Answer
$\lim\limits_{x\to-\infty}\frac{4x^3+6x^2-2}{2x^3-4x+5}=2$
Work Step by Step
$$A=\lim\limits_{x\to-\infty}\frac{4x^3+6x^2-2}{2x^3-4x+5}$$
Divide both numerator and denominator by $x^3$
$A=\lim\limits_{x\to-\infty}\frac{4+\frac{6}{x}-\frac{2}{x^3}}{2-\frac{4}{x^2}+\frac{5}{x^3}}$
$A=\frac{\lim\limits_{x\to-\infty}(4+\frac{6}{x}-\frac{2}{x^3})}{\lim\limits_{x\to-\infty}(2-\frac{4}{x^2}+\frac{5}{x^3})}$
$A=\frac{\lim\limits_{x\to-\infty}4+6\lim\limits_{x\to-\infty}\frac{1}{x}-2\lim\limits_{x\to-\infty}\frac{1}{x^3}}{\lim\limits_{x\to-\infty}2-4\lim\limits_{x\to-\infty}\frac{1}{x^2}+5\lim\limits_{x\to-\infty}\frac{1}{x^3}}$
$A=\frac{4+6\times0-2\times0}{2-4\times0+5\times0}$
$A=2$
