4_terehina-li-fik / posobie

Добавил:

Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.

Вуз:

Томский Государственный Университет

Предмет:

[НЕСОРТИРОВАННОЕ]

Файл:

[ 1 + (x) ] (x) = e:

kAK WIDNO, STRUKTURA PREDELA PREDPOLAGAET, ^TOBY OSNOWANIE STEPENI

BYLO PREDSTAWLENO W WIDE [1 + (x)]

GDE (x) - B.M.W., A POKAZATELX

.PDF

Скачиваний:

Добавлен:

30.05.2015

Размер:

2.22 Mб

Скачать

☆

<<< < Предыдущая 1 2 34 / 174 5 6 7 8 9 10 11 12 13 14 15 16 17 > Следующая >>>

1 +

! ;

;

= lim

x!0

29:

lim arcsin2 3x =

arcsin 3x

ln 2

2;7x

;

2;7x

;

= lim

!(3x)2

;

= lim

= 0:

;7x ln 2

x!0

x!0 7

ln 2

30:

lim

; cos 5x

x!0 ln (1 + x

arctg 6x)

25x2

;

cos 5x

(5x)2

= lim

ln(1 + x arctg 6x) x arctg 6x 6x

cos3 7x

31:

;

lim

1 + 3 sin 3x ;

(7x)

49x

1 ; cos3 7x = (1 ; cos 7x)(1 + cos 7x + cos2

7x)

3 = 3

q1 + 3 sin2 3x ;

3 sin2 3x

(3x)2

9x2

= 3

lim

49x2=2

x!0

3 9x2=2

+ x

32:

lim

0 p1 + 4x2

;

+ x = x(x + 1)

x (x + 1) ! 1 =

p1 + 4x2 ; 1 2x2

lim

= lim

2x2

x!0

33:

lim

;

x ;

2 = t

= x = t + 2

;

+ 2)

3(t + 2)

+ 12t

+ 8

+ 6t

lim

;

= lim

;

t2 + 4t

(t + 2)2

;

+ 4

;

= lim t(t

= lim t

+ 6t

+ 9t

+ 6t + 9)

+ 6t + 9 =

t!0

t2 + 4t

t!0

t(t + 4)

t!0

t + 4

pODOBNYE PREDELY MOVNO RE[ATX I DRUGIM SPOSOBOM. kAK IZWESTNO

IZ ALGEBRY, ESLI MNOGO^LEN OBRA]AETSQ W NOLX PRI ZNA^ENII x = x0 TO RAZLOVENIE NA MNOVITELI \TOGO MNOGO^LENA OBQZATELXNO BUDET SO-

DERVATX MNOVITELX (x ; x0): dRUGIE MNOVITELI MOVNO OPREDELITX,

NAPRIMER, RAZDELIW ISHODNYJ MNOGO^LEN NA WYRAVENIE (x ; x0) ILI

DRUGIMI PRIEMAMI. nAPRIMER:

34:

lim x2

;

3x + 2

x!2 x2

;

5x + 6

~ISLO x0

= 2 QWLQETSQ KORNEM KWADRATNYH TREH^LENOW. nAHODIM DRUGOJ

KORENX KAVDOGO TREH^LENA, LIBO PROSTO RE[IW KWADRATNYE URAWNENIQ

;

3x + 2 = 0 =)

x1 = 2 x2 = 1

;

5x + 6 = 0 =)

x1 = 2 x2 = 3

LIBO ISPOLXZUQ TOT FAKT, ^TO SWOBODNYJ ^LEN KWADRATNOGO URAWNE-

NIQ WLQETSQ PROIZWEDENIEM EGO KORNEJ. t.E. IZ PERWOGO URAWNENIQ

x2 = 2:

t K

. x1

= 2

= 1:

aNALOGI^NO DLQ WTOROGO URAWNENIQ

x2 = 6:

t K

. x1

= 2

= 3:

tAKIM OBRAZOM, SOGLASNO TEOREME wIETTA, MOVNO ZAPISATX

;

3x + 2 = (x ; x1)(x

; x2) = (x ; 2) (x ; 1)

