1.Вычислить предел последовательности: а) Lim 5x-7/2x
x→+∞

б) Lim 4x²+3x/2x⁵-7x³
x→+∞

в) Lim 7n²-6n+8/4n²+2n+6
n→+∞
г) Lim 8n³-3n+17/3n³-61n+1
n→+∞

д) Lim 14n⁵-23n²+4n-1/19n⁵+13n⁴-2n+8
n→+∞

е) Lim 6x²-17x+11/3x²-21x-17
x→+∞

ж) Lim (5x+1)²/ 7x²-x+10
x→+∞

з) Lim 3x²-7x²+17x-13/7x³-8x²+25
x→+∞

и) Lim 6x²-11n-3/7x²-x+5
x→+∞

к) Lim 8n²-3n/n-2
n→+∞

а) Lim (5x-7)/(2x) as x→+∞

To evaluate this limit, we need to focus on the terms with the highest power of x in the numerator and denominator. In this case, it is 5x in the numerator and 2x in the denominator.

Dividing every term in the numerator and denominator by x, we get:
Lim (5x/x - 7/x)/(2x/x) as x→+∞
Lim (5 - 7/x)/(2)

As x approaches infinity, 7/x approaches 0 since the denominator grows much faster than the numerator. Therefore, we can simplify the limit further:
Lim (5 - 0)/(2)
Lim 5/2

So, the limit of the sequence is 5/2.

б) Lim (4x²+3x)/(2x⁵-7x³) as x→+∞

In this example, we need to focus on the terms with the highest power of x in the numerator and denominator. It is 2x⁵ in the denominator.

Dividing every term in the numerator and denominator by x⁵, we get:
Lim ((4x²/x⁵) + (3x/x⁵))/((2x⁵/x⁵) - (7x³/x⁵)) as x→+∞
Lim (4/x³ + 3/x⁴)/(2 - 7/x²)

As x approaches infinity, 4/x³, 3/x⁴, and 7/x² approach 0 since the denominators grow much faster than the numerators.

Therefore, we can simplify the limit further:
Lim (0 + 0)/(2 - 0)
Lim 0/2
Lim 0

So, the limit of the sequence is 0.

в) Lim (7n²-6n+8)/(4n²+2n+6) as n→+∞

In this example, we again focus on the terms with the highest power of n in the numerator and denominator. It is 4n² in the denominator.

Dividing every term in the numerator and denominator by n², we get:
Lim ((7n²/n²) - (6n/n²) + (8/n²))/((4n²/n²) + (2n/n²) + (6/n²)) as n→+∞
Lim (7 - 6/n + 8/n²)/(4 + 2/n + 6/n²)

As n approaches infinity, 6/n and 8/n² approach 0 since the denominators grow much faster than the numerators.
Similarly, 2/n and 6/n² also approach 0.

Therefore, we can simplify the limit further:
Lim (7 - 0 + 0)/(4 + 0 + 0)
Lim 7/4

So, the limit of the sequence is 7/4.

г) Lim (8n³-3n+17)/(3n³-61n+1) as n→+∞

In this example, the terms with the highest power of n in the numerator and denominator are both n³.

Dividing every term in the numerator and denominator by n³, we get:
Lim ((8n³/n³) - (3n/n³) + (17/n³))/((3n³/n³) - (61n/n³) + (1/n³)) as n→+∞
Lim (8 - 3/n² + 17/n³)/(3 - 61/n² + 1/n³)

As n approaches infinity, 3/n², 17/n³, 61/n², and 1/n³ all approach 0 since the denominators grow much faster than the numerators.

Therefore, we can simplify the limit further:
Lim (8 - 0 + 0)/(3 - 0 + 0)
Lim 8/3

So, the limit of the sequence is 8/3.

д) Lim (14n⁵-23n²+4n-1)/(19n⁵+13n⁴-2n+8) as n→+∞

In this example, the terms with the highest power of n in the numerator and denominator are both n⁵.

Dividing every term in the numerator and denominator by n⁵, we get:
Lim ((14n⁵/n⁵) - (23n²/n⁵) + (4n/n⁵) - (1/n⁵))/((19n⁵/n⁵) + (13n⁴/n⁵) - (2n/n⁵) + (8/n⁵)) as n→+∞
Lim (14 - 23/n³ + 4/n⁴ - 1/n⁵)/(19 + 13/n + 2/n⁴ + 8/n⁵)

As n approaches infinity, 23/n³, 4/n⁴, 1/n⁵, 13/n, 2/n⁴, and 8/n⁵ all approach 0 since the denominators grow much faster than the numerators.

