Tính \(\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{x} - \dfrac{1}{3}} \right)\dfrac{1}{{{{\left( {x - 3} \right)}^3}}}\) bằng:

* Hướng dẫn giải

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{x} - \dfrac{1}{3}} \right)\dfrac{1}{{{{\left( {x - 3} \right)}^3}}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{3 - x}}{{3x{{\left( {x - 3} \right)}^3}}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{ - 1}}{{3x{{\left( {x - 3} \right)}^2}}} =  - \infty \\\left( {Do\,\,\mathop {\lim }\limits_{x \to 3} \left( {3x{{\left( {x - 3} \right)}^2}} \right) = 0;\,\,3x{{\left( {x - 3} \right)}^2} > 0\,\,khi\,\,x \to 3} \right)\end{array}\)

Chọn A.