;

5x + 6 = (x ;

(x ;

3):

pODSTAWLQEM \TI RAZLOVENIQ WMESTO KWADRATNYH TREH^LENOW I SOKRA-

]AEM SKOBKU (x

;

2) :

lim

;

2)(x

;

= lim x

; 1

;

x!2

(x ; 2)(x ; 3)

x!2 x ; 3

2 ; 3

x + 2 = t

35:

lim

px ; 6 + 2

x = t

;

x!;2

x + 2

t ! 0

6 + 2

8 + 2

= lim

;

= lim

;

= lim

;

t!0

; (t=8)

;2 3 (;8)

= lim

2 1 ; q1

= lim

t!0

pRI RE[ENII PREDELOW, SODERVA]IH RADIKALY, MOVNO ISPOLXZOWATX PRIEM UMNOVENIQ I DELENIQ \TIH WYRAVENIJ NA SOPRQVENNOE WYRAVE-

NIE, TAK, ^TOBY POLU^ILASX FORMULA RAZNOSTI KWADRATOW (SUMMY ILI

RAZNOSTI KUBOW).

lim p

36:

x2 ; 5

;

x!3 p5x

+ 1

;

= lim

;

; 5 + 2)

(

5x + 1 + 4)

(

5 ; 2)

(

x!3

(px2

5 + 2)

(p5x + 1

(p5x + 1 + 4)

;

5x + 1 + 4)

= lim

; 5

; 4)

(

5x + 1 + 4)

= lim

; 9)

(

(5x + 1

16)

(5x

15)

x!3

(px2

;

5 + 2)

;

x!3 (px2

;

5 + 2)

;

= lim

(x ;

= lim

(

3)(x + 3)

(

5x + 1 + 4)

(x + 3)

5x + 1 + 4)

x!3

(px2

5 + 2)

5(x

x!3

(px2

5 + 2)

(3 + 3)

16;

+ 4)

;

(

4 + 2) 5

4 5

rASSMOTRIM RE[ENIE PRIMERA 35 ISPOLXZUQ UMNOVENIE I DELENIE ^IS- LITELQ NA SOPRQVENNOE WYRAVENIE. pOSKOLXKU MY IMEEM KORENX TRETX- EJ STEPENI, TO SOPRQVENNYM BUDET TAKOE WYRAVENIE, POSLE UMNOVENIQ NA KOTOROE POLU^IM FORMULU SUMMY KUBOW

(a + b) (a2 ; ab + b2) = a3 + b3:

iTAK, SOPRQVENNYM QWLQETSQ NEPOLNYJ KWADRAT SUMMY ^ISEL a I b:

						3
						3							6 + 2								0
	35:					lim			px ;				6 + 2						=		0			!				=
	35:					lim			px ;										=		0							=
					x!;2							x + 2							3		0												3
					x!;2
						3																							2													2
							( px ; 6 + 2)											( q(x				; 6)							2	; 2			px		;		6 + 2					2	)
= lim																		( q(x																			6 + 2						)		=
																																							2

		x	!;	2				(x + 2)							3						6)				2		;					3								)
			!;					(x + 2)								( q(x ;					6)						;		2 px ; 6 + 2											)
															3								3							3
															3					6)			3			+ 2				3
= lim																( px ;				6)						+ 2												=
													3			( px ;				2									3							2

		x	!;	2																	;					2						; 6 + 2					)
			!;				(x + 2) ( q(x ; 6)														;					2		px				; 6 + 2					)
= lim																	(x ;			6 + 8)																		=
													3				(x ;			2									3							2

		x	!;	2			(x + 2) (								(x ; 6)						;					2						; 6 + 2					)
			!;				(x + 2) (						q		(x ; 6)						;					2		px				; 6 + 2					)
= xlim2													q						(x + 2)																			=
													3							2									3							2
															1

			!;				(x + 2) (						q
			!;				(x + 2) (								(x ; 6) ; 2 px ; 6 + 2 )
= xlim2																																	=
						3							2						3												2	)


