Вычислите пределы функций: lim(x->2) (x^2-x-2)/(x^2+x-6), lim(x->inf) (2x^2+3x-1)/(3x^2+2x+1), lim(x->3) (x^2-9)/(x-3), lim(x->5) (sqrt(x-1)-2)/(x-5), lim(x->-1) (x^3+1)/(x+1)

1. $\lim_{x \to 2} \left( \frac{x^2 - x - 2}{x^2 + x - 6} \right) = \lim_{x \to 2} \frac{(x - 2)(x + 1)}{(x - 2)(x + 3)} = \lim_{x \to 2} \frac{x + 1}{x + 3} = \frac{2 + 1}{2 + 3} = \frac{3}{5} = 0,6$ 2. $\lim_{x \to \infty} \left( \frac{2x^2 + 3x - 1}{3x^2 + 2x + 1} \right) = \lim_{x \to \infty} \frac{2 + \frac{3}{x} - \frac{1}{x^2}}{3 + \frac{2}{x} + \frac{1}{x^2}} = \frac{2 + 0 - 0}{3 + 0 + 0} = \frac{2}{3}$ 3. $\lim_{x \to 3} \left( \frac{x^2 - 9}{x - 3} \right) = \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 3 + 3 = 6$ 4. $\lim_{x \to 5} \left( \frac{\sqrt{x - 1} - 2}{x - 5} \right) = \lim_{x \to 5} \frac{(\sqrt{x - 1} - 2)(\sqrt{x - 1} + 2)}{(x - 5)(\sqrt{x - 1} + 2)} = \lim_{x \to 5} \frac{(x - 1) - 4}{(x - 5)(\sqrt{x - 1} + 2)} = \lim_{x \to 5} \frac{x - 5}{(x - 5)(\sqrt{x - 1} + 2)} = \lim_{x \to 5} \frac{1}{\sqrt{x - 1} + 2} = \frac{1}{\sqrt{5 - 1} + 2} = \frac{1}{2 + 2} = \frac{1}{4} = 0,25$ 5. $\lim_{x \to -1} \left( \frac{x^3 + 1}{x + 1} \right) = \lim_{x \to -1} \frac{(x + 1)(x^2 - x + 1)}{x + 1} = \lim_{x \to -1} (x^2 - x + 1) = (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3$ **Ответ:** 1) 0,6; 2) 2/3; 3) 6; 4) 0,25; 5) 3.