Answer
$$0$$
Work Step by Step
\begin{align*}
\lim _{x \rightarrow -\infty}\frac{5x-9}{4x^3+2x+7}&= \lim _{x \rightarrow -\infty}\frac{x}{x^3}\frac{5-9x^{-1}}{4+2x^{-2}+7^{-3}}\\
&=\lim _{x \rightarrow -\infty}\frac{x}{x^3}\lim _{x \rightarrow -\infty}\frac{5-9x^{-1}}{4+2x^{-2}+7^{-3}}\\
&=\frac{5}{4}\lim _{x \rightarrow -\infty} x^{-2} \\
&=0.\\
\end{align*}