\(\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{3{x^2}}}{{x - 3}}.\dfrac{{12x + 4}}{{2{x^3} - 6{x^2} + x - 3}}} \right)\) bằng:

* Hướng dẫn giải

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{3{x^2}}}{{x - 3}}.\dfrac{{12x - 4}}{{2{x^3} - 6{x^2} + x - 3}}} \right)\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{3{x^2}\left( {12x - 4} \right)}}{{\left( {x - 3} \right)\left( {2{x^3} - 6{x^2} + x - 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{3{x^2}\left( {12x - 4} \right)}}{{{{\left( {x - 3} \right)}^2}\left( {2{x^2} + 1} \right)}}\end{array}\)

Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 3} 3{x^2}\left( {12x - 4} \right) = 864\\\mathop {\lim }\limits_{x \to 3} {\left( {x - 3} \right)^2}\left( {2x + 1} \right) = 0\\{\left( {x - 3} \right)^2}\left( {2x + 1} \right) \ge 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to 3} \dfrac{{3{x^2}\left( {12x - 4} \right)}}{{{{\left( {x - 3} \right)}^2}\left( {2x + 1} \right)}} =  + \infty \)

Chọn A.