Therefore, we can simplify the limit further:
Lim (14 - 0 + 0 - 0)/(19 + 0 + 0 + 0)
Lim 14/19

So, the limit of the sequence is 14/19.

е) Lim (6x²-17x+11)/(3x²-21x-17) as x→+∞

In this example, the terms with the highest power of x in the numerator and denominator are both x².

Dividing every term in the numerator and denominator by x², we get:
Lim ((6x²/x²) - (17x/x²) + (11/x²))/((3x²/x²) - (21x/x²) - (17/x²)) as x→+∞
Lim (6 - 17/x + 11/x²)/(3 - 21/x - 17/x²)

As x approaches infinity, 17/x and 11/x² approach 0 since the denominators grow much faster than the numerators.

Therefore, we can simplify the limit further:
Lim (6 - 0 + 0)/(3 - 0 - 0)
Lim 6/3
Lim 2

So, the limit of the sequence is 2.

ж) Lim ((5x+1)²)/(7x²-x+10) as x→+∞

In this example, the terms with the highest power of x in the numerator and denominator are both x².

Expanding the numerator, we get:
Lim (25x² + 10x + 1)/(7x² - x + 10) as x→+∞

We can ignore the constant terms (10x and 1) when considering the limit.

Dividing every term in the numerator and denominator by x², we get:
Lim ((25x²/x²) + (10x/x²) + (1/x²))/((7x²/x²) - (x/x²) + (10/x²)) as x→+∞
Lim (25 + 10/x + 1/x²)/(7 - 1/x + 10/x²)

As x approaches infinity, 10/x and 1/x² approach 0 since the denominators grow much faster than the numerators.

Therefore, we can simplify the limit further:
Lim (25 + 0 + 0)/(7 - 0 + 0)
Lim 25/7

So, the limit of the sequence is 25/7.

з) Lim (3x²-7x²+17x-13)/(7x³-8x²+25) as x→+∞

In this example, the term with the highest power of x in the denominator is x³.

Dividing every term in the numerator and denominator by x³, we get:
Lim ((3x²/x³) - (7x²/x³) + (17x/x³) - (13/x³))/((7x³/x³) - (8x²/x³) + (25/x³)) as x→+∞
Lim (3 - 7/x + 17/x² - 13/x³)/(7 - 8/x + 25/x³)

As x approaches infinity, 7/x, 17/x², and 13/x³ all approach 0 since the denominators grow much faster than the numerators.
Similarly, 8/x and 25/x³ also approach 0.

Therefore, we can simplify the limit further:
Lim (3 - 0 + 0 - 0)/(7 - 0 + 0)
Lim 3/7

So, the limit of the sequence is 3/7.

и) Lim (6x²-11n-3)/(7x²-x+5) as x→+∞

In this example, the terms with the highest power of x and n are x² and n respectively.

As x approaches infinity, we can ignore the terms -11n and -3 since they are independent of x.
Dividing every term in the numerator and denominator by x², we get:
Lim (6 - 11n/x² - 3/x²)/(7 - 1/x + 5/x²) as x→+∞

As x approaches infinity, 11n/x² and 3/x² approach 0 since the denominators grow much faster than the numerators.
Similarly, 1/x and 5/x² also approach 0.

Therefore, we can simplify the limit further:
Lim (6 - 0 - 0)/(7 - 0 + 0)
Lim 6/7

So, the limit of the sequence is 6/7.

к) Lim (8n²-3n)/(n-2) as n→+∞

In this example, the terms with the highest power of n in the numerator and denominator are both n².

Dividing every term in the numerator and denominator by n², we get:
Lim ((8n²/n²) - (3n/n²))/((n/n²) - (2/n²)) as n→+∞
Lim (8 - 3/n)/(1 - 2/n²)

As n approaches infinity, 3/n and 2/n² approach 0 since the denominators grow much faster than the numerators.

Therefore, we can simplify the limit further:
Lim (8 - 0)/(1 - 0)
Lim 8/1
Lim 8

So, the limit of the sequence is 8.

I hope this explanation helps you understand how to calculate the limits of the given sequences. If you have any further questions, feel free to ask!