Реши предел функции 3x³ + x при x стремящемся к 0.

Привет! Конечно, помогу тебе разобраться с пределами функций. Вот решения: 1. $\lim_{x \to 0} \frac{3x^3 + x}{x} = \lim_{x \to 0} (3x^2 + 1) = 3(0)^2 + 1 = 1$ 2. $\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 2x - 3} = \lim_{x \to 3} \frac{(x - 3)(x + 3)}{(x - 3)(x + 1)} = \lim_{x \to 3} \frac{x + 3}{x + 1} = \frac{3 + 3}{3 + 1} = \frac{6}{4} = \frac{3}{2}$ 3. $\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} = \lim_{x \to 0} \frac{(\sqrt{x + 1} - 1)(\sqrt{x + 1} + 1)}{x(\sqrt{x + 1} + 1)} = \lim_{x \to 0} \frac{x + 1 - 1}{x(\sqrt{x + 1} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 1} + 1)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1} = \frac{1}{\sqrt{0 + 1} + 1} = \frac{1}{2}$ 4. $\lim_{x \to 2} \frac{x^2 - 4}{x + 2} = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x + 2} = \lim_{x \to 2} (x - 2) = 2 - 2 = 0$ 5. $\lim_{x \to 3} \frac{3 - x}{x^3 - 27} = \lim_{x \to 3} \frac{-(x - 3)}{(x - 3)(x^2 + 3x + 9)} = \lim_{x \to 3} \frac{-1}{x^2 + 3x + 9} = \frac{-1}{3^2 + 3(3) + 9} = \frac{-1}{27}$ 6. $\lim_{x \to 3} \frac{3x^2 - 11x + 6}{2x^2 - 5x - 3} = \lim_{x \to 3} \frac{(3x - 2)(x - 3)}{(2x + 1)(x - 3)} = \lim_{x \to 3} \frac{3x - 2}{2x + 1} = \frac{3(3) - 2}{2(3) + 1} = \frac{7}{7} = 1$ 7. $\lim_{x \to 5} \frac{x - 5}{2 - \sqrt{x - 1}} = \lim_{x \to 5} \frac{(x - 5)(2 + \sqrt{x - 1})}{(2 - \sqrt{x - 1})(2 + \sqrt{x - 1})} = \lim_{x \to 5} \frac{(x - 5)(2 + \sqrt{x - 1})}{4 - (x - 1)} = \lim_{x \to 5} \frac{(x - 5)(2 + \sqrt{x - 1})}{5 - x} = \lim_{x \to 5} -(2 + \sqrt{x - 1}) = -(2 + \sqrt{5 - 1}) = -(2 + 2) = -4$ 8. $\lim_{x \to \sqrt{5}} \frac{x^4 - 25}{x^2 - 5} = \lim_{x \to \sqrt{5}} \frac{(x^2 - 5)(x^2 + 5)}{x^2 - 5} = \lim_{x \to \sqrt{5}} (x^2 + 5) = (\sqrt{5})^2 + 5 = 5 + 5 = 10$ 9. $\lim_{x \to 3} \frac{x - 3}{x^2 - 9} = \lim_{x \to 3} \frac{x - 3}{(x - 3)(x + 3)} = \lim_{x \to 3} \frac{1}{x + 3} = \frac{1}{3 + 3} = \frac{1}{6}$ 10. $\lim_{x \to 5} \frac{x^2 - 8x + 15}{x^2 - 25} = \lim_{x \to 5} \frac{(x - 5)(x - 3)}{(x - 5)(x + 5)} = \lim_{x \to 5} \frac{x - 3}{x + 5} = \frac{5 - 3}{5 + 5} = \frac{2}{10} = \frac{1}{5}$ 11. $\lim_{x \to 2} \frac{3x^2 - 8x + 4}{5x^2 - 14x + 8} = \lim_{x \to 2} \frac{(3x - 2)(x - 2)}{(5x - 4)(x - 2)} = \lim_{x \to 2} \frac{3x - 2}{5x - 4} = \frac{3(2) - 2}{5(2) - 4} = \frac{4}{6} = \frac{2}{3}$ 12. $\lim_{x \to 6} \frac{x - 6}{\sqrt{x + 3} - 3} = \lim_{x \to 6} \frac{(x - 6)(\sqrt{x + 3} + 3)}{(\sqrt{x + 3} - 3)(\sqrt{x + 3} + 3)} = \lim_{x \to 6} \frac{(x - 6)(\sqrt{x + 3} + 3)}{x + 3 - 9} = \lim_{x \to 6} \frac{(x - 6)(\sqrt{x + 3} + 3)}{x - 6} = \lim_{x \to 6} (\sqrt{x + 3} + 3) = \sqrt{6 + 3} + 3 = \sqrt{9} + 3 = 3 + 3 = 6$ 13. $\lim_{x \to 0} \frac{x}{\sqrt{x + 4} - 2} = \lim_{x \to 0} \frac{x(\sqrt{x + 4} + 2)}{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{x(\sqrt{x + 4} + 2)}{x + 4 - 4} = \lim_{x \to 0} \frac{x(\sqrt{x + 4} + 2)}{x} = \lim_{x \to 0} (\sqrt{x + 4} + 2) = \sqrt{0 + 4} + 2 = \sqrt{4} + 2 = 2 + 2 = 4$ 14. $\lim_{x \to 2} \frac{4x^2 - 7x - 2}{5x^2 - 9x - 2} = \lim_{x \to 2} \frac{(4x + 1)(x - 2)}{(5x + 1)(x - 2)} = \lim_{x \to 2} \frac{4x + 1}{5x + 1} = \frac{4(2) + 1}{5(2) + 1} = \frac{9}{11}$ Надеюсь, тебе всё понятно! Если что, спрашивай.