48561662-Phương-phap-khử-dạng-vo-định

8/3/2019 48561662-Phng-phap-kh-dng-vo-nh

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etoanhoc.tk

Nhng dng v nh thng gp trong bi ton tm gii hn ca hm s

1

Gii hn dng v nh l nhng gii hn m ta khng thtm chng bng

cch p dng trc tip cc nh l v gii hn v cc gii hn c bn trnh by

trong Sch gio khoa. Do mun tnh gii hn dng v nh ca hm s, ta

phi tm cch khcc dng v nh bin i thnh dng xc nh ca gii

hn

Trong chng trnh ton THPT, cc dng v nh thng gp l :

0, , , 0. , 1

0

Sau y l ni dung tng dng c th.

I. GII HN DNG V NH 00

Gii hn dng v nh 00

l mt trong nhng gii hn thng gp nht

i vi bi ton tnh gii hn ca hm s. tnh cc gii hn dng ny,

phng php chung l s dng cc php bin i ( phn tch a thc thnh nhn

t, nhn c tv mu vi biu thc lin hp, thm bt, ) khcc thnh

phn c gii hn bng 0, a vtnh gii hn xc nh. Chnh cc thnh phn c

gii hn bng 0 ny gy nn dng v nh.

tnh gii hn dng v nh 00

, trc ht gio vin cn rn luyn cho

hc sinh knng nhn dng.

1. Nhn dng gii hn v nh 00

gii bi ton tm gii hn ca hm s, hc sinh cn xc nh gii hn

cn tm thuc dng xc nh hay v nh. Nu gii hn l v nh th phi xtxem n thuc dng v nh no c phng php gii thch hp. Bi vy vic

rn luyn knng nhn dng cho hc sinh c quan trng, gip hc sinh nh

hng c cch gii, trnh nhng sai xt c th mc phi.

i vi dng v nh0

0, vic nhn dng khng kh khn lm v hc sinh

thng gp gii hn :

0x x

f(x)lim

g(x) m

0 0x x x xlim f(x) = lim g(x) = 0


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2

Thc t hc sinh hay gp trng hp0x x

f(x)lim

g(x)m 0 0f(x ) = (x ) = 0g . Ngoi ra

trong mt sbi ton hc sinh phi thc hin cc php bin i chuyn v

dng v nh0

0 , sau mi p dng cc phng php khcc thnh phn cgii hn bng 0.

Khi ging dy, gio vin nn a ra mt sbi ton nhn mnh cho

hc sinh vic nhn dng nh :

0x x

f(x)lim

g(x) m

0x xlim f(x) 0

hoc

0x xlim g(x) 0

Trnh tnh trng hc sinh khng nhn dng m p dng ngay phng php gii.

V dp dng :

(Yu cu chung ca nhng bi tp l : Tnh cc gii hn sau).

V d 1 : 1 2x 2x - 2

L = limx +1

Bi gii :

1 2 2x 2=

x - 2 2 - 2L = lim 0

x +1 2 1

V d 2 : 2 2x 1 -x + 2

L = limx 1

Bi gii :

2 2x 1 -

x + 2

L = lim =x 1

v1

2 21

lim(x+2) = 1+2 = 3

lim(x - 1) = 1 - 1 = 0x

x

V d3 : 3 2x 11 3

L = limx 1 x 1

Bi gii :

2

2 2x 1 x 1

x 1 x 1=

1 3 x 3x +2L = lim lim

3 x 1 x 1 x 1

(x-1)(x 2) (x-2) 1-2 1lim lim(x 1)(x+1) (x+1) 1+1 2

etoanhoc.tk


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3

Dng v nh 00

c nghin cu vi cc loi c th sau :

2. Loi 1 :0x x

f(x)lim

g(x)m f(x), g(x) l cc a thc v f(x0) = g(x0) = 0

Phng php : Kh dng v nh bng cch phn tch c tv mu thnhnhn t vi nhn tchung l (x x0).

Gi s : f(x) = (xx0).f1(x) v g(x) = (x x0).g1(x). Khi :0 1 1

0 0 00 1 1x x x x x x

)

)

(x - x f (x) f (x)f(x)lim lim lim

g(x) (x - x g (x) g (x)

Nu gii hn 10 1

x x

f (x)lim

g (x)vn dng v nh 0

0th ta lp li qu trnh khn

khi khng cn dng v nh.

V dp dng :

V d4 :2

4 2x 2

2x - 5x +2L = lim

x +x - 6

Bi gii :

Ta phn tch c tv mu thnh nhn t vi nhn t chung : x - 22

4 2x 2 x 2

x 2=

2x - 5x +2 (x - 2)(2x - 1)L = lim lim

(x - 2)(x + 3)x +x - 6

2x - 1 2.2 1 3lim

x + 3 2 3 5

Vy 43

L5

V d 5 :2

5 x 2 2

x - 3x +2L = lim - 4x + 4x

Bi gii :

2

25 x 2 x 2

x 2

2

=

x - 3x +2 (x - 2)(x - 1)L = lim lim

(x - 2)- 4x + 4x - 1

limx - 2

x

( V gii hn ca t bng 1, gii hn ca mu bng 0)Vy 4L

etoanhoc.tk


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4

V d6 :2

2

3 n*

6 3 mx 1

+

+

x+x x +...+x - nL lim (m, n N )

x+x x +...+x - m

Bi gii : Ta sphn tch tv mu thnh nhn t vi nhn t chung : x

1 bng cch tch v nhm nh sau :

x + x2

+ x3

+ ... + xnn = (x1) + (x21) + (x3 - 1) + ...+ (xn - 1)

x + x2

+ x3

+ ... + xmm = (x1) + (x21) + (x3 - 1) + ...+ (xm - 1)

