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sequence of numbers.

Micromodule 12

BASIC THEORETICAL INFORMATION

SEQUENCE. THE LIMIT OF A NUMERICAL SEQUENCE.

THEOREMS ABOUT LIMITS

Sequence. The limit of a numerical sequence. Theorems about limits. Number e. Determinate and indeterminate forms.

Literature: [2, chapter 1], [3, chapter 3, §§ 3.1—3.8], [4, part 4, §§ 3.1— 3.3, 4.2—4.3], [6, chapter 4, § 3], [7, chapter 4, § 11], [10, chapter 3, § 2], [11, chapter 3, § 1], [12 chapter 2, § 1], [13].

12.1. Sequence

Definition 3.1. If the numbers xn correspond to each natural number according to the definite rule then the set of numbers {x1 , x2 , …, xn , …} is called the

We denote a sequence as {x n} ; xn is the term of sequence.


12.2. The Limit of a Numerical Sequence

Definition 3.2. Number a is called a limit of a sequence {x n } if for any
positive number ε > 0	such number					N (ε )	may be found, that for any
n > N (ε ) the inequality		xn − a	< ε will hold.


In this case we write down: lim x				n	= a or x		n	→ a for n → ∞.
		n→∞
Definition 3.3. The sequence {x n }						is said to be infinitesimal, if
			lim x = 0 .
			n→∞			n
Definition 3.4. The sequence {x n }						is said to be infinitely large if for any

number M > 0 there exists such number	N so that for n > N the inequality
x > M holds. In this case we write down:	lim x = ∞.
n	n→∞ n
	141

12.3. Theorems about Limits

We shall formulate properties of limits as theorems.

Theorem 3.1 The sequence can have only one limit.

Theorem 3.2 The sequence which has a finite limit, is bounded.

Theorem 3.3

If {xn} and {yn} have limits the following equalities are valid:

lim (x

± y

) = lim x

± lim y

; 2) lim x y

= lim x

lim y

;

n→∞ n

n→∞

n→∞ n

n→∞

lim Cyn = C lim yn , where С is constant;

n→∞

)

lim x

lim

n→∞ n

lim y

≠ 0

lim yn

n→∞ yn

(n→∞

n→∞

Theorem 3.4

Theorem 3.5

If x ≥ 0, then	lim x	n	= a ≥ 0.
n	n→∞	n
If for sequences {x }, {y } and {z }: x						≤ y	≤ z and	lim x = a,
	n		n	n	n	n	n	n→∞ n
lim zn = a , then lim yn exists and					lim yn = a .
n→∞	n→∞				n→∞

Theorem 3.6 Any monotone bounded sequence has a limit.

Theorem 3.7	For all elementary functions at any point where they are deter-
	mined, the equality		lim f (x) = f (lim x) is valid.
	mined, the equality		lim f (x) = f (lim x) is valid.
			x→a			x→a
Using Theorem 3.6. we can prove that there exists								+	1 n
Using Theorem 3.6. we can prove that there exists							lim 1	+	. We denote
this limit as e = 2,71828….							n→∞		n
this limit as e = 2,71828….
That is,

				1	n
		lim 1+		n	= e.
		n→∞		n

12.4. The Determinate and Indeterminate Expressions

While calculating the limits of variables it is necessary to take into account the following:

142

1)a sum and a product of the finite number of infinitesimal magnitudes, and also a product of an infinitesimal by a bounded magnitude are infinitesimal magnitudes;

2)a sum and a product of infinitely large magnitudes, and also a product of infinitely large magnitude by a nonzero constant are infinitely large magnitudes;

3)a quotient of a constant by an infinitely large magnitude is an infinitesimal, a quotient of nonzero constant by an infinitesimal is an infinitely large magnitude.

