1. **State the problem:** Find the limit $$\lim_{x \to -\infty} \frac{x^2}{1 - 8x^2}$$. 2. **Recall the rule for limits at infinity for rational functions:** When $x$ approaches infinity or negative infinity, the behavior of a rational function is dominated by the highest degree terms in numerator and denominator. 3. **Identify the highest degree terms:** Numerator highest degree term is $x^2$, denominator highest degree term is $-8x^2$. 4. **Divide numerator and denominator by $x^2$ to simplify:** $$\lim_{x \to -\infty} \frac{x^2}{1 - 8x^2} = \lim_{x \to -\infty} \frac{\cancel{x^2} \cdot 1}{\frac{1}{x^2} - 8 \cancel{x^2}} = \lim_{x \to -\infty} \frac{1}{\frac{1}{x^2} - 8}$$ 5. **Evaluate the limit:** As $x \to -\infty$, $\frac{1}{x^2} \to 0$, so $$\lim_{x \to -\infty} \frac{1}{0 - 8} = \frac{1}{-8} = -\frac{1}{8}$$. **Final answer:** $$\boxed{-\frac{1}{8}}$$