Answer
$$\frac{1}{2}$$
Work Step by Step
\begin{align*}
\lim _{x \rightarrow-\infty}\left(\frac{x^{2}+x-1}{8 x^{2}-3}\right)^{1 / 3}&=\lim _{x \rightarrow-\infty}\left(\frac{x^{2}/x^{2}+x/x^{2}-1/x^{2}}{8 x^{2}/x^{2}-3/x^{2}}\right)^{1 / 3}\\
&=\lim _{x \rightarrow-\infty}\left(\frac{1+1/x -1/x^{2}}{8 -3/x^{2}}\right)^{1 / 3}\\
&=\left(\frac{\lim _{x \rightarrow-\infty}(1)+\lim _{x \rightarrow-\infty}(1/x) -\lim _{x \rightarrow-\infty}(1/x^{2})}{\lim _{x \rightarrow-\infty}(8 ) -\lim _{x \rightarrow-\infty}(3/x^{2})}\right)^{1 / 3}\\
&=\left(\frac{1+0-0}{8-0}\right)^{1 / 3}\\
&=\frac{1}{2}
\end{align*}