In the examples of finding the limits sometimes indeterminate expressions occur: the ratio of two infinitesimal magnitudes; the ratio of two infinitely large magnitudes; a difference of two infinitely large magnitudes; a product of an infinitesimal by an infinitely large magnitude; an infinitesimal or infinitely large magnitude in an infinitesimal degree; a magnitude which tends to unit in an infinitely large degree. Symbolically the indeterminate expressions can be written down as:

∞

∞ − ∞, 0 ∞, 00 , ∞0 , 1∞.

∞ ,

Micromodule 12

EXAMPLES OF PROBLEMS SOLUTION

Еxample 1. Prove that

lim

n2 + 1

= 1.

n→∞

N (ε ) so that

for n > N (ε ).

Solution. Let us find such number

xn − 1

< ε

For this purpose we should solve the inequality

n2 + 1

−1

< ε.

(3.1)

As far as n N we get from the inequality (3.1):

n2 + 1

−1 < ε n2 + 1 <

(ε + 1)2

< ε2

+ 2ε n >

ε2 + 2ε

If to take

N (ε )

so that it would be equal to the nearest but greater than

natural

number

then

xn − 1

< ε

for all

n > N (ε ).

From here it

ε2 + 2ε

+ 1

= 1.

follows that lim

n→∞

7n4 − 5n2 + 4n

Еxample 2. Find

lim

n→∞ 10 + 2n − 3n4

143

Solution. We have an indeterminacy of a kind	∞		. We evaluate it by
	∞

dividing the numerator and the denominator by the greatest power of n:

7n4 − 5n2 + 4n

7 −

lim

= lim

n→∞ 10 + 2n − 3n4

n→∞

10 2

n4 +

− 3

lim 7 − lim

+ lim

n→∞

n→∞ n2

n→∞ n3

7 − 0 + 0

= −

7 .

lim 10

0 + 0 − 3

+ lim

− lim 3

n→∞ n4

n→∞ n3

n→∞

By means of division of the numerator and the denominator by the greatest power of n it is possible to prove the validity of such formula:

	a0n	m	+ a1n	m−1	+ …+ am	a0 / b0 ,		if m = k ;
								if	m < k;
lim						=	0,

n→∞ b0nk + b1nk −1 + …+ bk							∞,	if	m > k.

This formula is valid if we consider in the similar way limits of sequences where there is a value n and n → ∞ instead of x.

Еxample 3. Find

lim

n5 − n3

+ 4

n→∞ (n − 2)7 − 5

Solution. We have an indeterminacy of a kind ∞∞ . Then

−

− 2

lim

− n3

+ 4

= lim

(1− n

+ 4n

5 )

= lim

n→∞

3 n→∞

− 2)7

− 5

2 7

−

− 5n

− 7

−

	2
= lim	n5	= lim	1		= lim	1		=	1	= 0 .
= lim	3	= lim	3	− 2	= lim		1	=	∞	= 0 .
n→∞	3	n→∞	3	− 2	n→∞		1		∞
	n7		n7	5		n35

144

Here we have used limits

lim n p

= 0, if

p < 0 ;

n→∞

lim n p

= ∞, if

p > 0.

n→∞

Еxample 4. Find lim

n!− (n + 1)!

n→∞ (n + 2)!− n!

Solution. We have an indeterminacy of a kind ∞

. Then

∞

lim

n!− (n + 1)!

= lim

n!− n!(n + 1)

n→∞

(n + 2)!− n!

n→∞ n!(n + 1)(n + 2) − n!

= lim

n!(1− n − 1)

= lim

−n

= lim

−n

= 0

n→∞ n!((n + 1)(n

+ 2) − 1)

n→∞ (n + 1)(n + 2) −1

n→∞ n2 + 3n + 1

Еxample 5. Find lim 1+ 3 + 5 + …+ (2n −1) .

n→∞ 1+ 4 + 7 + …+ (3n − 2)

∞

. We use the formula

Solution. We have an indeterminacy of a kind

a1 + an

∞

Sn =

for sum of n terms of arithmetical progression:

1+ 3 + 5 + …+ (2n − 1) =

1+ 2n − 1 n = n2 ,

n(3n − 1)

