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Национальный авиационный университет

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Number			A is called a limit of a function y = f (x) if x →∞, if for any
small number			ε > 0	such number M = M (ε) > 0 may be found, that for any x
such as	x	> M (ε),		the following inequality is fulfilled	f (x) − A	< ε.

We denote it as:

lim f (x) = A or f (x) → A if x → ∞.

x→∞

A function y = f (x) is called infinitely large if M > 0 such number δ = δ(M ) > 0 may be found,х–a < δ, х ≠ a, the following inequality is fulfilled

x → a, if for any number that for any x such as

f (x) > M .

We denote it as:

A function α(x) is called infinitesimal if x → a, if

lim α(x) = 0.

x→a

lim f (x) = ∞.

x→∞

For example, the function y = 1x is infinitely large if х → 0, and is infinitesimal if х→±∞ .

13.4. One-sided limits

Number B is called a limit from the right of a function f (x) if x → a, if for any very small number ε > 0 such number δ = δ(ε) > 0 may be found, that for any x such as 0 < х – a < δ, the following inequality is fulfilled

f (x) − B < ε.

We denote it as:

lim f (x) = B, or f(a + 0) = B.

x→a+0

Number C is called a limit from the left of a function f (x) if x→ a, if for any small number ε > 0 such number δ = δ(ε) > 0 may be found, that for any x such as 0 < a – x < δ, the following inequality is fulfilled

f (x) − C < ε.

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We denote it as:

lim f (x) = C, or f(a – 0) = C.

x→a−0

A limit from the right and a limit from the left are called one-sided limits of a function.

If a function f (x) has a limit at point a, then

lim

f ( x ) = f ( a − 0 ) =

f ( a + 0 ).

x → a

13.5. Properties of limits

Property 1. Suppose each function

f (x)

and g(x) has a finite limit at

point а, then:

1) lim( f (x) ± g(x)) = lim

f (x) ± lim g(x) ;

x→a

lim

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

x→a

f (x)

lim f (x)

lim

x→a

(if lim g(x) ≠0 );

lim g

(x)

x→a

g(x)

x→a

lim

(

f (x) g(x) =

lim g (x)

lim f (x) x→a

;

x→a

)

(x→a

)

lim cf (x) = c lim

f (x)

( с is a constant)..

x→a

Property 2. Suppose the functions

g(x),

f (x) and h(x) are defined at

some neighbourhood of point x0 , except may be, the point x0.

Suppose

lim g(x) = lim h(x) = A and g(x) ≤ f (x) ≤ h(x) , then

x→ x0

lim f (x) = A .

x→x0

Property 3. (about limit of monotone function). If a function f (x) is mo-

notone and bounded for x < x0

or for x > x0 ,

then it has limit from the right or

limit from the left.

f (x) = f (x0 − 0) or

f (x) = f (x0 + 0).

That is,

lim

x→ x0 −0

x→ x0 +0

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Micromodule 13

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Find a function inverse to the function

y =

ex − e− x

Solution. We can solve the given equation with respect to x :

ex − e − x = 2y, i.e. e2x − 2yex −1 = 0.

This equation is square with respect to ex : ex = y +

y2 +1 .

We accept only one value of a root, because ex

> 0. After taking logarithms

of both parts we have x = ln ( y +

y2 +1). This function is inverse to the initial

one. We can rewrite it in a usual form, denoting the argument as

and the

function as y : y = ln (x +

x2 +1).

Example 2. Find a domain of definition of a function

y =

sin (2x −1) +

20 − x − x2 .

Solution. The given function is the sum of two functions. Therefore we find

the

domain of definition

for each

function

separately:

y =

sin(2x −1) and

y =

20 + x − x2 . For the first function it must be

sin (2x −1) ≥ 0,

i.e.

2x −1

[2kπ; (2k +1)π] , whence

x kπ +

;

kπ +

. For the second function the

domain of definition is the set of numbers

x [−4; 5]. Hence, the domain of

definition of the given function is the set of intervals

kπ +

; kπ +

∩[−4; 5], k Z

π + 1

1 ; 5 .

−π + 1 ;

−

1 ;

π +

2 2

2 2 2

Example 3. Find out which of the given functions is even, odd or neither even, nor odd.

а) f (x) = x sin 2x; b) f (x) = ln 22+− xx ; c) f (x) = x2 − 3x + 8 . Solution. а) We obtain:

f (− x) = − x sin(−2x) = (− x) (− sin 2x) = x sin 2x ; f (− x) = f (x) , hence, the given function is even;

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b) Substituting − x into the formula b) we get:

f (− x) = ln	2 + (− x)	= ln	2 − x	= ln(2 − x) − ln(2 + x) = − ln	2 + x	= − f (x),
	2 − (− x)		2 + x
					2 − x

i.e. the function is odd;

c) Substituting − x into the formula c) we get:

f (− x) = (− x)2 − 3(− x) + 8 = x2 + 3x + 8.

