Trong các giới hạn sau, giới hạn nào bằng \( + \infty \).
Giải chi tiết:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( { - 4{x^2} + 7x + 1} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^2}\left( { - 4 + \dfrac{7}{x} + \dfrac{1}{{{x^2}}}} \right) = - \infty \\\mathop {\lim }\limits_{x \to - \infty } \left( {1 - {x^3} - {x^4}} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^4}\left( {\dfrac{1}{{{x^4}}} - \dfrac{1}{x} - 1} \right) = - \infty \\\mathop {\lim }\limits_{x \to - \infty } \left( {2{x^3} + {x^5} + 7} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^5}\left( {\dfrac{2}{{{x^2}}} + 1 + \dfrac{7}{{{x^5}}}} \right) = - \infty \\\mathop {\lim }\limits_{x \to - \infty } \left( { - 4{x^3} + 2{x^2} + 3} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^3}\left( { - 4 + \dfrac{2}{x} + \dfrac{3}{{{x^3}}}} \right) = + \infty \end{array}\)
Chọn D.