Giải tích 11
Bài 6b (SGK-T133):Tính: \[\mathop {\lim }\limits_{x \to – \infty } ( – 2{x^3} + 3{x^2} – 5)\]
Giải: $\mathop {\lim }\limits_{x \to – \infty } ( – 2{x^3} + 3{x^2} – 5) = \mathop {\lim }\limits_{x \to – \infty } {x^3}\left( { – 2 + \frac{1}{x} – \frac{5}{{{x^2}}}} \right) = + \infty $