The analytical expression of the function has changed, i.e. it is neither even, nor odd.

Example 4. Sketch the graph of the function

y = 2sin ( 3x + 6)+ 1 .

Solution. We can write down the given function as y = 2sin 3( x + 2)+ 1.

It is easy to find out that the period of this function is equal to 23π .

In comparison with the graph of a function y = sin x the graph of the given function is removed two units to the left along the axis Ox, the amplitude of the

graph of the given function is twice greater and the graph moves one unit upward along the axis Oy (Fig. 3.8).

As far as the function is periodic the graph can be extended both sides along the axis Ox.

2sin(3x+6)+1

2sin(3x+6)

2sin(3x)

sin(3x)

sin(x)

Fig. 3.8

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Micromodule 13

CLASS AND HOME ASSIGNMENT

Find a domain of definition of the functions:

y =

3x − 2

2. y = 3x − 2 − x2 .

3. y =

x + 4

x + 1

x − 1

y = lg

2x − 1

5. y =

x − 2

y = lg(lg(x2

+ 3x + 3)) .

x + 3

lg x

y =

log1 ( x2

− 4x + 3) .

( log2

y = lg log1

(x −1)) .

(

y =

x −

10. y =

− x .

11. y = arcsin

)

3x − 1 .

Evaluate limits.

12.	lim		3x5 + 2x −1				.
			8x5 − 2(x − 1)4 − 2
	x→∞
14.	lim		7x2 + 4x + 1	.

	x→∞		−2x6 − x4 + 1
16.	lim	x3 − 5x2 + 2x + 2			.

	x→1		x2 + 5x − 6
18.	lim	3 7 + x2 − 3 1+ 7x				.
			x2 −1
	x→1
20.	lim (		x2 + 2x − 3 − x) .
	x→∞

Sketch graphs of the functions:

13.	lim		5x3 + 5x + 3		.

	x→∞ 4x2 − 2x4 + 3
15.	lim		2x2 − 5x + 2 .
	x→ 1		8x3 − 4x2
	2
17.	lim		5 − x −	2x + 2			.

	x→1		x2 − 4x + 3
19.	lim	3 x + 6 −		x + 2		.
			x + 14 − 4
	x→2

21. lim (3 x3 + x − 3 x3 − 1) .

x→∞

22.	y =	x − 2	. 23.	y = x2 − 4x .				24. y = x2 +			x + 2		.	25.	y =	1	.


		x − 3														x2 + 4

26.	y = 4 − 2x .			27.	y =	4 − 2			x	.		28.		y =	3 − x2 + 2x .

29.	y = log2 (1− x) .			30.	y = 3−		x	2x .				31.		y = 10lgsin x .


32.	y = 1/ ctg x .			33.	y = arcsin x + 2arccos x .							34.		y = cos3x .

Answers

12.3/8. 2. ∞ . 13. 0. 14. –1,5. 15. –5/7. 16. 0,125. 17. –5/24. 18. –4/3.

19.∞ , if x → −∞ ; 2, if x → +∞ . 20. 0.

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Micromodule 13

SELF-TEST ASSIGNMENTS

13.1. Find the limits of functions

13.1.1. а)

lim

2x2 − x3

;

lim

2x2 + 9x − 5

;

3 8x9 − 7x7 + 5x

x→∞

x→−5 x2 + 3x − 10

lim

4x − 3 −

2x + 3

x→3

2x2 − 5x − 3

13.1.2. а)

lim

4x2 − 2x −

x + 1

;

lim

3x2 − 7x + 2

;

3x − 5

+ 2x − 1

x→∞

x→ 3

lim

4x2 + 3x −1

x→−1 5x + 6 −

3x + 4

13.1.3. а)

lim

3x3 − 2x2 + 5

;

lim

2x2 − 3x − 2

;

x→∞ 4 + x − 2x2

x→2

3x2 − 7x + 2

lim

2x + 2 − 1

x → −

4x2 − 1

13.1.4. а) lim

3 8x3 + 3x2 − 3 4x2 + 7x

;

lim

x2 + 2x − 3

;

x→∞

3x + 5

x→−3 2x2 + 5x − 3

lim

6x − 1 −

12x − 3

x→ 1

3x2 − 4x + 1

13.1.5. а)

lim

12x − 7

;

lim

x2 + x −12

;