Khi :22

2 2x 1 x 1

3 n3 n

6 3 m 3 m

1 - 1)+( - 1)+

+ 1 - 1)+( - 1)lim lim

(x- )+(x x +...+(x - 1)x+x x +...+x - nL

x+x x +...+x - m (x- )+(x x +...+(x - 1)

x 1

n-1 n-2

m-1 m-2

1 1 + (x + 1) +...+ ( )

1 1 + (x + 1) +...+ ( )

lim

(x- ) 1

(x- ) +1

x + x +...+ x +

x + x +...+ x

n-1 n-2

m-1 m-2x 1

1 + (x + 1) +...+ (x + x +...+ x +1)lim

1 + (x + 1) +...+ (x + x +...+ x +1)

n-1 n-2

m-1 m-2

1 + (1 +1) +...+ (1 + 1 +...+ 1 +1)1 + (1 +1) +...+ (1 + 1 +...+ 1 +1)

n(n + 1)

1 2 3 ... n n(n + 1)2m(m + 1)1 2 3 ... m m(m + 1)

2

Vy 6n(n + 1)

Lm(m + 1)

V d7 :4 3 2

7 4 3 21

2x - 5x +3x + x - 1L lim

3x - 8x + 6x - 1x

Bi gii :3 2

7 3 2x 1

3 2 2

3 2 2

4 3 2

4 3 2 x 1

x 1 x 1

=(x-1)(2x - 3x +1)

L = lim(x-1)(3x - 5x +x+1)

2x - 3x +1 (x-1)(2x - x -1)= =

3x - 5x + x +1 (x-1)(3x - 2x -1)

2x - 5x +3x + x - 1lim

3x - 8x + 6x - 1

lim lim

2

2x 1 x 1

x 1

2x - x -1 (x -1)(2x+1)= lim = lim

3x - 2x -1 (x -1)(3x+1)

2x+1 2.1+1 3= lim = =3x+1 3.1+1 4

etoanhoc.tk


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5

Vy 73

L =4

Kt lun:

Phng php gii bi tp loi ny l phn tch a thc thnh nhn t

vi nhn tchung l x - x0. Yu cu i vi hc sinh l :

Phi nm vng cc phng php phn tch a thc thnh nhn t, cc

hng ng thc, cng thc phn tch tam thc bc hai, a thc bc ba thnh nhn

t:2

0

0

cf(x) = ax + bx + c = (x - x ) ax -

x

, ( f(x0) = 0)

Ngoi cc hng ng thc ng nh, hc sinh cn nhcc hng ng thc

bxung l : an - bn = (a - b)(an -1+ an - 2b ++ abn - 2+ bn - 1), *n N

an

+ bn

= (a + b)(an -1

- an - 2

b +- abn - 2+ bn - 1), n l s tnhin l.

hc sinh d nh, cn ly cc trng hp c thnh : n = 2, 3, 4 v

trng hp c bit : xn - 1 = (x - 1)(xn - 1+ xn - 2++ x + 1).

Tutheo c im tng bi m bin i mt cch linh hot kh dng

v nh. Trong qu trnh thc hnh, nhiu khi sau cc bin i khcc thnh

phn c gii hn bng 0 ta vn gp gii hn dng v nh 00

mi ( thng l

n gin hn so vi gii hn ban u). Ti y ta tip tc qu trnh khn

khi gii hn cn tm khng cn dng v nh 00

th thi.

Bi tp tluyn

1)

3

4x 1

x 3x 2

lim x 4x 3

2) x 0(1 x)(1 2x)(1 3x) 1

lim x

3)100

50x 1

x 2x 1lim

x 2x 1

4)

n 1

2x 1

x (n 1) nlim

(x 1)

3. Loi 2 :0x x

f(x)lim

g(x)m f(x), g(x) cha cc cn thc cng bc v f(x0)=g(x0)= 0

Phng php :Nhn c tv mu vi biu thc lin hp tng ng ca

biu thc cha cn thc (gi tt lphng php nhn lin hp hay dng biu

thc lin hp) trc cc nhn t x - x0 ra khi cc cn thc, nhm khcc

thnh phn c gii hn bng 0. Biu thc cha cn thc c thl t, mu hay c

etoanhoc.tk


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6

tv mu ca phn thc cn tm gii hn ). Lu l c thnhn lin hp mt

hay nhiu ln kh dng v nh.

Cc cng thc thng c s dng khi nhn lin hp l :

3 32 23 33 3

( A B)( A B) = A - B , (A 0, B 0)

( A B)( A A B+ B ) =A B

Gio vin cn cho hc sinh thy c hai cng thc ny xut pht t hai

hng ng thc sau hc sinh d nh:

2 2

2 2 3 3

(a - b)(a + b) = a - b

(a b)(a ab + b ) = a b

V dp dng:

V d 8 : 8 2x 23x - 2 - x

L = limx - 4

Bi gii :Nhn c tv mu vi biu thc lin hp tng ng, ta

c :

8 2 2x 2 x 2

3x - 2 - x ( 3x - 2 - x)( 3x - 2 + x)L = lim lim

x - 4 (x - 4)( 3x - 2 + x)

2

2x 2 x 2

x 2

3x - 2 - x (x - 2)(-x + 1)lim lim

(x - 4)( 3x - 2 + x) (x - 2)(x + 2)( 3x - 2 + x)

x + 1 2 + 1 1lim

16(x + 2)( 3x - 2 + x) (2 + 2)( 3.2-2+2)

Vy8

1L =

16

etoanhoc.tk


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V d 9 : 91

x+2 1L lim

x+5 2

x

Bi gii :

