosx

f(x

0()

≤

解

f

(2)cosxxx(22(2cos

2k

2π1kkcosx332

2k

241kkZcosx33

f(

22k3

2π4π2k33

kZ

f(x)

sinsinax2cosaRx2x

ax

a

x(2cos)

g()

x(2

gx)

2xcosxxcosx2