4 16x4 − 3x3 + 3

2x2 − 7x + 3

x→∞

2x2 − 1

x→3

lim

4x − 5 −

x + 1

x→2

2x2 − 5x + 2

13.1.6. а)

lim

3x2 − 5x + 2 ;

lim

6x2 − 5x − 4

;

x→∞

1+ 4x − 7x3

x→− 1

2x2 + 3x + 1

lim

3x − 4 −

x→2

2x2 − x − 6

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13.1.7. а) lim	5x2 − 3 + 7 4x4 − 2x2 + 3	; b)	lim	x2 − 6x + 5	;

x→∞	3x − 2		x→5	2x2 − 9x − 5

(

)

− 1

lim

x + 1

27x3 +

x→− 2

13.1.8. а)

lim

3 2x4 − 5x3 + 1

;

lim

2x2 + x − 3

;

x5 − 3x4 + 7x3

4x2 − 9

x→∞

x→− 3

lim

2x −

3x − 8

x2 − 9x + 8

x→8

13.1.9. а)

lim

5 + x − 3x3

;

lim

5x2 + 8x + 3

;

x→∞ 7x2 − x3

x→− 3

5x2 − 7x − 6

lim

3x + 1 −

2x + 3

x→2

x2 + 2x − 8

13.1.10. а)

lim

7x2 − x + 5 ;

x → ∞ 2x3 − x2

c) lim 3 − 2x + 9 .

x→0

4x2 − 3x

13.1.11. а)

3 x5 − 2x2

lim

;

x→∞

3x7 + 5x4 − 2x

lim

x2 + 3 − 5x −1

x2 + x − 2

x→1

13.1.12. а)

lim

x2 − 3x + 5 − 2x

;

3 x3 + 3x2

x→∞

lim

12x2 − 1 − 4x2 + 1

2x −

x→ 2

13.1.13. а)

lim

5 3x7 − 2x5

;

x→∞

2x5 − 4x2 + x

lim

3x + 4 −

6x + 6

x→−

9x2 − 4


b) lim			x3		−1	;
b) lim					4x − 1	;
	x→1 5x2 −				4x − 1
b)	lim		2x2		− 3x − 2			;
b)	lim			8x3 + 1				;
	x→− 1			8x3 + 1
	2
b)	lim			x3 − 1			;
b)	lim	5x2 − 2x − 3					;
	x→1	5x2 − 2x − 3
b)	lim		3x2		+ 11x − 4				;
b)	lim			2 + 21x +			68		;
	x→−4 x			2 + 21x +			68

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13.1.14. а) lim	3 x5 − 2x2
		;

x→∞ 4	3x7 + 5x4 − 2x

c) lim

x2 + 3 −

5x −1

x2 + x − 2

x→1

13.1.15. а)

lim

3 x2 − 3x + 5 − 2x

;

x→∞

3 x3 + 3x2

lim

12x2 − 1 − 4x2 + 1

;

2x −

x→ 2

13.1.16. а)

5 3x7 − 2x5

lim

;

x→∞

2x5 − 4x2 + x

lim

3x + 4 −

6x + 6

x→−

9x2 − 4

13.1.17. а)

lim

x2 − 2x −

x + 2

;

3x − 4

x→∞

lim

2x + 5 −

4x + 1

x→2

x2 + 3x − 10

13.1.18. а)

lim

x4 − 2x3 + 3

;

x→∞

5x4 + x2

lim

16x2 − 1

12x + 4 −

4x +

x→−

b)	lim			2x2	− 3x − 2			;
b)	lim			8x3 + 1				;
	x→−	1		8x3 + 1
		2
b)	lim			x3	− 1	;
b)	lim	5x2 −			2x − 3	;
	x→1	5x2 −			2x − 3
b)	lim			3x2	+ 11x − 4				;
b)	lim								;
	x→−4 x2 + 21x + 68
b)	lim			2x2	− x − 3				;
b)	lim		8x12 − 10x −				3		;
	x→ 3		8x12 − 10x −				3
	2
b)	lim			2x2 + 5x + 3				;
b)	lim			4x2	+ 3x − 1			;
	x→−1			4x2	+ 3x − 1

13.1.19. а)

lim

3 x6 − 3x +

3 x4 − x3

2x2 + 5x + 3

;

lim

x→∞

x2 + 7x − 11

x→−1 4x2 + 3x − 1

lim

5x + 3 −

3 − 2x

x→0

5x2 − 6x

13.1.20. а)

lim

3 x2 − 3x +

x2 + 3x

;

lim

6x2 − x − 2 ;

x→∞

3 8x3 + 3x2 + 2

x→ 2

3x2 + x − 2

lim

10x − 1 −

5x + 4

x→1

2x2 + x − 3

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13.1.21. а)

lim

3 3 − x3 + 2x + 5

; b)

lim

4x2 − 4x −15

;

