Đáp án:
\( - \dfrac{1}{{18}}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{{{\left( {2 - x} \right)}^3}{{\left( {3 - \left| x \right|} \right)}^3}}}{{\left( {2{x^4} + 4{x^2} + 1} \right){{\left( {2 - 3x} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{{\left( {\dfrac{2}{x} - 1} \right)}^3}{{\left( {\dfrac{3}{x} - \left| 1 \right|} \right)}^3}}}{{\left( {2 + \dfrac{4}{{{x^2}}} + \dfrac{1}{{{x^4}}}} \right){{\left( {\dfrac{2}{x} - 3} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{{\left( { - 1} \right)}^3}.{{\left( { - \left( { - 1} \right)} \right)}^3}}}{{2.{{\left( { - 3} \right)}^2}}} = - \dfrac{1}{{18}}
\end{array}\)
