Bài tập 4.30 trang 166 SBT Toán 11

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{2{x^4} + 15x + 6}}{{{x^3} - 5x + 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^4}\left( {2 + \frac{{15}}{{{x^3}}} + \frac{6}{{{x^4}}}} \right)}}{{{x^3}\left( {1 - \frac{5}{{{x^2}}} + \frac{2}{{{x^3}}}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {2 + \frac{{15}}{{{x^3}}} + \frac{6}{{{x^4}}}} \right)}}{{1 - \frac{5}{{{x^2}}} + \frac{2}{{{x^3}}}}} =  - \infty 
\end{array}\)

Chọn D.

-- Mod Toán 11 HỌC247