			!;				( q(x ; 6)							; 2 px ; 6 + 2
=											1																	=					1								=				1	=	1	:
			3							2			3											2						3				3										4+2 2+4


		( q(;2; 6)									;2 p;2;6+2															)				p64+2 p8+4																	12	35

37:

lim

sin 5x

;

= t

= lim

sin 5(t +

)

x = t +

et+

;

sin(5t + 5 )

(et

sin 5t

= lim

;

e (1

; et)

e (1 ; et)

t!0

sin 5t 5t

= lim

;5t

e (;t)

t!0

;

= t

38:

lim

cos 5x

;

cos 3x

= t +

sin2 x

cos(5t + 5 )

cos(3t + 3 )

cos 5t

cos 3t

= lim

;

= lim

(

;

t!0

sin2(t + )

t!0

;

sin t)2

= 1 ; cos 3t

(3t)2

cos 3t 1 ;

9t2

cos 5t 1

;

25t2

sin t t

(1 ;

9t2

25t2

16t2

= lim

2 ) ; (1 ; 2 )

= lim

= 8:

t!0

;x2

39:

lim

; pcos x = lim

1 ; q1

;

=2) = lim

x!0

x sin 3x

x!0

x 3x

x!0

3x2

1.3.4. nEOPREDELENNOSTX WIDA

(0 1)

dLQ RASKRYTIQ DANNOJ NEOPREDELENNOSTI NEOBHODIMO SNA^ALA PRIWES-
TI EE K WIDU	0	! ILI	1! ^TO OBY^NO MOVNO SDELATX, UBRAW ODIN IZ
	0		1
MNOVITELEJ W ZNAMENATELX KAK OBRATNU@ WELI^INU, T.E.
(0 1) = 1	=	1!	(0 1) =	0		=	00!
					1
		1	1
0

rASKRYTIE POLU^A@]IHSQ NEOPREDELENNOSTEJ UVE RASSMOTRENO RANEE. pRIWEDEM PRIMERY.

sin x

40:

lim(x

;

= (0

) = lim(x

;

x ; 1 = t

cos

2 x

;

= lim

x = t + 1

= lim

x!1

cos

t !t0

t!0

cos

(t + 1)

= lim

;

; sin

cos

2 t + 2

2 t

;2 t

; 1

41:

lim x

= (

0) = lim

x!1

;

x!1

= x

! 1 ) x ! 0

= lim

= 1:

x!1

; 1 x

42:

lim (n2 + 5n)

sin

= (

0) =

2n2

n!1

;

n ! 1 )

! 0

;

2n2

= lim (n2

+ 5n)

n!1

;

2n2

sin 3 2n2

;

2n2

n;+ 5n

= lim

;2n2

n!1 3 ; 2n2

n!1

1.3.5. nEOPREDELENNOSTX WIDA (11)

dANNU@ NEOPREDELENNOSTX WSEGDA RACIONALXNO RASKRYWATX, ISPOLXZUQ FORMULU 2-GO ZAME^ATELXNOGO PREDELA

STEPENI DOLVEN BYTX WELI^INOJ, OBRATNOJ \TOJ B.M.W.

= (11) = lim 8 1 + ;2

;2 5x

43: lim

; 3x!

3x !

!1 <

xlim

1 + ;2

; 2

= e

= e(;10=3):

lim

5x = lim

;10x

= ;10

x!1

dLQ

FORMIROWANIQ 2-GO ZAME^ATELXNOGO

PREDELA MY OSNOWANIE STEPE-

NI ZAPISALI W WIDE

1 + ;3x2

! GDE

;3x2! - B.M.W.

w POKAZATELE STEPE-

NI UMNOVILI I RAZDELILI NA WELI^INU, OBRATNU@ \TOJ B.M.W. dALEE

WYDELILI ^ISLO e I NA[LI PREDEL OSTAW[EGOSQ W POKAZATELE STEPENI

WYRAVENIQ.

3(2x2+x)

x2+5

44: xlim 1 + x2 + 5!

=xlim

+ 5!

