Trả lời bởi giáo viên
\mathop {\lim }\limits_{x \to - \infty } \left( { - {x^3} + 3x} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^3}.\left( { - 1 + \dfrac{3}{{{x^2}}}} \right)
Vì \mathop {\lim }\limits_{x \to - \infty } {x^3} = - \infty \mathop {\lim }\limits_{x \to - \infty } \left( { - 1 + \dfrac{3}{{{x^2}}}} \right) = - 1 < 0
\Rightarrow\mathop {\lim }\limits_{x \to - \infty } {x^3}.\left( { - 1 + \dfrac{3}{{{x^2}}}} \right)= + \infty
\\Rightarrow\mathop {\lim }\limits_{x \to - \infty } \left( { - {x^3} + 3x} \right)= + \infty
Hướng dẫn giải:
Đưa {x^3} ra ngoài.
\mathop {\lim }\limits_{x \to - \infty } {x^{2k + 1}} = - \infty ,k \in \mathbb{Z}
\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = - \infty ,k < 0 = > \mathop {\lim }\limits_{x \to - \infty } k.f\left( x \right) = + \infty