1+ 4 + 7 + …+ (3n − 2) =

1+ 3n − 2

n =

Then

lim

1+ 3 + 5 + …+ (2n − 1)

= lim

2n2

= lim

1+ 4 + 7

+ …+ (3n − 2)

3n − 1

n→∞

n→∞ n(3n − 1)

n→∞

Еxample 6. lim (

n2 − 2n − 3 − n) .

n→∞

Solution. We have an indeterminacy of a kind (∞ − ∞). Let us multiply the numerator and the denominator of the given fraction by the expression conjugate

to the numerator		n2 − 2n − 3 + n . Then
lim ( n	2	− 2n − 3 − n) = lim	( n2	− 2n − 3 − n)( n2 − 2n − 3 + n)	=
				n2 − 2n − 3 + n
n→∞		n→∞

145

= lim

n2 − 2n − 3 − n2

= lim

−2n − 3

n→∞ n2 − 2n − 3 + n

∞

−2 −

−2

= lim

= −1

1+ 1

∞

n→∞

1−

−

+ 1

Micromodule 12

CLASS AND HOME ASSIGNMENT

Using the definition of a limit prove that

lim x

= a .

n→∞

lim

3n − 5

= 3 .

lim

n2 + 5

= 1 .

lim 6n2 + 7

= 6 .

n→∞

2n − 1 2

n→∞

n→∞ 5n2 + 6

lim an = 0 , if

< 1 .

lim

2n + 3

2 .

n→∞

3n + 2

Evaluate limits of sequences:

6. а) lim

3n − 2

;

b) lim

n3 + 1

;

c) lim

100n2 + n + 1 .

n→∞ n + 1000

n→∞

2n + 11

n→∞

4n3 + 5

7.	lim		4n4 − 5n2 + 4				.
						3)
n→∞ (n3 − n + 2)(2n +
9.	lim	n! (n2 + 1)			.
			(n + 2)!
n→∞
11.	lim		3 + 5 + …+ (2n + 1) .
	n→∞		2n2 − 3n + 4
13.	lim		3n + 2n	.

	n→∞		3n + 4n
15.	lim ( 4n2 − n − 2 − 2n) .
	n→∞

			13		14
8.	lim	n 3		+ 3n		4 + 2	.
				16
n→∞
				n 5 −1

10.	lim			(n + 2)!+ n!				.
			(n
	n→∞			+ 2)!− (n + 1)!
12.	lim		3n + 1			.

	n→∞ 3n+1 + 1
			1	−1
14.	lim		2n		.
			1
	n→∞
			4n −1

146

Answers

6. а) 3; b) ∞; c) 0. 7. 2. 8. ∞. 9. 1. 10. 1. 11. 1/2. 12. 1/3. 13. 0. 14. 1/2.

15.–1/4.

Micromodule 12

SELF-TEST ASSIGNMENTS

Evaluate limits of sequences:

12.1. а) lim	(2 + n)3 − (3 + n)3 + 1		;
	(2 + n)2	+ (3 + n)2
n→∞

lim

(n + 2)!+ (n + 1)!

;

(n + 3)!

n→∞

12.2. а)

lim

3n3 − 2(n + 1)2 + 2

;

(2n +

1)3 + n −1

n→∞

lim

1+ 5 + …+ (4n − 3)

;

n→∞

1+ 7 + …+ (6n − 5)

12.3. а)

lim

5n4 − 3n + 1

;

2 − 7n − 2n4

n→∞

lim

2 + 4 + 6 + ... + 2n

;

1+ 3 + 5 + ... +

(

2n +

)

n→∞

12.4. а)

lim

1000n4 − 3n + 1

;

2 − 7n − n5

n→∞

lim

n5 + 2n6 + 1

;

n→∞

n10 − 4

12.5. а)

lim

(2n)4 − 3n3 + n

;

4n + n3

n→∞ 100 +

lim

(n + 2)!+ (n + 1)!

