\(\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - {{\cos }^2}3x}}{{2{x^2}}}\) bằng giá trị nào sau đây?

* Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - {{\cos }^2}3x}}{{2{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}3x}}{{2{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}3x}}{{9{x^2}.\dfrac{2}{9}}} = \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin 3x}}{{3x}}} \right)^2}.\dfrac{9}{2} = \dfrac{9}{2}\).

Chọn C.