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.