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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?

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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.

lim_{x → -∞} (sqrt(1 + 4x^6)) / (2 - x^3)

Discover how to evaluate the limit lim_{x → -∞} (sqrt(1 + 4x^6)) / (2 - x^3) using dominant term approximation. A detailed step-by-step solution explores polynomials and limits at infinity, suitable for high school calculus students.

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