Trả lời bởi giáo viên
Đáp án đúng: b
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {3^ + }} 2x - 1 = 5 > 0\\\mathop {\lim }\limits_{x \to {3^ + }} 3 - x = 0\\3 - x < 0\,khi \,x \to {3^ + }\\ \Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{2x - 1}}{{3 - x}} = - \infty \end{array}\)
Hướng dẫn giải:
\(\begin{array}{l}\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L > 0;\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right) = 0\\g\left( x \right) < 0\,khi\,x \to {x_0}\end{array} \right.\\ \Rightarrow \mathop {\lim }\limits_{x \to {x_0}} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = - \infty \end{array}\)
