1. The problem is to find the limit: $$\lim_{x \to -\infty} \frac{1+x^6}{1+x^4}$$. 2. When $x$ approaches negative infinity, the highest powers of $x$ dominate the behavior of the function. 3. The numerator is $1 + x^6$ and the denominator is $1 + x^4$. 4. Since $x^6$ grows faster than $x^4$ as $x \to -\infty$, the term $x^6$ dominates the numerator and $x^4$ dominates the denominator. 5. We can factor out $x^6$ from numerator and $x^4$ from denominator to analyze the limit: $$\frac{1+x^6}{1+x^4} = \frac{x^6(\frac{1}{x^6} + 1)}{x^4(\frac{1}{x^4} + 1)} = x^{6-4} \cdot \frac{\frac{1}{x^6} + 1}{\frac{1}{x^4} + 1} = x^2 \cdot \frac{\frac{1}{x^6} + 1}{\frac{1}{x^4} + 1}$$ 6. As $x \to -\infty$, $\frac{1}{x^6} \to 0$ and $\frac{1}{x^4} \to 0$, so the fraction tends to $\frac{0 + 1}{0 + 1} = 1$. 7. Therefore, the limit behaves like $x^2 \cdot 1 = x^2$ as $x \to -\infty$. 8. Since $x^2$ tends to $+\infty$ as $x \to -\infty$, the limit is $+\infty$. Final answer: $$\lim_{x \to -\infty} \frac{1+x^6}{1+x^4} = +\infty$$.