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Национальный университет кораблестроения им. адм. Макарова

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Vtoroy_modul_po_analizu_files / ФОРМУЛА ТЕЙЛОРА И ЕЕ ПРИЛОЖЕНИЯ К ВЫЧИСЛЕНИЮ ПРЕДЕЛОВ ФУНКЦИЙ

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<<< < Предыдущая 1 23 / 43 4 > Следующая >>>

f (x) = x + x2 +

5x3

+ o x3 +

+ −x −

−

+ o x3

−

+ o x3

lim

f (x)

= lim

+ o x3

5x3

x→0 g (x)

x→0

+ o(x3)

tg xe−x2

− ln(ch2 x)

lim

arctg (x cos) − tg x

x→0

tg x arctg x

o x3

cos x x

cos x o x2

x cos x = x

1 −

+ o x2

= x −

+ o x3

arctg t = t −

, t → 0.

tg x = x +

+ o x3

+ o t3

# arctg (x cos x) =

(x + o(x))3

= x −

+ o x3

−

+ o x3

= x −

+ o x3

−

+ o x

+ o x3

= x −

+ o x3

g (x) = x −

5x3

7x3

+ o x3

−

x +

+ o x3

= −

+ o x3


o x3	'	o x2

	x # !

x2 $

$( et = 1+t+o(t) −x2

t " xe−x2

= x

1 − x2 + o x2

= x − x3

+ o x3

tg t = t +

t → 0,

+ o

(x + o(x))3

tg xe−x

x − x3 + o x3

+ o x3

x3 + o x3

= x −

2x3

= x − x3 + o x3

+ o x3

)

o x4 *

ch2 x = 2 ln (ch x) = 2 ln 1 +

+ o x4

+ o x2

= 2

+ o x4

−

+ o x4

−

= 2

+ o x4

= x2

+ o x4

2x3

f (x) = x−

+o x3

−

x2 −

+o x4

= −

+o x3

lim

f (x)

= lim

−

+ o x3

7x3

x→0 g (x)

x→0 −

+ o(x3)

1/4

√

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

1 − x − 2 cos x

lim

√

− (x + 1) ch

x→0

exp

√ x

1−4x

" # $

o x3

√

# & o x

# &

1−4x

o x2

√

= 1 +

3t2

+ o t2 , t → 0,

1 − t

√

3 (4x)2

1 − 4x

= x

1 +

+ o x2

= x + 2x2 + 6x3 + o x3

exp

√1 − 4x

= exp

x + 2x2

+ 6x3 + o x3

x + 2x2

+ o x2

= 1 + x + 2x2 + 6x3 + o x3

(x + o(x))

+ o x3

x2 + 4x3 + o x3

= 1+ x + 2x2 + 6x3 +

+ o x3

= 1 + x +

5x2

49x3

+ o x3 .

" # $

' x + 1 1 x → 0 ! # & o x3 * ch √5x = 1 + 5x22 + o x3 .

√

(x + 1) ch

= (x + 1) 1 +

+ o x3

+ o x3 .

g (x) = 1 + x +

= 1 + x +

5x2

5x3

5x2

49x3

+ o x3 −

5x + o x3

= 17x

+ o x3

− 1 + x + 5x +

" # $

o x3 '

! #

# &

e # ! #

ln (e (1 + 2x)) = ln e + ln (1 + 2x) =

= 1 + 2x −

(2x)2

(2x)3

+ o x3 = 1 + 2x −2x2 +

8x3

+ o x3 .

(1 + t)1/4 = 1 +

· −

+ o t3

3t2

7t3

= 1 +

−

+ o t3

128

8x3

1/4

(ln (e (1 + 2x)))1/4 =

1 + 2x − 2x2 +

+ o x3

2x − 2x

8x3

+ o x

−

3 2x − 2x2 + o x2

= 1 +

+ o x3 =

7 (2x + o(x))3

128

()

2x3

3x2

3x3

7x3

= 1 +

−

+ o x3

−

+ o x3

√

1 − x = 1 −

−

+ o x3

cos x = 1 −

+ o x3

f (x) = (ln (e (1 + 2x)))1/4 + √1 − x − 2 cos x =

= 1

−2+

+ 2

−2

+ −

−

3x2

−

+x2+

2x3

3x3

7x3

−

+ o x3 =

x3 + o x3

+ o x3

lim

f (x)

= lim

x3 + o x3

17x3

x→0 g (x)

x→0

+ o(x3)

136

− 2

√

ch x + cos 2+x2

1 + 3x4

lim

x arctg x − exp

x→0

+ 1

1+x2

# $ % &

% ' (

) ( x

( * #

% & $ % & (

% & $ ( (


( + x · arctg x x2			x2	x2
			1+x2

1 − exp	x2	x2 x → 0 ,
	2
	1+x

( ( )

o x3 o x3

) % '

o x

( x

o x4

arctg x = x − x33

+ o x4

o x3 .

