Đáp án:
5A
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
\mathop {\lim }\limits_{x \to - 2} \dfrac{{x\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{{x^2}\left( {x + 2} \right) + \left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \dfrac{{x\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \dfrac{{x\left( {{x^2} - 2x + 4} \right)}}{{{x^2} + 1}}\\
= \dfrac{{ - 2\left( {4 + 4 + 4} \right)}}{{4 + 1}} = - \dfrac{{24}}{5}\\
\to C\\
B4:\\
\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{1}{x} + 3}}{{\sqrt {2 + \dfrac{3}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{3}{{\sqrt 2 }} = \dfrac{3}{{\sqrt 2 }}\\
\to C\\
B5:\\
\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2 - \dfrac{1}{{{x^2}}}}}{{\dfrac{3}{{{x^2}}} - 1}} = \dfrac{2}{{ - 1}} = - 2\\
\to A
\end{array}\)
