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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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(x − x )k + o((x − x )n) x → x . **"

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k=0

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x → x0 n3

2 **" n3

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4	. f (x)

x0 (n + 1)3

x - ξ 5 0 5 x x0 x < ξ < x0 x0 < ξ < x"

)

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f k (x

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(n + 1) !

k=0

(x) =

f (n+1)(ξ)

(x − x0)n+1

(n+1)!

f n (x0) !" f (x) " # $" "

	n		(x − x		)k + o((x − x )n)				x → x , %
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x → 0.

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k=0

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x2k+1

x → 0.

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k=0

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x2n

− . . . + (−1)n

+ o x2n+1

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x2k

2n+1

x → 0.

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(−1)

+ o x

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k=0

x2n+1

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− . . . + (−1)n

+ o x2n+2

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(2n + 1)!

x2k+1

(−1)k

x → 0.

sin x =

+ o x2n+2

k=0

(2k + 1)!

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α (α − 1)

x2 +

α (α − 1) (α − 2)

+ . . . +

α (α − 1) . . . (α − (n − 1))

xn + o(xn)

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x → 0, α / N, α = 0,

k=0

= 1, Ck =

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, k = 1, 2, . . . ; ")

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x→0 g (x)		x→0 bxn + o(xn)			b

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1		1		a
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							x → 0
2x + 3x + o x			− x + 3x		2	+ o x
	2		3			3		= x + o x3	.
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o x3 − o x3 = o x3

3x + 5x2 + x4 − o x4 1 + 5x − x3 + o x3 x → 0

% ! &

''( ) !

! *

3x + 5x	+	x	4	+ o x	4	+
2	+ 25x	+				−
+ 15x				o x	4
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= 3x + 20x			4
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+ , o x4 x → 0

-! #

! !$


o xk		k

	% # . #$

f (x) = ex + x2|x| o(xn) /

# - x0 ! , ! , " "

% g (x) = x2|x| g (0) = g (0) = g (0) = 0 g (0)

" % 2 # . #$ g (x) o(xn) * g (x) = o(x) n = 13

g (x) = o x2 n = 23 n ≥ 3

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#$ &'4(

f (x) = 1 + x + o(x) n = 1 3 f (x) = 1 + x + x22 + o x2

n = 23 n ≥ 3 "

# -

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+ !

% # . #$

√

f (x) = ex ·

1 + x o x2

5$ #$ -

ex 1

√

1 + x

1 x → 0

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f (x)

1 + x +

+ o x2

1 +

−

+ o x2

! 6

f (x) = 1 +

−

+ o x2

+ x 1 +

+ o(x) +

7x2

, x → 0.

(1 + o(1)) = 1 +

+ o x2

f (x) = sin x · ln (1 + x) o x5

sin x x ln (1 + x) x x → 0 sin x

ln (1 + x) o x4

f (x) = x −

+ o x4

x −

−

+ o x4

= x x −

−

+ o x4

−

x −

+ o x2

−

= x2

−

+ o x5

, x → 0.

" " # $

! % #

x + 2x2 + 3x3 + o x3 2 = x2 + 2x2 2 + 3x3 2 +

+ 2 x · 2x2 + x · 3x3 + 2x2 · 3x3 +o x3 x + 2x2 + 3x3 + o x3 .

f (x) = ex−x2 o x3

f (x) " & ''(')* + x − x2 x x → 0 , -

%
et = 1 + t +	t2	+	t3	+ o t3 % t = x − x2	→ 0
	2
x → 0 %		6

			x − x2 2		x − x2 3
f (x) = 1 +	x − x2	+		+		+ o x3	.
f (x) = 1 +	x − x2	+	2	+	6	+ o x3	.
			2		6

f (x) = 1 + x − x2 − 5x3 + o x3 , x → 0.

2 6

f (x) = esin ln(1+2x) o x3

/ f (x) "

ln (1 + t)

t − t2

+ t3

+ o t3

% t = 2x → 0 x → 0

(2x)

+ o x3

= 2x−2x2 +

8x + o x3

ln (1 + 2x) = 2x−

! u = 2x − 2x2 +

8x3

+ o x3

u → 0

x → 0 sin u = u − u3 + o u3

, %

8x3

sin ln (1 + 2x) = 2x − 2x2 +

+ o x3

−

(2x + o(x))3 =

0 y = 2x − 2x2 +

= 2x − 2x2 +

4x3

+ o x3 .

+ o x3

4x3

y → 0 x → 0

ey = 1 + y +

+ o y3

4x3

f (x) = 1+ 2x − 2x2 +

+ o x3

2x − 2x2 + o x2

= 1 + 2x −

4x3

(2x + o(x))3 + o x3

+ o x3

!" #

f (x) = exp

sin

2x − 2x2 +

+ o x3

8x3

= exp

2x − 2x2 +

+ o x3

−

(2x + o(x))3 + o x3

= exp 2x − 2x2 +

+ o x3

4x3

= 1 + 2x − 2x2 +

+ o x3

2x − 2x2 + + o x2

4x3

(2x + o(x))3 + o x3

= 1 + 2x −

+ o x3

, x →

$ !

f (x) = arcsin x3

o x5

ln(1+x2)

1 + x

→ 0

arcsin x

arcsin x3

o x7

1 + x2

& o x6

" ' #

x3 + o x7

x + o x5

f (x) =

x2 −

+ o(x6)

1 −

+ o(x4)

− −

−

x + o x5

+ o x4

+ o x2

x + o x5

1 +

−

+ o x4

= x+

−

+ o x5

! n" #$

$ ϕ (x) = Axm, m N" '

! $

( $ f (x) = tg x o x6 "

' "

tg x	=	cos x		!	cos x · tg x =	sin x"
			sin x
y	=		tg x
(

(

o x6

)

1 −

+ o x5

ax + bx3 + cx5 + o x6

x −

+ o x6

120

)

x :

a = 1;

x3 : −

+ b = −

;

x5 :

−

+ c =

120

* a = 1, b =

, c =

15 " #

tg x = x +

x5 + o x6

, x → 0.

+ !

(

y = th x o x6

" +

th x = x − 13 x3 + 152 x5 + o x6 , x → 0.

( $

f(x) = arcsin x o x6 "

, (

o x6 )

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