lim x-> -infinity [3x^2 + sqareroot (x^6 + x^4 + x^2 + 1)]/[2x^3 -7]



We are tasked with evaluating the limit of the following expression as \( x \to -\infty \):

\[
\lim_{x \to -\infty} \frac{3x^2 + \sqrt{x^6 + x^4 + x^2 + 1}}{2x^3 - 7}
\]

### Step-by-Step Solution:

1. **Simplify the dominant terms:**
   For large negative \( x \), the highest power of \( x \) in both the numerator and denominator will dominate the behavior of the expression.

   - The **numerator**: 
     \[
     3x^2 + \sqrt{x^6 + x^4 + x^2 + 1}
     \]
     The dominant term inside the square root is \( x^6 \), so:
     \[
     \sqrt{x^6 + x^4 + x^2 + 1} \approx \sqrt{x^6} = |x^3| = -x^3 \quad (\text{since } x \to -\infty)
     \]
     Hence, the numerator becomes approximately:
     \[
     3x^2 - x^3
     \]

   - The **denominator**:
     \[
     2x^3 - 7
     \]
     As \( x \to -\infty \), the dominant term is \( 2x^3 \), so the denominator becomes approximately:
     \[
     2x^3
     \]

2. **Rewrite the expression:**
   Now, the limit can be approximated by simplifying the dominant terms:
   \[
   \lim_{x \to -\infty} \frac{3x^2 - x^3}{2x^3}
   \]

3. **Divide every term by \( x^3 \):**
   \[
   = \lim_{x \to -\infty} \frac{x^3 \left(\frac{3}{x} - 1\right)}{2x^3}
   \]
   Simplifying:
   \[
   = \lim_{x \to -\infty} \frac{\frac{3}{x} - 1}{2}
   \]

4. **Evaluate the limit:**
   As \( x \to -\infty \), \( \frac{3}{x} \to 0 \), so the expression simplifies to:
   \[
   \frac{0 - 1}{2} = \frac{-1}{2}
   \]

Thus, the value of the limit is:
\[
\boxed{-\frac{1}{2}}
\]

Would you like more details or have any questions?

### Related Questions:
1. How does the behavior of higher powers of \( x \) affect limits at infinity?
2. What is the general strategy for solving limits involving square roots as \( x \to \infty \) or \( x \to -\infty \)?
3. Can L'Hopital's Rule be used here? Why or why not?
4. How would the limit change if \( x \to \infty \) instead of \( x \to -\infty \)?
5. What happens if the degree of the numerator is less than the degree of the denominator in such limits?

**Tip:** When evaluating limits at infinity, always focus on the highest powers of \( x \) in the numerator and denominator to simplify the expression.

lim x-> -infinity [3x^2 + sqrt(x^6 + x^4 + x^2 + 1)]/[2x^3 -7]

Discover how to solve the limit lim x → -∞ [3x^2 + sqrt(x^6 + x^4 + x^2 + 1)]/[2x^3 - 7] using dominant term simplification and algebraic techniques. A detailed step-by-step guide for understanding limits at infinity, suitable for advanced high school or undergraduate students.

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