x-23x-1 -15. limx→∞ 3x+4

15. $$lim_{x \to \infty} (\frac{x-2}{3x+4})^{3x-1}$$

$$lim_{x \to \infty} (\frac{x-2}{3x+4})^{3x-1} = lim_{x \to \infty} (\frac{x(1-\frac{2}{x})}{3x(1+\frac{4}{3x})})^{3x-1} = lim_{x \to \infty} (\frac{1-\frac{2}{x}}{3(1+\frac{4}{3x})})^{3x-1} = lim_{x \to \infty} \frac{(1-\frac{2}{x})^{3x-1}}{(3(1+\frac{4}{3x}))^{3x-1}} = \frac{lim_{x \to \infty} (1-\frac{2}{x})^{3x-1}}{lim_{x \to \infty} (3(1+\frac{4}{3x}))^{3x-1}}$$

$$lim_{x \to \infty} (1-\frac{2}{x})^{3x-1} = lim_{x \to \infty} ((1-\frac{2}{x})^{\frac{x}{-2}})^{\frac{-2(3x-1)}{x}} = e^{lim_{x \to \infty} \frac{-6x+2}{x}} = e^{-6}$$

$$lim_{x \to \infty} (3(1+\frac{4}{3x}))^{3x-1} = 3^{lim_{x \to \infty} (3x-1)} \cdot lim_{x \to \infty} (1+\frac{4}{3x})^{3x-1} = \infty \cdot lim_{x \to \infty} ((1+\frac{4}{3x})^{\frac{3x}{4}})^{\frac{4}{3x}(3x-1)} = \infty \cdot e^{lim_{x \to \infty} \frac{12x-4}{3x}} = \infty \cdot e^4$$

Тогда:

$$lim_{x \to \infty} (\frac{x-2}{3x+4})^{3x-1} = \frac{e^{-6}}{\infty \cdot e^4} = 0$$

Ответ: 0