1+x2

1 + x2 = x2

1 − x2 + o x3

= x2 − x4

+ o x5

exp

= exp

− x4 + o x5

1 + x2

=1 + x2 − x4 + o x5 + 12 x2 + o x3 2 + o x5 =

=1 + x2 − 12 x4 + o x5 .

													1		x2	+ o x2	2
	( .																2
1	( .													2
1	x2 + o x3	2
		2											(
2													(
2
o x4
																		x4
		x4							x4									x4
g (x) = x2 −			+o x5		−	1 + x2 −						+ o x5				+1 =			+o x5	.
g (x) = x2 −		3	+o x5		−	1 + x2 −				2		+ o x5				+1 =		6	+o x5	.
	/

	)			%			'
o x5 .							x2		x4
				ch x = 1 +				+			+ o x5					.
				ch x = 1 +			2	+		4!	+ o x5					.

2+2xx2 x x → 0

o x4

o x3

= x 1 −

+ o x3

= x −

+ o x4

2 + x2

1 +

cos

= cos

x −

+ o x4

2 + x2

= 1 − 2

x + o x2

+ o x5

x − 2 + o x4

+ 24

= 1 −

13x4

+ o x5

2 6 1 + 3x4

= 2 1 + 3x

+ o x5

= 2 + x4 + o x5

13x4

f (x) = 1 +

+ o x5

+ 1 −

+ o x5

−

2 + x4 + o x5

= −

+ o x5

−

5x4

lim

f (x)

= lim

+ o x5

−

x→0

g (x)

x→0

+ o(x5)

e√

1−2x

e(x3+x2−1)

x+1

lim

− sin

x→0 sh

1 +

# $

# & )

* o x3 + # # o x3

−

1 x 2

1 x

ln 1 +

+ o x3

−

1 x 2

1 x

sh ln 1 +

+ o x3

1 x

+ o(x)

+ o x3

& #

g (x) = 2

1 x

+ o(x)

+ o x3

* o x3

√

(2x)

(2x)2

(2x)3

1 − 2x

= 1 −

−

+ o x3 .

+ o x3

= 1 − x −

−

√

exp

− 2x

= exp

1 − x −

−

+ o x3

= e · exp

−x −

−

+ o x3

= e 1 − x +

−

+ o x3

x +

+ o x2

−

(x + o(x))

+ o x3

= e 1 − x +

+ o x3

+ x

+ o x

−

ex3

+ o x3

= e − ex −

+ o x3

= 1 − x + x2

− x3 + o x3

1 + x

x3 + x2 − 1

e x3

+ x2 − 1

1 − x + x2 − x3 + o x3

x + 1

=e x3 (1 + o(1)) + x2 (1 − x + o(x)) −

−1 − x + x2 − x3 + o x3 = −e + ex + ex3 + o x3 .

ex3

f (x) = e − ex −

+ o x3

−e + ex + ex3

+ o x3

5ex3

+ o x3 .

f (x)

5ex3

lim

= lim

= 20e.

x→0

g (x)

+ o(x3)

ln (1 − 3x)2/3 − (1 + 3x)2/3 + ch √

−

lim

sin (sin x) − arctg (arctg x)

x→0+0

o x3

sin t = t −

+ o t3

arctg t

= t −

+ o t3

, t → 0.

(x + o(x))3

sin (sin x) =

x −

+ o x3

−

+ o x3

= x −

+ o x3

(x + o(x))3

arctg (arctg x) =

x −

+ o x3

−

+ o x3

= x −

2x3

+ o x3

x −

2x3

g (x) =

+ o x3

− x −

+ o x3

+o x3

o x3

" !

(1 − 3x)2/3 − (1 + 3x)2/3 =

= −2·

· −

= −4x−

8 3

(3x)−2·

(3x)

+o x

x +o x

√

8x = 1 +

24 · 30

+ o x3

f (x) = ln −4x −

= 1 + 4x +

8x2

32x3

+ o x3 .

8x3

+ o x3 + 1 + 4x +

8x2

32x3

+ o x3

−

= ln 1−

+ o x3

= −

x3+o x3

f (x)

−4588 x3 + o x3

= −

lim

x→0+0 g (x)

x→0+0

+ o(x3)

lim

arctg 3 + x2 − arctg (2 + cos x)

ln (1 + x) − ex + 1

x→0

o x2

g (x) = x−

+o x2

−

1 + x +

+ o x2

+1 = −x2 +o x2

→ 0

arctg

3 + x2

o(x)

arctg 3 + x2 =

+ o(x) .