;

n→∞

(n + 3)!

lim

n 3

+ 2n 4 + 1

;

n→∞

n 4

− 2

lim

− 2n

+ 2n

n→∞ 5n

lim

− n

5 + 3

;

n→∞

(n − 1)8

lim (

n2 − 4n + 3 −

n2 + 1) .

n→∞

lim

+ 2n

4 + 1

;

n→∞

(n + 3)8

( n3 + 3 −

n3 − 2 ).

lim n2

n→∞

lim

−

+ ... +

(−1)

n−1

;

n→∞

lim (

n2 − n + 3 −

n2 + n) .

n→∞

lim

+ 2n4 + 5

;

n→∞

n 8

− 2

− 1

lim

n→∞ 5n + 1

147

12.6. а)

(3n)4

+ 5n3 − 2n

n4 + 5 +

3 7 − 2n3

lim

;

b) lim

;

n→∞ (4n)3 − n − 6

n→∞

3 3 − 8n6 + n −1

− 1

;

lim

...

n→∞

8n + 1

12.7. а)

lim

(2n −1)2 (n − 3) + 6n3 − n

;

lim

n3 − 3 +

3 5n5

;

(n − 1)3 + n +

4 2n5 − 7

n8 + 5

n→∞

...

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

lim

;

lim

(

+ 2

)

!−

(

)

n→∞

... +

n→∞

n + 1 !

12.8. а)

lim

(3n + 2)2 (n − 1) + 3n2 + 1

;

lim

n4 − 2 + 3 n7

;

(n − 2)3 +

2n +

4 2n5 − 7

n→∞

n16 + 1

...

(n + 3)!+ (n + 2)! .

lim

;

lim

n→∞

... +

n→∞

(n + 3)!− (n + 2)!

12.9. а)

lim

(4n −1)2 (n + 2) + 2n2 − 1

;

lim

n5 + 1 +

4 n9

;

(2n −1)3

+ n2

n→∞

n→∞ 4 2n11 − 7 n9 + 1

...

(−1)n

1−

(n −1)!− (n − 2)! .

lim

;

lim

(−1)

(

n − 1 !+

(

n − 2

)

n→∞

)

1−

... +

12.10. а)

lim

(3n − 1)2 (n − 2) − 5n3

;

lim

n − 3 + 3 3n4

;

8(n − 1)3 +

4n +

4 2n3 − 3

6n4

n→∞

... +

(n + 2)!+ n! .

lim

;

lim

(−1)n

n→∞

... +

n→∞

(n + 2)!− n!

1−

148

12.11. а)

lim

(2n −1)2 (n −1) − 5n3

8(n + 1)5 + n + 1

;

n→∞

...

lim

;

(−1)n

n→∞

1−

... +

12.12. а)

lim

(2n + 1)2 (n + 2) + 3n3

5(n − 3)4

;

n→∞

... +

c) lim

;

(−1)n

n→∞

1−

... +

12.13. а)

lim

(3n − 1)(n −1)2 + 2n3

10(n + 1)3 + 1

;

n→∞

lim

1+ 2 + 3 − 4 + ...− 2n ;

n→∞

n2 + 1

12.14. а)

lim

(2n − 1)(n − 3)2 + 7n3

14(n − 1)3 + 2

;

n→∞

lim

1+ 3 + 5 + ... + (2n −1) ;

n→∞

2n2 + 3

12.15. а)

lim

(2n −1)(n − 3)2 − n3

;

11(n − 2)3 + 22

n→∞

lim

1+ 4 + 7 + ... + (3n − 2) ;

n→∞

2n2 + 1

12.16. а)

lim

(4n −1)(n + 1)2 + 3n3

6(n + 2)3 + 5

;

n→∞

lim

1+ 5 + 9 + ... + (4n − 3) ;

n→∞

3n2 + 1

3	n4 + 7 − 6	5n7
b) lim			;
	5 n7 + 3 +
n→∞		n

2(n + 2)!+ 3 n! d) lim ( ) .

n→∞ n + 2 !− n!