91 1

( x+2 1)( x+2 1) ( x+5 2)x+2 1L lim limx+5 2 ( x+5 2)( x+5 2) ( x+2 1)

x x

1 1

(x + 2 - 1)( x+5 2) (x + 1)( x+5 2)= lim lim

(x + 5 - 4)( x+2 1) (x + 1)( x+2 1)x x

1

x+5 2 1 5 2= lim 2

x+2 1 1 2 1x

Vy L9 = 2

V d 10 :n

*10 m1

x - 1L lim , (m, n N )

x - 1

x

Bi gii :n

10 m1

n-1 n-2 m-1 m-2n n n n m m m

m-1 m-2 n-1 n-2m m m m n n n1

x - 1L lim

x - 1

( x - 1) ( x ) +( x ) +...+ x +1 ( x ) +( x ) +...+ x +1=lim

( x - 1) ( x ) +( x ) +...+ x +1 ( x) +( x) +...+ x +1

x

x

m mm-1 m-2 m

n n1 n-1 n-2 n

(x - 1)( x + x +...+ x+1)=lim

(x - 1)( x + x +...+ x+1)

x

m mm-1 m-2 m

n n1 n-1 n-2 n

x + x +...+ x+1 m=lim

nx + x +...+ x+1

x

Vy 10m

L =n

Kt lun:

Phng php dng biu thc lin hp l phng php ch yu c s

dng tnh cc gii hn c cha cn thc cng bc. C thxem y l thut

ton c bn cho php tnh c kh nhiu gii hn ca hm s cha cn thc,phng hng r rng, d hiu.Vic xc nh biu thc lin hp l khng qu

etoanhoc.tk


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8

kh khn i vi hc sinh. Tuy nhin gio vin cn rn luyn knng xc nh

v nhn biu thc lin hp khi tnh gii hn. Theo cch ny, nhiu bi ton tuy

gii c nhng phi qua cc php bin i di dng vi biu thc cng knh.

Nu dng cc gii khc nh thm bt, i bin s cho li gii ngn gn hn.

Bi tp tluyn

1)3

x 1

x x 3lim

x 1

2)

2

3x 2

x 4lim

2 3x 2

3)2 2x a

x b a blim

x a

4)

3 23

2x 1

x 2 x x 1lim

x 1

5)n

x 0

1 axlim

x

6)

n n

x 0

a x alim

x

4. Loi 3:0x x

f(x)lim

g(x)m f(x) cha cc cn thc khng cng bc v f(x0)=g(x0)= 0

Phng php : S dng thut ton thm bt i vi f(x) c thnhn

biu thc lin hp. Chng hn nh :

0 0

m n

m n0 0 0

x x x x

u(x) v(x)f(x)L= lim = lim ,( u(x ) v(x ) = 0,g(x ) = 0)

g(x) g(x)

Ta bin i :

0 0

0 0

m nm n

x x x x

m n

x x x x

u(x) - c + c - v(x)u(x)- v(x)L lim lim

g(x) g(x)u(x) - c v(x) - c

= lim limg(x) g(x)

Ti y cc gii hn0 0

m n

1 2x x x x

u(x) - c v(x) - cL lim , L lim

g(x) g(x) u tnh c

bng cch nhn lin hp.

V dp dng :

V d 11 :3

11 2x 1

x+3 x+7L lim x 3x+2

etoanhoc.tk


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9

Bi gii :

x 1 x 1

x 1 x 1

3 3

11 2 2

3

2 2

lim lim

lim lim

x+3 x+7 ( x+3 2) + (2 x+7)L

x 3x+2 x 3x+2

x+3 2 2 x+7

= x 3x+2 x 3x+2

23 3 3

2 2 23 3x 1 x 1

(2 x+7) 4 2 x+7 ( x+7)( x+3 2)( x+3+2)= lim lim

(x 3x+2)( x+3+2) (x 3x+2) 4 2 x+7 ( x+7)

2 2 23 3x 1 x 1

x+3 4 8 (x+7)= lim lim

(x 3x+2)( x+3+2) (x 3x+2) 4 2 x+7 ( x+7)

x 1 x 1 23 3

x 1 1 x= lim lim

(x 1)(x 2)( x+3+2) (x 1)(x 2) 4 2 x+7 ( x+7)

x 1 x 1 23 3

1 1= lim lim

(x 2)( x+3+2) (x 2) 4 2 x+7 ( x+7)

23 3

1 1= (1 2)( 1+3+2) (1 2) 4 2 1+7 ( 1+7)

1 1 1=

4 12 6

Vy 111

L6

V d 12 :3

12 2

0

1+2x - 1+3xL lim

xx

Bi gii :

33

12 2 20 0

1+2x - (x+1) + (x+1) - 1+3x1+2x - 1+3xL lim lim

x x

x x

3

2 20 0

1+2x - (x+1) (x+1) - 1+3x=lim +lim

x x

x x

etoanhoc.tk


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0 2

2 23 3 3

0 2 2 23 3

1+2x - (x+1) 1+2x +(x+1)=lim

x 1+2x +(x+1)

(x+1) - 1+3x (x+1) ( 1) 1+3x ( 1+3x)+lim

x (x+1) ( 1) 1+3x ( 1+3x)

x

x

x

x

2 3

2 2 2 23 30 0

2 23 30 0

(1+2x) - (x+1) (x+1) - (1+3x)lim lim

x 1+2x +(x+1) x (x+1) (x 1) 1+3x ( 1+3x)

- 1 x+3lim lim

1+2x +(x+1) (x+1) (x 1) 1+3x ( 1+3x)

x x

x x

2 23 3

- 1 0+3

1+2.0 +(0+1) (0+1) (0 1) 1+3.0 ( 1+3.0)1 1

12 2

Vy 121

L2

Kt lun :

Phng php chung tnh cc gii hn ca biu thc cha cc cn thckhng cng bc l thm, bt mt lng no , tch thnh nhiu gii hn ri

nhn lin hp. Cn lu l c ththm bt mt hng s( thng chn l u(x0)

hoc v(x0)) hay mt biu thc. Vic thm bt da trn c im tng bi v

phi tht tinh t. Thut ton thm bt cn c p dng hiu qui vi cc

dng v nh khc.