1 + (3 + x2)2

1 +

! "#

arctg

3 + x2

= arctg 3 +

+ o x2

(arctg (2 + cos x))

− sin x

1 + (2 + cos x)2

− sin x

−x + o(x)

= −

+o(x) .

5 + 4 + o(x) + 1 + o(x)

5 + 4 cos x + cos2 x

arctg (2 + cos x) = arctg 3 −

+ o x2

f (x) = arctg 3 +

+ o x2

−

arctg 3 −

+ o x2

lim

f (x)

= lim

+ o x2

= −

x→0 g (x)

x→0 −x2 + o(x2)

− arctg

√

arcsin

√

1 + x 2

lim

(cos 2x)ctg x − (1 − th 5x)2/5

x→0

o x2

(2x)2

ln cos 2x = ln 1−

+ o x3

= ln 1−2x2

+ o x3

= −2x2+o x3

(cos 2x)ctg x = exp

cos x · ln cos 2x

sin x

(1 + o(x)) · −2x2 + o x3

= exp

−2x + o x2

(x + o(x2))

= 1+

−2x + o x2

+ 2

−2x + o x2

= 1−2x+2x2 +o x2

(1 − th 5x)2/5 =

1 − 5x + o x2

2/5 =

=1 + 25 −5x + o x2 − 253 −5x + o x2 2 + o x2 =

=1 − 2x − 3x2 + o x2 .

g (x) = 1 − 2x + 2x2 + o x2 −

− 1 − 2x − 3x2 + o x2 = 5x2 + o x2 .

o x2 ' ( ) )

) o(x)

arcsin

√

−

1 −

√

+ o(x)

√

1 +

√

+ o(x) .

2 1 −

√

+ o(x)

4 2

arcsin

√

+ o x2

√

arctg 1+x

√ √

√

1− 2x+ o(x) .

x+ o(x)

√

arctg

1 + x

√

−

+ o x2

f (x) =

√

+ o x2

−

9x2

−

√

−

+ o x2

9x2

lim

f (x)

= lim

+ o x

5x2 + o(x2)

x→0 g (x)

x→0

f (x) g(x)

f (x) g (x)

	√				1

					x4
lim	1		− x2
				.
		cos x
x→0

o x4

! "#$

f (x) =

1 −

x2 −

x4 + o x4

1 − 1 x2

1 x4 + o(x4) =

−

= 1

−

+ o x4

1 − −

+ o x4

−

+ o x2

+ o x4

5x4

−

= 1−

+ o x4

1 +

+ o x4

+o x4 .

−

lim f (x)g(x) = lim

1 −

+ o x4

= exp

x→0

		6	√		5x
			√		5
lim			− 3	cos 2x	arcsin	x .
lim	3 + x2		− 3	cos 2x		x .
x→0	3 + x2

' arcsin5 x = x5 + o x5

5x (

o x4

)

= 2 1 − x

+ x

+ o x4

3 + x2

1 +

√

cos 2x

= 3 1 − 2x2 +

+ o(x4) =

2x4

= 1 +

−2x2 +

+ o x4

−

−2x2

+ o x2

+ o x4

= 1 −

2x2

2x4

+ o x4

f (x) = 2 1 −

+ o x4 −

2x2

2x4

4x4

− 1

−

+ o x4

= 1 +

+ o x4

4x4

lim (f (x))g(x) = lim

x5+o(x5)

= e

1 +

+ o x4

x→0

lim

1 − cos x

(1+x)3

1+3x

x→0

− cos x

x → 0 +

o x3

1+o(x)

o x4

" o x4

1 − cos x

−

o x

24·30

+ o x

+ o x6

f (x) =

−

24·30

+ o x

1 −

12·30

+ o x

+ o(x6)

1 +

+ o(x4)

= 1 −

12 · 30

+ o x4

−

+ o x4

= 1−

+o x2

− cos x =

−

+ o x4

−

+ o x4

1 −

+ o x2

f (x) =

= 1 −

+ o x .