4 16x4 − 7x3 + x2 − 4

2x2 − 7x + 5

x→∞

x→ 5

lim

x + 3 − 5x − 1

x→1

x3 − 1

13.1.22. а)

lim

2x2 − 4x + 5

;

lim

2x2 + 7x − 4

;

x→∞ x4 + 3x3 − x2 + 3

x→−4 3x2 + 10x − 8

lim

4x2 + 3 −

2x + 3

2x2 + 3x −

x→ 1

13.1.23. а)

lim

5x − 4 −

2x + 3

;

lim

5x2 − 9x − 2 ;

x→∞

5x2 − 3x + 5

x→2

x3 − 8

lim

6x + 7 −

2x + 3

x→−1

2x2 + x − 1

13.1.24. а)

lim

4 x12 − 4x5 + 3 x3 + 5x2

;

lim

4x2 + 4x − 3

;

x→∞

x3 − 3x2 + 7

x→−

2x2 + 7x + 6

lim

x + 1 − 2x − 2

x→3

x3 − 27

lim

3x3 −8x2 + 7

lim

3x2 −8x −3

13.1.25. а) x→∞ 20x2 −4x + 3 ;

x→3 2x2 −9x + 9 ;

lim

16x2 − 1

8x + 3 − 1

x→−

;

13.1.26. а) lim

3 8x3 − 3x +

x2 + 7x

;

lim

5x2 − 11x + 2

;

x→∞

3x − 5

10x

+ 3x − 1

x→ 5

lim

3x − 4 −

4 − x

x→2

2x2 − 3x − 2

13.1.27. а) lim

5 x4 − 3x3 + 5 + x − 1

;

lim

3x2 − 8x + 5 ;

x→∞

3 2x2 − 3x

x→ 5

3x2 + x −10

lim

3x + 4 −

2x + 3

x→−1

11x2 + 10x −1

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13.1.28. а) lim	x5 − 17x2		;		b) lim	3x2 − 5x − 2	;
13.1.28. а) lim		7	;		b) lim	3x2 + 4x + 1	;
x→∞ x3 + 2x2 +		7			x→− 1	3x2 + 4x + 1
					3
c) lim	3x + 2 −	6x		.
c) lim				.
x→ 2	3x2 − 5x + 2
3


	3	x2 − 7x + 4 x3 + 2x2
13.1.29. а) lim
		5 x4 + 5x2 − 1
	x→∞
c)	lim	x2 + 2 −	6x − 3		.
		x2 − 4x − 5
	x→5
13.1.30. а)	lim	3x11 − 5x10 + 7x3			;

	x→∞ 7x6 + 12x10 + 5x11
c)	lim	7x + 2 −	3x + 6	.

	x→1	x2 + 4x − 5

13.2. Sketch graphs of the functions.

; b) lim	2x2	+ 9x + 9	;
	x2	+ x − 6
x→−3

b) lim 2x2 + 5x − 3 ;

x→ 12 6x2 − x − 1

13.2.1. а) y =

;

b) y = 10lgcos 2x ;

c) y

;

x − 2

x2 + 4x +

13.2.2. а) y =

;

b) y = 6 − 4x ;

c) y =

;

x + 1

x2 − 2x + 3

− 1

d) y = 2 x .

d) y = 2x2 .

13.2.3. а) y = ln(e − x) ;	b)	y = sin x + cos x ;
13.2.4. а) y = log2 x2 ;	b)	y = sin2 x ; c) y =

c) y =

x + 2 ; x + 1

	x + 1		1
	x + 1		; d) y = 2x .
	x −1		; d) y = 2x .
	x −1
	d)		y = 2arcsin x .

13.2.5. а) y = log3 (

− 1) ;

y = tg2 x ;

c) y =

x + 2

;

y = 2arc tg x .

x − 4

13.2.6. а) y =

;

y = ctg3 x ;

c) y =

x − 2

;

d) y = arccos 2x .

x2 + 2

x + 1

|x|

13.2.7. а) y =

;

y = sin(2x −

1) ;

c) y =

;

y = 2

x2 −1

x + 3

13.2.8. а) y = log2 x4 ;

b) y = 2cos2 x ; c) y =

x − 3

;

y = 3sin x .

x + 2

13.2.9. а) y = log2 x + log3

x ;

b) y =

sin x ; c) y =

1− x

;

d) y = 5cos x .

x + 1

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