Answer
$$-\frac{5}{2}$$
Work Step by Step
\begin{align*}
\lim _{x \rightarrow-\infty} \frac{\sqrt[3]{x}-5 x+3}{2 x+x^{2 / 3}-4}&=\lim _{x \rightarrow-\infty} \frac{\frac{1}{x^{2 / 3}}-5+\frac{3}{x}}{2+\frac{1}{x^{1 / 3}}-\frac{4}{x}}\\
&=\frac{\lim _{x \rightarrow-\infty}\frac{1}{x^{2 / 3}}-\lim _{x \rightarrow-\infty}(5)+\lim _{x \rightarrow-\infty}\frac{3}{x}}{\lim _{x \rightarrow-\infty}(2)+\lim _{x \rightarrow-\infty}\frac{1}{x^{1 / 3}}-\lim _{x \rightarrow-\infty}\frac{4}{x}}\\
&=-\frac{5}{2}
\end{align*}