3	2n4 + 1	+ 5	n7
b) lim				;
	5 n8 + 2
n→∞		+	n

3(n + 1)!+ (n − 1)! d) lim ( ) .

n→∞ n + 1 !− (n − 1)!

3	n7 + 5 + 4	3n9
b) lim			;
	5 n7 + 2 +
n→∞		n

d)	lim	(n + 2)!+ 5 n! .
	n→∞	2(n + 2)!− n!
b)		5 n7 + 4 2n11
	lim					;
		5 n7 + 1 +
	n→∞		n5
d)	lim	(n + 3)!+ 5 (n + 2)! .
	n→∞	3(n + 3)!− (n + 2)!
b)	lim	4 n11 + 4 3n13
				;
		5 n16 + 1 +
	n→∞		n
d)	lim	(n + 3)!− 4 (n + 2)! .
	n→∞	2(n + 3)!+ (n + 2)!
		3 n10 + 4 2n13
b)	lim				;

	n→∞	5 n16 + 1 + 2
d)	lim	(n + 3)!− 5 (n + 2)! .
	n→∞	3(n + 3)!+ (n + 2)!

149

12.17. а)

lim

(n + 2)3 − (n − 1)3

;

n→∞

3n2 + 2n − 1

lim

1+ 6 + 11+ ... + (5n − 4)

;

n→∞

4n2 + 1

12.18. а)

lim

(n + 3)3 − (n − 2)3

;

n→∞

5n2 + 2n − 6

lim 1+ 7 + 13 + ... + (6n − 5) ;

n→∞

5n2 + 1

12.19. а)

lim

(n + 1)4 − (n −1)4

;

n→∞

2n3 + n + 1

lim

+ ... +

n − 1

;

n→∞

12.20. а)

lim

(n + 2)4 − (n − 2)4

;

n→∞

4n3 + 1

lim

+ ... +

3n − 2

;

n→∞

12.21. а)

lim

5n(n − 1)3 − 2n

;

2 −

8n2 − 3n4

n→∞

lim

2 + 6 + 10 + ... + (4n − 2)

;

3 + 5 +

... +

(

)

n→∞

2n + 1

12.22. а)

lim

(2n + 1)(2n −1)2 + 5n3

;

3(n − 2)3 + 1

n→∞

lim

1+ 3 + 5 + ... + (2n − 1) ;

n→∞

(2n + 3)(n + 1)

12.23. а)

lim

(2n − 3)2 (n − 3) − 6n3

;

2(2n − 1)3

n→∞

... +

lim

;

(−1)n

n→∞

1−

... +

150

3 n7 + 4 5n9

lim

;

n→∞

7 n16 − 1 + 3

lim

(n − 1)!− 2 (n − 2)! .

(

)

− 2)!

n→∞

n −1 !+ (n

3 n5 + 4 3n7

lim

;

n→∞

5 n8 − 1 + 1

lim

n!− 2 (n − 2)! .

n→∞

n!+ (n − 2)!

3 n4 + 5 6n6

lim

;

n→∞

5 n6 −1 + 2

lim

(n + 1)!− 2 (n − 1)! .

(

)

−1)!

n→∞

n + 1 !+ (n

5 n4 + 5 3n3

lim

;

n→∞

7 n6 + 1 − 1

lim

(n + 1)!− 4 (n − 1)!

(

)

−1)!

n→∞

n + 1 !+ (n

( n3 + 2 − n3 − 4 );

lim n2

n→∞

lim

n7 + 4n8 + 2

n→∞

(n + 1)16

lim

4 n7 + 5 4n11

;

n→∞

3 n7 + 1 + n7

lim

1+ 4n+ 2 .

n→∞

5 − 3 4n

n + 2 + 7 3n4

lim

;

n3 − 9 3n4

n→∞

lim

(n + 4)!+ (n + 2)! .

n→∞

(n + 4)!− (n + 3)!

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