Answer
$$2$$
Work Step by Step
Given $$ \lim _{x\rightarrow -\infty}\frac{4x^3+6x^2-2}{2x^3-4x+5}$$
Then
\begin{aligned}
\lim _{x\rightarrow -\infty}\frac{4x^3+6x^2-2}{2x^3-4x+5} &=
\lim _{x\rightarrow-\infty}\frac{4\frac{x^3}{x^3}+6\frac{ x^2}{x^3}-\frac{2}{x^3}}{2\frac{x^3}{x^3} -4\frac{x}{x^3}+\frac{5}{x^3}} \\
&=
\lim _{x\rightarrow-\infty}\frac{4 +\frac{ 6}{x}-\frac{2}{x^3}}{2 -\frac{4}{x^2}+\frac{5}{x^3}} \\
&= \lim _{x\rightarrow -\infty}\frac{4+0-0}{2-0 +0}\\
&=2
\end{aligned}