>1 + x2

x2+5

lim

1 +

= e

x2 + 5!

x!1

= e6:

lim

3(2x2 + x)

= lim

6x2

= 6

x!1

+ 5

x!1 x2

(

;x2

45: lim 1

= (11) = lim 8

1+ ;1

;x2

!1 <

lim

1 +

;

= e

x!1

= e0 = 1:

lim

;2x

= lim

= 0

x!1

x(x3)

x +4

46: xlim 1+ x2 + 4!

=(11)=xlim

> 1+ x2 + 4!

x2+4

lim

1 +

= e

x2 + 4!

x!1

= e+1 = +1:

x x3

= lim x4

lim

= +

x!1 x2 + 4

x!1 x2

= (11) =

sin 3x

47:

lim (1 + sin 3x)ln(1;5x)

x!0

ln(1

; 5x) ;5x

(

) ;

5x = e;3=5:

(1 + 3x)3x

= lim

1 + arcsin2 p

x!0

48: lim

p1+7x;1 = (11) =

arcsin2 p

)2 = x

x 2

(

)

= lim

(1 + x)x

= e

p1 + 7x ; 1

2 7x

x!0

(x "ln x +x

#) =

49: xlim!1 fx [ln(x + 3) ; ln x]g = xlim!1

= xlim

1 +

= (0

) = xlim

8ln

1 + 3

x9 =

!1 (

x!)

= (11) = ln lim

1 +

= ln e3 = 3:

5x + 4

3x;1

50:

xlim

= (11) =

;

5x + 4=1 +

5x + 4

;

1= 1+ (5x+4);(5x;2) = 1+

5x ; 2

5x ;

5x ; 2

; 2

5x6; 2

(3x

;

= lim

1 +

9(5x ; 2) 4

!1 <

;

5x ; 2

lim

1 +

= e

x!1

5x ; 2

= e0 9:

lim

6 (3x

; 1)

= lim

6 3x

x!1 (5x ; 2) 4

x!1

5x 4

dLQ FORMIROWANIQ WTOROGO ZAME^ATELXNOGO

PREDELA W DANNOM PRIMERE

NEOBHODIMO W OSNOWANII STEPENI PRIBAWITX I OTNQTX EDINICU I SDE- LATX NEOBHODIMYE PREOBRAZOWANIQ, ^TOBY POLU^ITX BESKONE^NO MALU@ DOBAWKU K EDINICE. zATEM SFORMIROWATX POKAZATELX STEPENI, KAK \TO DELALOSX W PREDYDU]IH PRIMERAH.

51: xlim

0xx22 ; 42xx ++

211x

= (11) =

;

; 2x + 1

= 1 + x2

; 2x + 1

1 = 1 +

2x ; 1

x2 ; 4x + 2

;

x2 ; 4x + 2 ;

; 4x + 2

(2x

1) x

2x 1

x ;

4x + 2

x ; 4x + 2

;

4x + 2

1 +

;

= lim

;

4x + 2

xlim

1 +

2x ; 1

= e

= e2:

;

4x + 2

lim

(2x ; 1) x = lim

= 2

x!1 x2 ; 4x + 2

x!1 x2

pRI WY^ISLENII PREDELOW PODOBNOGO WIDA MOVNO PREDWARITELXNO UPROSTITX WYRAVENIE, ZAMENQQ BESKONE^NO BOLX[IE MNOGO^LENY BOLEE PROSTYMI, WYDELIW W NIH GLAWNYE ^LENY. nAPRIMER,

52: xlim

0xx22

;

42xx ++

211x = (11) =

;

; 2x + 1

= 1 + x

; 2x + 1

;

1 = 1+

2x ; 1

= 1+

; 4x + 2

x2 ;4x+2

1 + 2

= lim

= e2

!1 <

w SLEDU@]IH PRIMERAH DLQ WYDELENIQ EDINICY W OSNOWANII STEPENI UDOBNO WWESTI ZAMENU PEREMENNOJ.