Bi tp tluyn

1)3

x 0

1 x 1 xlim

x 2)

3

2x 2

x 11 8x 43lim

2x 3x 2

3)n m

x 0

1 ax 1 bxlim

x

4)

3 2

x 0

2x 1 x 1lim

sinx

5)3

4x 7

x 2 x 20lim

x 9 2

6)

3

2x 0

1 4x 1 6xlim

x

5. Gii hn dng v nh 00

ca hm slng gic

etoanhoc.tk


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11

Phng php :Thc hin cc php bin i i sv lng gic s

dng cc kt qu gii hn c bn sau y :

+)x 0 x 0

sinx xlim 1, lim 1

x sinx

+)x 0 x 0 x 0

sinax sinax sinaxlim lim( .a) =a.lim =a

x ax ax

+)x 0 x 0 x 0 x 0 x 0

sinax sinax bx ax sinax bx ax alim lim( . . ) lim .lim .lim

sinbx ax sinbx bx ax sinbx bx b

+)x 0 x 0 x 0 x 0

tgax sinax a sinax alim lim( . ) lim .lim a

x ax cosax ax cosax

Trong qu trnh bin i, hc sinh cn vn dng linh hot cc cng thc lng

gic, thm bt, nhn lin hp

V dp dng

V d 13 : 13 x 01+sinax - cosax

L lim1- sinbx - cosbx

Bi gii :

13x 0 x 0

1+sinax - cosax 1- cosax+sinaxL lim lim

1- sinbx - cosbx 1- cosbx - sinbx

2

x 0 x 02

ax ax axax ax ax 2sin sin cos2sin +2sin cos 2 2 22 2 2=lim limbx bx bx bx bx bx2sin - 2sin cos 2sin sin - cos2 2 2 2 2 2

x 0 x 0

ax ax axsin sin cos a2 2 2=lim .lim

bx bx bx bsin sin - cos

2 2 2

Vy 13a

Lb

V d 14 : 14 2x 0

1 cosaxL lim

x

Bi gii :

etoanhoc.tk


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12

2 2

22 2 2

14 2 2x 0 x 0 x 0 x 0

ax ax ax2sin sin sin

1 cosax a a a2 2 2L lim lim lim . limax axx x 2 2 2

2 2

Vy2

14

aL

2

V d 15 : 15 201 xsinx - cos2x

L limsin xx

Bi gii :

15 2 20 0

1 xsinx - cos2x (1 - cos2x) xsinxL lim lim

sin x sin xx x

2

2 20 0 0

0 02

2sin x xsinx sinx(2sinx x) 2sinx xlim lim lim

sinxsin x sin xx x

lim 2 lim 2 1 3sin x sin x

x x x

x x

Vy L15 = 3

V d 16 : *16 2x 01- cosx.cos2x...cosnx

L lim (n N )x

Bi gii :

16 2x 0

2x 0

1- cosx.cos2x...cosnxL lim

x1-cosx+cosx-cosxcos2x+...+cosx.cos2x...cos(n-1)x-cosx.cos2x...cosnx

limx

2x 0

2 2 2x 0 x 0 x 0...

1-cosx+cosx(1- cos2x)+...+cosx.cos2x...cos(n-1)x(1- cosnx)lim

x1-cosx cosx(1-cos2x) cosx.cos2x...cos(n-1)x(1- cosnx)

lim lim limx x x

Theo kt qubi 14 ta c :

2

2x 0

1

2

1-cosxlim

x

2

2 2x 0 x 0 x 0

.cosx(1-cos2x) 1-cos2x 2

lim lim cosx lim2x x

etoanhoc.tk


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13

2x 02

2x 0 x 0 x 0 x 0. ... .

cosx.cos2x...cos(n-1)x(1- cosnx)lim

x1- cosnx n

lim cosx lim cos2x lim cos(n-1)x lim2x

Do

2 2 2 2 2 2

16

1 2 n 1 2 ... n n(n+1)(2n+1)

L ...2 2 2 2 12

Trong bi tp ny ta s dng thut thm bt :

cosx, cosxcos2x,, cosxcos2xcos(n - 1)x

bin i v tnh gii hn cho. C th nhn thy thut thm bt ng vai tr

quan trng trong knng bin i i vi bi tp ny.

V d 17 :2

17 2x 01 x cosxL lim

x

Bi gii :

2 2

17 2 2x 0 x 0

(1 x cosx 1 x 1) (1 cosx)L lim lim

x x

22 2 2

2 2 2x 0 x 0 x 0 x 02 2

x( ( 2

(

2sin1 x 1 1 cosx 1 x 1) 1 x 1)lim lim lim lim

x x xx 1 x 1)

2

22

2x 0 x 0 x 0 x 02 2 2

x x

2 2 .(

2sin sin1 x 1 1 1lim lim lim lim

x 2xx 1 x 1) 1 x 12

1 11

2 2

Vy L17 = 1.

Kt lun :

kh dng v nh i vi hm slng gic, hc sinh cn nm vng

v vn dng linh hot cc php bin i i s, lng gic cng nh p dng

cc gii hn c bn. y chc gii hnx 0

sinxlim 1

x c s dng trc tip,

cc kt qucn li khi lm bi phi chng minh li.

etoanhoc.tk


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vn dng gii hnx 0

sinxlim 1

x , cn a hm s cn tnh gii hn v

dng :0 0 0x x x x x x

sinf (x) f (x) tgf (x)lim , lim , lim

f (x) sinf (x) f (x) vi

0x xlim f (x) 0

bng cch

thm, bt, i bin hay nhn, chia ng thi vi mt lng thch hp no .

