Question

Вычислите:

Answer 1

$\lim_{x \to \infty} \frac{3x^3-4x^2+8}{-5x^3+2x^2+x}= \lim_{x \to \infty} \frac{\frac{3x^3}{x^3}-\frac{4x^2}{x^3}+\frac{8}{x^3}}{\frac{-5x^3}{x^3}+\frac{2x^2}{x^3}+\frac{x}{x^3}}=\frac{3-0+0}{-5+0+0}=-\frac{3}5$

Answer 2

$\lim_{x \to \infty} \frac{3 x^{3}-4 x^{2} +8 }{-5 x^{3}+2 x^{2} +x } = \lim_{x \to \infty} \frac{x^3 \cdot (3-\frac{4}{x} +\frac{8}{x^3}) }{x^3 \cdot (-5+\frac{2}{x} +\frac{1}{x^2})} = \lim_{x \to \infty} \frac{3-\frac{4}{x} +\frac{8}{x^3}}{-5+\frac{2}{x} +\frac{1}{x^2}}=- \frac{3}{5}$