53:

lim (5

;

2x)

= (11) =

x!2

;

2 = t

x = t + 2

;

! 0

;

2x = 5

; 2(t + 2) = 1

; 2 = p4 + 2t ; 2 = 2(

1 + t=2

; 1)

= t=2

( 2)

= lim (1

;

2t) (t=2) = lim (1

;

2t) t =lim

8(1+( 2t))(;2t) 9

;

= e;4:

t!0

;

54:

lim

(x 1)3

;

= 1

x!1

;

1 = t 2

x = t + 1

= 1 4t t

1 4t

= 4 2(t + 1) (t + 1)

;

(x ; 1)3

(

4t)

;t3

= lim [1 + (

;

4t)]t3 = lim 8[1 + (

;

4t)](

;

4t) 9

>( 4)

= lim 8[1 + (

:;2

;

4t)](

;

4t) 9 t

= e;4=0 = e

= 0:

z A M E ^ A N I E.

pRI WY^ISLENII PREDELOW, PODOBNYH RASSMOTREN-

NYM W DANNOM PARAGRAFE, NEOBHODIMO WSEGDA PREVDE WSEGO UBEDITXSQ W

NALI^II NEOPREDELENNOSTI WIDA (11) :

w ZAKL@^ENIE PRIWEDEM PRIME-

RY PREDELOW, KOTORYE NE IME@T OTNO[ENIQ KO WTOROMU ZAME^ATELXNOMU

PREDELU, TAK KAK ONI NE SODERVAT NEOPREDELENNOSTI DANNOGO WIDA.

55:

nlim

0n2 ;22n+1

15n =

lim

;2n+1

= lim

= 1

3n ;

n!1 3n2 ;2

n!1 3n2

= lim

5n =

= 0:

n!1

x2 ; 2x + 1

5x=

;1=3+1=+

58:

lim

= lim

!;1

3x2

;

!;1

1.4. nEPRERYWNOSTX FUNKCII

1.4.1. oDNOSTORONNIE PREDELY

dO SIH POR MY RASSMATRIWALI PREDEL FUNKCII W TO^KE, POLAGAQ, ^TO ON NE ZAWISIT OT TOGO, S KAKOJ STORONY MY PODHODIM K TO^KE x0: sU]ESTWUET, ODNAKO, MNOGO PREDELOW, W KOTORYH \TO QWLQETSQ SU]EST- WENNYM, I WELI^INA PREDELA ZAWISIT OT TOGO, SLEWA ILI SPRAWA OT TO^KI x0 MY NAHODIMSQ.

pUSTX x STREMITSQ K TO^KE x0 OSTAWAQSX WSE WREMQ MENX[E x0: zA-

PISYWAETSQ \TO TAK:
x ! x0 ; 0	(x < x0):

sOOTWETSTWENNO PREDEL FUNKCII PRI x ! (x0 ;0) NAZYWAETSQ PREDELOM FUNKCII SLEWA ILI LEWOSTORONNIM PREDELOM I OBOZNA^AETSQ

lim f(x):

x!x0;0

aNALOGI^NO: PUSTX x STREMITSQ K TO^KE x0 OSTAWAQSX WSE WREMQ BOLX[E x0: zAPISYWAETSQ \TO TAK:

x ! x0 + 0 (x > x0):

sOOTWETSTWENNO PREDEL FUNKCII PRI x ! (x0 +0) NAZYWAETSQ PREDELOM

FUNKCII SPRAWA ILI PRAWOSTORONNIM PREDELOM I OBOZNA^AETSQ

lim

f(x):

x!x0+0

pRIMERY.

1: y = arctg

x ; 2

nAJDEM PREDELY \TOJ FUNKCII SLEWA I SPRAWA OT TO^KI x = 2:

lEWOSTORONNIJ PREDEL:

lim arctg

= arctg

= arctg(

) =

x ; 2

2 ; 0 ; 2

x!2;0

<<< < Предыдущая 1 2 34 / 174 5 6 7 8 9 10 11 12 13 14 15 16 17 > Следующая >>>

Соседние файлы в папке 4_terehina-li-fik

#
30.05.2015516.46 Кб13main.pdf
#
30.05.2015444.12 Кб13main_uch_posobie.pdf
#
30.05.20152.22 Mб14posobie.PDF
#
30.05.2015700.17 Кб14programma.pdf