Trong khi gii bi tp, hc sinh c th gp kh khn, lng tng a vccdng trn. Gio vin cn khc phc bng cch cho hc sinh lm cc bi tp nh :

2

20 1

sinx sin(x 1)lim , lim , ...

1 cosx x 3x+2x x

Bi tp tluyn

Tnh cc gii hn sau :

1)0

1+sinx 1 sinxlim

tgxx

2)0

(a+x)sin(a+x) asinalim

xx

3)x 0

1 cosxcos2xco3xlim

1 cosx

4)

2

20

2sin x+sinx 1lim

2sin x 3sinx+1x

5)3

3x

4

1 cotg xlim

2 cotgx cotg x

6)

3

x 0

1 cosx cos2x cos3xlim

1 cos2x

6. Gii hn dng v nh 00

ca hm sm v lgarit.

Phng php : Thc hin cc php bin i v s dng cc gii hn c

bn sau y :

+)x

x 0

1lim 1

x

e

+) x 0ln(1 x)

lim 1x

Cc gii hn trn u c tha nhn hoc chng minh trong Sch gio khoa.

Ngoi ra gio vin cn a ra cho hc sinh hai gii hn sau :

etoanhoc.tk


15/30


15

+)x xlna

x 0 x 0

a 1 1lim lim .lna lna

x x.lna

e

( V

xlna

x 0

1lim 1

xlna

e

)

+)x 0 x 0 x 0

a .log (1 x) ln(1 x) ln(1 x)1

lim lim lim lnax x.lna lna x

V dp dng :

V d 18 :ax bx

18 x 0L lim

x

e e

Bi gii :

x 0 x 0

ax bxax bx

18

1) 1)lim lim

( (L

x x

e ee e

ax bx

x 0 x 0

ax bx

x 0 x 0

( 1) ( 1)lim lim

x x

( 1) ( 1)a. lim b. lim

ax bx

a b

e e

e e

Vy L18 = a - b.

Trong bi tp ny s dng gii hn c bn ta thc hin thm bt 1v tch thnh hai gii hn. Cn nhn mnh cho hc sinh khi x 0 th ax 0 ,

do vyax bx

x 0 x 0

( 1) ( 1)lim 1, lim 1

ax bx

e e

.

V d 19 :sin2x sinx

19 x 0L lim

sinx

e e

Bi gii :

sin2x sinxsin2x sinx

19 x 0 x 0

1) 1)( (L lim limsinx sinx

e ee e

sin2x sinx

x 0 x 0

sin2x sinx

x 0 x 0

1 1lim lim

sinx sinx

1 1lim .2cosx lim

sin2x sinx

e e

e e

x 0 x 0 x 0

sin2x sinx

. (2cosx)

1 1

lim lim limsin2x sinx

2 1 1

e e

etoanhoc.tk


16/30


16

Vy L19 = 1.

V d 20 :x 2

20 x 2

2 xL lim

x 2

Bi gii :x 2 x 2

20 x 2 x 2

4) 4)2 x (2 (xL lim lim

x 2 x 2

x 2x 2

x 2 x 2 x 2 x 2

x 2

x 2 x 2

1) 2)(x+2)4 4

14

4(2 (x2 xlim lim lim lim

x 2 x 2 x 2 x 2

2lim lim (x+2) 4ln2 4

x 2

Vy L20 = 4ln2 - 4

V d 21 :2

23 2x

21 2x 0

1 xL lim

ln(1+x )

e

Bi gii :

22 33 2 2x2 2x

21 2 2x 0 x 0

( 1)1 x 1) (1 xL lim lim

ln(1+x ) ln(1+x )

ee

2 23 32 2x 2 2x

2 2 2x 0 x 0 x 0

( 1) 11 x 1) ( 1 x 1lim lim lim

ln(1+x ) ln(1+x ) ln(1+x )

e e

3 32 23

32 2 23

2 2

2 2

2 2

x 0 x 02

2x( ( ) 1

( ) 1

.1 2x

ln(1+x )

1 x 1)( 1 x 1 x )lim lim

( 1 x 1 x )ln(1+x ) 2x

e

2

2

232 2 23

2

2

2x

x 0 x 0 x 02.

( ) 1

1 2x

ln(1+x )

xlim lim lim

2x( 1 x 1 x )ln(1+x )

e

22

2 232 23

2

2

2x

x 0 x 0 x 0 x 02. .

( ) 1

x 1ln(1+x )

2xln(1+x )

1 7.1 1.( 2)

3 3

1lim lim lim lim2x1 x 1 x

e

Vy 217

L3

Kt lun :

etoanhoc.tk


17/30


17

tnh cc gii hn dng v nh ca hm sm v lgarit, hc sinh thc

hin cc php bin i p dng cc gii hn c bn. Yu cu hc sinh phi

thnh tho cc php ton v lu tha v lgarit.

s dng cc gii hn c bn, bng cch thm, bt, nhn lin hp,

hc sinh phi bin i hm s cn tm gii hn v mt trong cc dng :

0 0 0 0

f(x) f(x)a

x x x x x x x x

ln 1+f(x) log 1+f(x)1 a 1lim , lim , lim , lim

f(x) f(x) f(x) f(x)

e

vi

0x xlim f (x) 0

Bi tp tluyn

Tnh cc gii hn sau :

1) 2)

x x

x xx 0

5

4 3lim9

3)x 0

2

2

x3 cosx

xlim

4)

x

3 4x 0

(1 )(1 cosx)lim

2x 3x

e

5)x 0

1 1 xlim .ln

x 1 x

6)sin2x sinx

2x 0lim

5x + tg x

e e

II. GII HN DNG V NH

Gii hn dng v nh

c dng l :

