1. **State the problem:** Find the limit $$\lim_{x \to -\infty} \frac{1 - x^2}{x^2 + 3}$$. 2. **Recall the rule for limits at infinity:** When $x$ approaches $\pm \infty$, the highest power terms dominate the behavior of rational functions. 3. **Identify the highest powers:** The numerator's highest power term is $-x^2$ and the denominator's highest power term is $x^2$. 4. **Divide numerator and denominator by $x^2$ to simplify:** $$\lim_{x \to -\infty} \frac{\frac{1}{x^2} - \frac{x^2}{x^2}}{\frac{x^2}{x^2} + \frac{3}{x^2}} = \lim_{x \to -\infty} \frac{\frac{1}{x^2} - 1}{1 + \frac{3}{x^2}}$$ 5. **Use cancellation notation to show division:** $$\lim_{x \to -\infty} \frac{\cancel{\frac{1}{x^2}} - 1}{1 + \cancel{\frac{3}{x^2}}}$$ 6. **Evaluate the limit:** As $x \to -\infty$, $\frac{1}{x^2} \to 0$ and $\frac{3}{x^2} \to 0$, so the expression becomes $$\frac{0 - 1}{1 + 0} = \frac{-1}{1} = -1$$. **Final answer:** $$\boxed{-1}$$