Question

Решите вариант 5.10 класс. Даю 40 баллов

Answer 1

$1)lim_{-3 }( \frac{1}{3} {x}^{2} - 2x + 1 ) = \\ = \frac{1}{3} \times 9 + 6 + 1 = 10 \\ 2)lim_{3} \frac{ {x}^{2} + x - 12}{x - 3} = \\ = lim_{3} \frac{ (x - 3)(x + 4)}{x - 3} = 7 \\ 3) lim_{ \infty } \frac{ 2{n}^{2} + 9}{ {n}^{2} + n - 8 } = \\ = lim_{ \infty } \frac{ 2 + \frac{9}{ {n}^{2} } }{ 1 + \frac{1}{n} - \frac{8}{ {n}^{2} } } = 2 \\ 4)lim_{ \infty } \frac{ 3x + 5}{x} = \\ =lim_{ \infty }( 3 + \frac{5}{x} ) = 3$