\(\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^4} + 15x + 6}}{{{x^3} - 5x + 2}}\) bằng
A. 2 | B. 3 | C. \( + \infty \) | D. \( - \infty \) |
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^4} + 15x + 6}}{{{x^3} - 5x + 2}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^4}\left( {2 + \frac{{15}}{{{x^3}}} + \frac{6}{{{x^4}}}} \right)}}{{{x^3}\left( {1 - \frac{5}{{{x^2}}} + \frac{2}{{{x^3}}}} \right)}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( {2 + \frac{{15}}{{{x^3}}} + \frac{6}{{{x^4}}}} \right)}}{{1 - \frac{5}{{{x^2}}} + \frac{2}{{{x^3}}}}} = - \infty
\end{array}\)
Chọn D.
-- Mod Toán 11