0x x(x )

f(x)L lim

g(x)

trong :0 0x x x x

(x ) (x )

f(x) g(x)lim lim

kh dng v nh ny, phng php thng thng l chia c tv mu

cho lu tha bc cao nht ca tv mu ca phn thcf(x)

g(x) . C thnh sau :

1) Nu f(x), g(x) l cc a thc c bc tng ng l m, n th ta chia c

f(x), g(x) cho xk

vi k = max{m, n}

m m 1m m 1 1 0

n n 1xn n 1 1 0

a x +a x +...+a x+aL lim

b x +b x +...+b x+b

vi *m na ,b 0, m,n N

Khi xy ra mt trong ba trng hp sau :

etoanhoc.tk


18/30


18

+) m = n (bc ca tv mu bng nhau), chia c tv mu cho xn ta

c:0m 1 1nn 1

m m

0n 1 1 n nnn 1

x x

m

n

aa aa ax xx

bb b b b

x xx

alim lim

b

+ +...+ +L

+ +...+ +

+) m > n (bc ca t ln hn bc ca mu, k = m), chia c tv mu cho

xmta c :

nm nm n+1

nm n

0m 1 1mm 1

m

x x0n 1 1m

m

b

xb

x

aa aa

ax xxlim limbb b

x x x

+ +...+ +L

+...+ ++

+) m < n (bc ca t nhhn bc ca mu, k = n), tng tnh trn ta c

:

0m m 1n m nn m+1

x0n 1

n n

aaa...

x xxlim 0bb

b ...x x

L

Hc sinh cn vn dng kt qu :

0 0 0 0x x x x x x x x

1 1lim f (x) lim 0, lim f (x) 0 lim

f (x) f (x)

Sau khi xt ba trng hp ny, hc sinh cn trt ra nhn xt kt qu gii

hn cn tm da vo bc ca tv mu. Lu l c th chia tv mu cho xh

vi

h min{m, n}.

2) Nu f(x), g(x) l cc biu thc c cha cn thc th ta quy c ly gi

tr mk

( trong k l bc ca cn thc, m l sm cao nht ca cc s hng

trong cn thc) l bc ca cn thc . Bc ca t ( mu) c xc nh l bc

cao nht cc biu thc trn t( di mu). Sau ta p dng phng php kh

nh vi trng hp f(x), g(x) l cc a thc. Qua hc sinh c th ddng

phn on kt qu gii hn dng

cn tm.

V dp dng :

V d 22 :3 2

22 3x

2x 3x 1L lim

5x 6

etoanhoc.tk


19/30


19

Bi gii : Chia c tv mu cho x3ta c :

3

3

3 2

22 3x x

3 12

2x x6 55

x

2x 3x 1L lim lim

5x 6

Vy 222

L5

. Ta c thtrnh by theo cch sau :

33 3

333

3 2

22 3x x x

3 1 3 1x 2 2x x 2x x6 56 5x 5xx

2x 3x 1L lim lim lim

5x 6

V d 23 :2 2

2x23 3x (2x 1)(3x x+2)2x+1 4xlimL

Bi gii :

2 2 4 2

2 2x x23

3x (2x 1)(3x x+2) 12x (2x+1)(3x x+2)

2x+1 4x 4x (2x+1)lim limL

3 2

3 2

2 3

x x

5x x+2

4x

5 1 24

4 1xx x4 8 28+x

4xlim lim8x

Vy23

1

2L

V d 24 : 24 5x(x 1)(x 2)(x 3)(x 4)(x 5)

L lim(5x 1)

Bi gii :

24 5x

(x 1)(x 2)(x 3)(x 4)(x 5)L lim

(5x 1)

5 5x

1 2 3 4 51 1 1 1 1

x x x x x 1

515

x

lim

Vy 24 51L 5

etoanhoc.tk


20/30


20

V d 25 : 25 x 2x+3

L limx 1

Bi gii :

Chia c tv mu cho x ta c :

25 x x2 2

3x+3 xL lim limx 1 x 1

x

1+

V phi a x vo trong cn bc hai nn ta xt hai trng hp :

*) 2x x > 0 x x

Khi :x + x + x +2 2

22

1

3 3 3x x xlim lim lim 1

x 1 x 1

x x x

1+ 1+ 1+

1

*) 2x x < 0 x x

Khi , ta c :x x x2 2

22

1

3 3 3

x x xlim lim lim 1x 1 x 1

xx x

1+ 1+ 1+

1

Vx x2 2

1, 1lim limx+3 x+3

x 1 x 1

nn khng tn ti

x 2

x+3lim

x 1

V d 26 :32 2

26 5x 4 4 4

9x 1 x 4L lim

16x 3 x 7

Bi gii : Chia c tv mu cho x ta c :

32 2

32 2

26 5 5x x4 44 4 4 4

9x 1 x 49x 1 x 4 x xL lim lim

16x 3 x 7 16x 3 x 7

x x

2

33

x 4 4

55

9x 1 1 4

x x xlim16x 3 1 7

x x x

Tng tBi 25, ta xt hai trng hp :

etoanhoc.tk


21/30


21

*) 4 42 , xx x > 0 x x x

Khi :

2

33 2 3

+26 4x x4 4

5

545

3

54

9x 1 1 4 19

9 0 3x x x x xL lim lim216 0316x 3 1 7

xx xxx

1 4x

1 7

16 x

*) 4 42 , xx x < 0 x x x

Khi ta c :

22

44 4

2233 3

33 3

26 x x x4 44 44 55 5

55 5

1

xx

3

xx

9x 1 1 49x 1 1 4 1 49

x xx x xx xL lim lim lim1 716x 3 1 7 16x 3 1 7 16x xx x xx x

2

4

33

44 5

5

x

1

x

3

x

1 49

9 0 3x x21 7 16 0

16x x

lim

V +26 26L L nn ta c : 26

3L

2

Kt lun :

So vi dng v nh 00

, dng v nh

dtm hn. Hc sinh cn xc

nh ng dng v ch cn quan tm n bc ca tv mu t phn on

kt qu gii hn cn tm. Ch i vi gii hn dng

ca hm sc cha

cn thc ta khng nhn lin hp. y l im khc bit cn phn bit trnh

nhm ln.

