7.a) lim n→∞ 2n³ + n² n³ - 4n + 1 ; 6) lim n→∞ n³ + n² - n + 1 5n³ - 4n + 17 ; B) lim n→∞ 3n⁴ + 1 n³ + 2n² . 8.a) lim n→∞ 2n² - 3n + 1 n² - 4n ; 6) lim n→∞ n³ + n² + 1 4n² + 7 ; B) lim n→∞ 3n⁴ + 1 n⁴ + 2n² - 1 . 9.a) lim n→∞ 3n - n² 2n + 1 ; 6) lim n→∞ 5 - 3n - n² n² + n + 1 ; B) lim n→∞ 1 n³ + 8 .

7.a)

\[\lim_{n \to \infty} \frac{2n^3 + n^2}{n^3 - 4n + 1} = \lim_{n \to \infty} \frac{\frac{2n^3}{n^3} + \frac{n^2}{n^3}}{\frac{n^3}{n^3} - \frac{4n}{n^3} + \frac{1}{n^3}} = \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{1 - \frac{4}{n^2} + \frac{1}{n^3}} = \frac{2 + 0}{1 - 0 + 0} = 2\]

Ответ: 2

7.б)

\[\lim_{n \to \infty} \frac{n^3 + n^2 - n + 1}{5n^3 - 4n + 17} = \lim_{n \to \infty} \frac{\frac{n^3}{n^3} + \frac{n^2}{n^3} - \frac{n}{n^3} + \frac{1}{n^3}}{\frac{5n^3}{n^3} - \frac{4n}{n^3} + \frac{17}{n^3}} = \lim_{n \to \infty} \frac{1 + \frac{1}{n} - \frac{1}{n^2} + \frac{1}{n^3}}{5 - \frac{4}{n^2} + \frac{17}{n^3}} = \frac{1 + 0 - 0 + 0}{5 - 0 + 0} = \frac{1}{5}\]

Ответ: 1/5

7.в)

\[\lim_{n \to \infty} \frac{3n^4 + 1}{n^3 + 2n^2} = \lim_{n \to \infty} \frac{\frac{3n^4}{n^3} + \frac{1}{n^3}}{\frac{n^3}{n^3} + \frac{2n^2}{n^3}} = \lim_{n \to \infty} \frac{3n + \frac{1}{n^3}}{1 + \frac{2}{n}} = \frac{\infty + 0}{1 + 0} = \infty\]

Ответ: ∞

8.a)

\[\lim_{n \to \infty} \frac{2n^2 - 3n + 1}{n^2 - 4n} = \lim_{n \to \infty} \frac{\frac{2n^2}{n^2} - \frac{3n}{n^2} + \frac{1}{n^2}}{\frac{n^2}{n^2} - \frac{4n}{n^2}} = \lim_{n \to \infty} \frac{2 - \frac{3}{n} + \frac{1}{n^2}}{1 - \frac{4}{n}} = \frac{2 - 0 + 0}{1 - 0} = 2\]

Ответ: 2

8.б)

\[\lim_{n \to \infty} \frac{n^3 + n^2 + 1}{4n^2 + 7} = \lim_{n \to \infty} \frac{\frac{n^3}{n^2} + \frac{n^2}{n^2} + \frac{1}{n^2}}{\frac{4n^2}{n^2} + \frac{7}{n^2}} = \lim_{n \to \infty} \frac{n + 1 + \frac{1}{n^2}}{4 + \frac{7}{n^2}} = \frac{\infty + 1 + 0}{4 + 0} = \infty\]

Ответ: ∞

8.в)

\[\lim_{n \to \infty} \frac{3n^4 + 1}{n^4 + 2n^2 - 1} = \lim_{n \to \infty} \frac{\frac{3n^4}{n^4} + \frac{1}{n^4}}{\frac{n^4}{n^4} + \frac{2n^2}{n^4} - \frac{1}{n^4}} = \lim_{n \to \infty} \frac{3 + \frac{1}{n^4}}{1 + \frac{2}{n^2} - \frac{1}{n^4}} = \frac{3 + 0}{1 + 0 - 0} = 3\]

Ответ: 3

9.a)

\[\lim_{n \to \infty} \frac{3n - n^2}{2n + 1} = \lim_{n \to \infty} \frac{\frac{3n}{n} - \frac{n^2}{n}}{\frac{2n}{n} + \frac{1}{n}} = \lim_{n \to \infty} \frac{3 - n}{2 + \frac{1}{n}} = \frac{3 - \infty}{2 + 0} = -\infty\]

Ответ: -∞

9.б)

\[\lim_{n \to \infty} \frac{5 - 3n - n^2}{n^2 + n + 1} = \lim_{n \to \infty} \frac{\frac{5}{n^2} - \frac{3n}{n^2} - \frac{n^2}{n^2}}{\frac{n^2}{n^2} + \frac{n}{n^2} + \frac{1}{n^2}} = \lim_{n \to \infty} \frac{\frac{5}{n^2} - \frac{3}{n} - 1}{1 + \frac{1}{n} + \frac{1}{n^2}} = \frac{0 - 0 - 1}{1 + 0 + 0} = -1\]

Ответ: -1

9.в)

\[\lim_{n \to \infty} \frac{1}{n^3 + 8} = \frac{1}{\infty + 8} = 0\]

Ответ: 0