The given limit is:

$\lim_{{x \to -\infty}} \frac{\sqrt{1 + 4x^6}}{2 - x^3}$

Step-by-Step Solution:

1. Focus on dominant terms: When $x \to -\infty$, the highest powers of $x$ will dominate both the numerator and the denominator. Let's analyze the dominant terms.

   • In the numerator $\sqrt{1 + 4x^6}$, the term $4x^6$ will dominate over $1$.
   • In the denominator $2 - x^3$, the term $-x^3$ will dominate over 2.

2. Simplify the numerator: As $x \to -\infty$: $\sqrt{1 + 4x^6} \approx \sqrt{4x^6} = 2|x^3| = 2(-x^3) \quad (\text{since } x^3 \text{ is negative for large negative } x)$ Therefore, $\sqrt{1 + 4x^6} \approx -2x^3$.

3. Simplify the denominator: As $x \to -\infty$, the denominator $2 - x^3$ can be approximated as: $2 - x^3 \approx -x^3$

4. Form the limit: Now, we substitute these approximations into the original limit: $\lim_{{x \to -\infty}} \frac{-2x^3}{-x^3} = \lim_{{x \to -\infty}} 2 = 2$

Final Answer:

$\boxed{2}$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What happens to the limit if the term $4x^6$ is replaced with $x^4$?
2. How would you solve a similar limit for $x \to +\infty$?
3. What techniques can be used for limits involving higher-order polynomials?
4. How does the square root affect the behavior of the limit as $x \to -\infty$?
5. What are other common methods to evaluate limits at infinity?

Tip: When finding limits as $x \to \infty$ or $x \to -\infty$, focus on the highest degree terms to simplify the process.