Vi gii hn khi x , cn lu hai kh nng x v x

trong php ly gii hn c cha cn bc chn. Nu hc sinh khng n vn

ny th rt d mc phi sai lm. Hn na trng hp ny cn lin quan ti bi

ton tm tim cn ca hm s cha cn thc.

Bi tp tluyn

1)

2 3

2 2x

2x 3 4x+7lim

3x 1 10x 9

2)20 30

50x

(2x 3) (3x+2)lim

(2x+1)

etoanhoc.tk


22/30


22

3)2 n

n+1xn 2

(x+1)(x 1)...(x 1)lim

(nx) 1

4)2

x 2

x 2x 3xlim

4x 4 x+2

5)34 5 2

4x 4 3

x 1 x 2lim

x 1 x 2

6)3

3 4x

ln(1 x x)lim

ln(1 x x)

III. GII HN DNG V NH

Dng tng qut ca gii hn ny l :

0x x(x )

lim f(x) g(x)

trong 0 0x x x x

(x ) (x )

lim f(x) lim f(x)

Phng php ch yu kh dng v nh ny l bin i chng v dng

v nh 0 ,0

bng cch i bin, nhn lin hp, thm bt,

V dp dng :

V d 27 : 227 xL lim x x x

Bi gii :

Nhn v chia biu thc lin hp tng ng l : 2x x+x , ta c :

2 2

227 x 2x

( x x x)( x x+x)L lim x x x

x x+xlim

2 2

2 2x x

x

x x+x x x+x

x x xlim lim

V x nn chia c tv mu cho x ta c :

2x x

1 1

21x x+x 1 1x

xlim lim

Vy 271

L2

Trong v dny, bng cch nhn lin hp, ta chuyn gii hn cn tmt dng sang dng

.

etoanhoc.tk


23/30


23

V d 28 : 28 xL lim x+ x x

Bi gii :

28 x x

( x+ x x)( x+ x x)L lim x+ x x lim

x+ x x

x x

x+ x x xlim lim

x+ x x x+ x x

x x 1

x 1 1lim lim

2x+ x x 1+ 1x

( chia c tv mu cho x )

Vy 281L2

V d 29 : 229 xL lim x x 3 x

Bi gii : Trong v dny cn lu khi x cn xt hai trnghp x v x

+) Khi x

th :2 2x x 3 xx x 3

Do 2xlim x x 3 x

+) Khi x th gii hn c dng . Ta kh bng cch nhn linhp bnhthng

2 22

x x 2( x x 3 x)( x x 3 x)lim x x 3 x lim

x x 3 x

2 2

x x x2 2 2

31

x x 3 x x 3lim lim lim

x x 3 x x x 3 x x x 31

x

x

Khi x th x < 0, do 2x x

etoanhoc.tk


24/30


24

x x2

2

3 31 1 1 1x xlim lim

2 21 3x x 3 1 11x xx

Vy2

xlim x x 3 x ,

2

x

1lim x x 3 x 2

Qua v dny mt ln na nhn mnh cho hc sinh ch vi gii hn khi

x cn xt x v x i vi hm s cha cn thc bc chn.

V d 30 : 3 3 2 230 x

L lim x 3x x 2x

Bi gii :V hm s cn tm gii hn cha cc cn thc khng

cng bc nn ta thm bt c thnhn lin hp.

3 33 2 2 3 2 230 x x

)L lim x 3x x 2x lim ( x 3x x ( x 2x x)

x x

3 2 23

1 2lim limx 3x x x 2x x G G

+)

3 2 3

213 2 3 2 2

3 3

3 23

x x

x 3xG

x 3x x x 3x x

xlim x 3x x lim

2 2

3 2 3 2 23 3

3 3

x x

3 3 31

33 3x 3x x x 3x x1 1 1

x xx

lim lim

+)2 2

2

2

x x2

x 2x x

x 2x xG lim x 2x x lim

2x x

2 2 21

22x 2x x1 1

xx

lim lim

Vy L30 = G1 - G2 = 2

Vd 31 : *31 m nx 1m n

L lim , (m, n N )1 x 1 x

Bi gii :

etoanhoc.tk


25/30


25

31 m n m nx 1 x 1

m n m 1 n 1L lim lim

1 x 1 x 1 x 1 x 1 x 1 x

1 2m nx 1 x 1

m 1 n 1lim lim G G

1 x 1 x 1 x 1 x

+)2 m 1

1 m mx 1 x 1

m 1 m (1 x x ... x )G lim lim

1 x 1 x 1 x

2 m 1

mx 1

(1 x) (1 x ) ... (1 x )lim

1 x

m 2

m 1x 1

(1 x) 1 (1 x) ... (1 x ... x )lim

(1 x)(1 x ... x )

m 2

m 1x 1

1 (1 x) ... (1 x ... x ) 1 2 ... m 1 m 1lim

1 x ... x m 2

Tng tta tnh c 2n 1

G2

Vy 31 1 2m 1 n 1 m n

L G G2 2 2

Trong bi tp ny ta s dng thut ton thm, bt tch gii hn cn tmthnh hai gii hn v tnh cc gii hn ny bng cch bin i v dng

0

0. Vic

thm bt biu thc phi tinh tv ph thuc vo c im tng bi.

