We are tasked with evaluating the following limit as $x \to -\infty$:

$\lim_{x \to -\infty} \frac{\sqrt{3x^4 + x}}{x^2 - 8}$

Step 1: Analyze the behavior of the function as $x \to -\infty$

As $x \to -\infty$, both the numerator and the denominator grow large. Let’s handle each separately.

Numerator: $\sqrt{3x^4 + x}$

For large negative $x$, the term $3x^4$ will dominate the term $x$. So, we approximate the square root:

$\sqrt{3x^4 + x} \approx \sqrt{3x^4} = \sqrt{3} |x^2| = \sqrt{3}x^2 \quad (\text{since } x^2 \text{ is positive for both positive and negative } x).$

Denominator: $x^2 - 8$

As $x \to -\infty$, the term $x^2$ dominates the constant $-8$, so:

$x^2 - 8 \approx x^2.$

Step 2: Simplify the limit

Now, we simplify the limit by substituting these approximations:

$\lim_{x \to -\infty} \frac{\sqrt{3x^4 + x}}{x^2 - 8} \approx \lim_{x \to -\infty} \frac{\sqrt{3} x^2}{x^2}.$

This simplifies to:

$\lim_{x \to -\infty} \sqrt{3} = \sqrt{3}.$

Final Answer:

$\lim_{x \to -\infty} \frac{\sqrt{3x^4 + x}}{x^2 - 8} = \sqrt{3}.$

Would you like further details on any of the steps?

Here are 5 related questions to deepen understanding:

1. How does the limit change if $x \to \infty$?
2. What would happen if the numerator had a higher degree than the denominator?
3. Can we apply L'Hôpital's rule here? Why or why not?
4. How do we handle similar limits involving cube roots or other radicals?
5. What is the difference between limits at $x \to -\infty$ and $x \to \infty$?

Tip: Always analyze the dominant terms in the numerator and denominator when taking limits at infinity!