Answer
$$\lim_{x\to-\infty}\frac{x^2-4x+8}{3x^3}=0$$
Work Step by Step
*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\times\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$
$$A=\lim_{x\to-\infty}\frac{x^2-4x+8}{3x^3}$$
Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x^3$ in this case:
$$A=\lim_{x\to-\infty}\frac{\frac{1}{x}-\frac{4}{x^2}+\frac{8}{x^3}}{3}$$
$$A=\frac{0-0+0}{3}=0$$
