Answer
$$\lim_{x\to-\infty}\frac{\sqrt[3]x-5x+3}{2x+x^{2/3}+4}=-\frac{5}{2}$$
Work Step by Step
We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator.
Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$
$$A=\lim_{x\to-\infty}\frac{\sqrt[3]x-5x+3}{2x+x^{2/3}+4}=\lim_{x\to-\infty}\frac{x^{1/3}-5x+3}{2x+x^{2/3}+4}$$
The highest power of $x$ in the denominator here is $x$, so we divide both numerator and denominator by $x$: $$A=\lim_{x\to-\infty}\frac{x^{-2/3}-5+\frac{3}{x}}{2+x^{-1/3}+\frac{4}{x}}=\lim_{x\to-\infty}\frac{\frac{1}{x^{2/3}}-5+\frac{3}{x}}{2+\frac{1}{x^{1/3}}+\frac{4}{x}}$$
$$A=\frac{0-5+0}{2+0+0}=-\frac{5}{2}$$
