Answer
$$\lim_{x\to-\infty}\Big(\frac{x^2+x-1}{8x^2-3}\Big)^{1/3}=\frac{1}{2}$$
Work Step by Step
We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator.
Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$
$$A=\lim_{x\to-\infty}\Big(\frac{x^2+x-1}{8x^2-3}\Big)^{1/3}$$
The highest power of $x$ in the denominator here is $x^2$, so we divide both numerator and denominator by $x^2$: $$A=\lim_{x\to-\infty}\Big(\frac{1+\frac{1}{x}-\frac{1}{x^2}}{8-\frac{3}{x^2}}\Big)^{1/3}=\Big(\frac{\lim_{x\to-\infty}1+\lim_{x\to-\infty}\frac{1}{x}-\lim_{x\to-\infty}\frac{1}{x^2}}{\lim_{x\to-\infty}8-\lim_{x\to-\infty}\frac{3}{x^2}}\Big)^{1/3}$$
$$A=\Big(\frac{1+0-0}{8-0}\Big)^{1/3}=\Big(\frac{1}{8}\Big)^{1/3}=\frac{1}{\sqrt[3]8}=\frac{1}{2}$$
