Giải chi tiết giúp mình với

Đáp án + Giải thích các bước giải:

$\begin{array}{l} \mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} - 2{x^5}}}{{5{x^4} + x + 4}}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^5}\left( {\dfrac{3}{x} - 2} \right)}}{{{x^4}\left( {5 + \dfrac{1}{{{x^3}}} + \dfrac{4}{{{x^3}}}} \right)}}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x\left( {\dfrac{3}{x} - 2} \right)}}{{\left( {5 + \dfrac{1}{{{x^3}}} + \dfrac{4}{{{x^3}}}} \right)}}\\ = \mathop {\lim }\limits_{x \to - \infty } x.\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\dfrac{3}{x} - 2} \right)}}{{\left( {5 + \dfrac{1}{{{x^3}}} + \dfrac{4}{{{x^3}}}} \right)}}\\ \left\{ \begin{array}{l} \mathop {\lim }\limits_{x \to - \infty } x = - \infty \\ \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\dfrac{3}{x} - 2} \right)}}{{\left( {5 + \dfrac{1}{{{x^3}}} + \dfrac{4}{{{x^3}}}} \right)}} = \dfrac{{ - 2}}{5} < 0 \end{array} \right.\\ \Rightarrow \mathop {\lim }\limits_{x \to - \infty } x.\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\dfrac{3}{x} - 2} \right)}}{{\left( {5 + \dfrac{1}{{{x^3}}} + \dfrac{4}{{{x^3}}}} \right)}} = + \infty \\ \Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^4} - 2{x^5}}}{{5{x^4} + x + 4}} = + \infty \end{array}$

`#Pô`