Kt lun :

i vi dng v nh , ta phi tuvo c im tng bi m vn

dng linh hot cc knng thm bt, nhn lin hp, phn tch thnh nhn t

bin i v kh dng v nh. Ta thng chuyn chng vcc dng v nh d

tnh hn l0

0,

.

Bi tp tluyn

1)xlim x x x x

2) 2 23 3

xlim (x 1) (x 1)

3)x

lim x x x x x x

4) 3 2

xlim x 1 x

etoanhoc.tk


26/30


26

5) xlim ln(5x 8) ln(3x 5)

6) 5x

lim (x 1)(x 2)...(x 5) x

IV. GII HN DNG V NH 0.


0

x(x x )

lim f (x).g(x)

trong 0 0

x x(x x ) (x x )

lim f (x) 0, lim g(x)

kh dng v nh ny, ta thng tm cch chuyn chng v dng gii

hn khc dtnh hn nh0

0,

bng cch nhn lin hp, thm bt, i bin

V dp dng :

V d 32 : 232x

L lim x x 5 x

Bi gii : Ta kh dng v nh ny bng cch nhn lin hp

a v dng v nh

2 2

232 2x x

x( x 5 x)( x 5 x)L lim x x 5 x limx 5 x

etoanhoc.tk


27/30


27

2 2

2 2 2x x x

x(x 5 x ) 5x 5lim lim lim

x 5 x x 5 x x 5 x

x

2x x > 0 x x

Do :2x x

2

5 5 5lim lim

25x 5 x1 1

xx

Vy 325

L2

V d 33 : 33x 1

xL lim(1 x)tg

2

Bi gii : t t 1 x ta c :

x 1 t 0

33t 0 t 0

(1 t) tL lim t.tg lim t.tg

2 2 2

t 0 t 0 t 0

t

t t 2 22lim t.cotg lim limt t2

tg tg2 2

Vy33

2L

Bi tp tluyn

1) 2

xlim x 4x 9 2x

2) 2 4 4

xlim x 3x 5 3x 2

3) 2 33xlim x 4x 5 8x 1

4)x

4

lim tg2x.tg x4

5) 2 2x a

xlim a x tg

2a

6)1 1

2 x x

xlim x e e 2

V. GII HN DNG V NH 1


etoanhoc.tk


28/30


28

0

g(x)

x xlim f(x)

, trong

0 0x x x xlim f(x) 1, lim g(x)

Hai gii hn c bn thng c s dng khi tnh gii hn dng v nh 1 l :

+)

x

x

1lim 1 x e

(1)

+) 1x

xlim 1 x e

(2)

Trong qu trnh vn dng, hc sinh bin i v dng

0x x

f(x)

1lim 1f(x)

e

nu0x x

limf(x)

0

1

g(x)

x xlim 1 g(x) e

nu

0x xlim g(x) 0

bin i gii hn cn tm, hc sinh vn dng mnh sau (da vo

tnh lin tc ca hm sm).

Nu hai hm s f(x), g(x) thomn cc iu kin :1)

0x xlim f (x) a 0

2)0x x

lim g(x) b

th 0

g(x) b

x xlim f (x) a

Hai gii hn c bn v mnh trn l c stnh cc gii hn dng v

nh 1

V dp dng

V d 34 : 1

x34

x 0L lim 1+ sin2x

Bi gii :

sin2x1 1 sin 2x 1 x.

x sin 2x x sin 2x34x 0 x 0 x 0L lim 1+ sin2x lim 1+ sin2x lim 1+ sin2x

etoanhoc.tk


29/30


29

Ta c : 1

sin2x

x 0lim 1+ sin2x e

( hc sinh d hiu nn t t = sin2x)

x 0 x 0

sin2x sin2xlim 2lim 0

x 2x

Do :

sin2x1 x

2sin2x

34x 0

L lim 1+ sin2x e

V d 5 :4 3x

35x

x 1L lim

x 2

Bi gii : s dng gii hn c bn ta bin i :x 1 1

1x 2 (x 2)

4 3x(x 2).4 3x

(x 2)

35x x

x 1 1L lim lim 1

x 2 (x 2)

V

(x 2)

x

x x x

1lim 1 e

(x 2)

4

34 3x 3x 4 xlim lim lim 32(x 2) x 2

1x

nn 335L e

Bi 36 :tg2 y

4

36t 0

L lim tg y4

Bi gii : t y x , x y 0

4 4

.Ta c :

21 tg y

tg2 y2tgy4

36t 0 t 0

1 tgyL lim tg y lim

4 1 tgy

2

22tgy 1 tg y

.1 tg y 1 tgy 1 tgy 2tgy

2tgy 2tgy

t 0 t 0

2tgy 2tgylim 1 lim 1

1 tgy 1 tgy

etoanhoc.tk


30/30


V

1 tgy

2tgy

y 0

2tgylim 1 e

1 tgy

v

2

y 0 y 0

2tgy 1 tg y. 1 tgy 1

1 tgy 2tgylim lim

nn 136

L e

Kt lun :

Vi dng v nh 1 , vic nhn dng khng kh khn i vi hc sinh.

Tuy nhin, lm c bi tp, hc sinh phi vn dng tt cc knng a

cc gii hn cn tm v mt trong hai gii hn c bn (1) v (2). Hai knng

ch yu c s dng l i bin v thm bt.

Bi tp tluyn

1) 2cotg x

2

x 0lim 1 x

2)

1

sinx

x 0

1 tgxlim

1 sin x

3)

2x

2

2x

x 3lim

x 2

3)

cot g x

x 1lim 1 sin x

5)2

1

x

x 0lim(cos2x)

6)

x

x

1 1lim sin cos

x x

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