+ o(x4)

1 + o(x2)

& ' o x2


ln	(1 + x)3				= 3 ln (1 + x) − ln (1 + 3x) =
		1 + 3x

		x2					9x2
= 3		x −		+ o x2		− 3x −		+ o x2	= 3x2 + o x2	.
			2				2

lim 1 −

+ o x2

= e−

lim (f (x))g(x) =

3x2+o(x2)

x→0

ln2(x+√

)

x tg x

1+x2

lim

x→0

ch x − exp

−

* +

x tg x = x x +

+ o x3

= x2 +

+ o x4

, o x4 #

ch x− exp −

= 1 +

+ o x4

−

− 1 −

+ o x4

= x2

−

+ o x4

x2 +

+ o x4

1 +

+ o x2

$ f (x) = x2 − x124

+ o(x4) =

1 − x122 + o(x2) =

5x2

= 1 +

+ o x2

1 +

+ o x2

= 1 +

+ o x2

o x2

x + √

1 + x2

o(x)#

ln x + 1 + x2 = ln (x + 1 + o(x)) = x + o(x) .

5x2

lim (f (x))g(x) = lim

x2+o(x2)

= e

1 +

+ o x2

x→0

()

((

etg x −

√

x2 ctg x4

lim

1 + 2x

2x ch x − ln (1 + sh x)2

x→0

x2 cos x4

1 + o x4

1 + o x2

x2 ctg x4 =

sin x4

x4 + o(x4)

x2 + o(x2)

o x2

−

2x ch x − ln (1 + sh x)2 = 2x 1 +

+ o x3

− 2 ln 1 + x +

+ o x4

+ x3 + o x4

−

4 2

x +

+ o x

− 2

+ o x4

−

(x + o(x))4

5x4

+ o x2

−

= x2 +

o x4

tg x

√

−

1 + 2x

= exp

x +

+ o x4 − 1 + x −

−

5x4

+ o x4 =

= 1 + x +

+ o x3 +

x +

+ o x3 +

−

x + o x2

+ o x2

+ o x4

5x4

− 1 + x −

−

+ o x4

= x2 + x4 + o x4

f (x) =

1 + x2 + o x2

5x2

1 +

+ o(x2)

−

5x2

1 + x2 + o x2

+ o x2

= 1 +

+ o x2

1+o(x2)

x2+o(x2)

lim (f (x))g(x)

lim

1 +

+ o x2

x→0

lim

cos x + x2 ·

3 x +

arctg x

x→+0

! arctg

x → 0

arctg

3π

x → 0

2x3

" o x3

f (x) = 1 −

√

+ o x3

· 3 1 + 8x =

= 1 −

1 +

4x3

+ o x3

+ o(x) = 1 +

+ o x3

3π

2x3

lim (f (x))g(x)

lim 1 +

+ o x3

e2π .

x→+0

2+cos x

lim

1 −

+ ch x

x→+0

2+cos x

x → 0

o x4

f (x) =

−

+ o x4 +

+ 1 +

+ o x4

= 1 −

+ o x4

1 −

+ o x4

e−

lim (f (x))g(x)

lim

x→+0

log

(2x2 − 20x + 53)

(x − 5)2n+1

x0 = 2 y = 5x3−6x2+12x o (x − 2)3n+2

y =

= x sh2 x o (x)2n

x0 = −1 y = (x + 1) ln

x2+2x+2

1−2x−x2

o (x + 1)

−

x +

(sin x + cos x)

2n+1

x +

−4x+4

= 2

y =

√

o((x − 2) )

3 7−3x

−2

cos (x + 2) · cos (x + 3)

(x + 2)2n

= 3 y = sh (x − 3) ch (x − 4) o (x − 3)2n

(2x + 3) e4x2+4x−3

o x − 1 2n+1

= −3 y = (x + 5) e(x2+6x+8)(x2+6x+10)

o (x + 3)4n+3

x−2

x0 = 2 y =

(x−3)2

o((x − 2) )

= 1 y =

x−1

(x − 1)2n

(4x−2x2−1)2

= −1 y =

2x2

+2x−7

o((x + 1)n)

x2+x−2

x0 = −1

3x+9

y =

−x2−x+2

o((x + 1) )

= −1 y =

x+2

(x + 2)3n+2

x(x2+3x+3)

x0 = −1 y = (x + 2) ln (x + 3) o((x + 1)n)

= 1 y = x2 − 2x − 1 ln √

x2 − 2x + 5

o (x − 1)2n+1

x0 =

log5

(x2−x+

)

3+2x

5−2x

o x −

= π

y =

x2 − πx

x − π

2n+1

cos2 x o

−1

5x2

+10x−3

1 + sin

πx

x+1

o (x + 1)2n

<<< < Предыдущая 1 23 / 43 4 > Следующая >>>

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15.02.20151.19 Mб7ФОРМУЛА ТЕЙЛОРА И ЕЕ ПРИЛОЖЕНИЯ К ВЫЧИСЛЕНИЮ ПРЕДЕЛОВ ФУНКЦИЙ.pdf