\(\mathop {\lim }\limits_{x \to - \infty } \left( { - {x^3} + 3x} \right)\) bằng
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.
\(\mathop {\lim }\limits_{x \to - \infty } {x^{2k + 1}} = - \infty ,k \in \mathbb{Z}\)
\(\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = - \infty ,k < 0\) \( = > \mathop {\lim }\limits_{x \to - \infty } k.f\left( x \right) = + \infty \)