The given limit expression is:

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

To solve this limit, let's analyze it step by step:

Step 1: Simplify by identifying dominant terms

As $x \to +\infty$, the dominant term in the numerator is $x^2$, and in the denominator, it is $\sqrt{3x^4} = x^2\sqrt{3}$.

Thus, we can approximate:

$\lim_{x \to +\infty} \frac{x^2 - 8}{\sqrt{3x^4 + x}} \approx \lim_{x \to +\infty} \frac{x^2}{\sqrt{3x^4}} = \lim_{x \to +\infty} \frac{x^2}{x^2\sqrt{3}} = \frac{1}{\sqrt{3}}$

Step 2: More rigorous analysis

To be more precise, let's factor out the highest power of $x$ from both the numerator and the denominator. We do this by factoring $x^2$ from the numerator and $x^2$ (inside the square root) from the denominator:

$= \lim_{x \to +\infty} \frac{x^2(1 - \frac{8}{x^2})}{x^2\sqrt{3 + \frac{1}{x^3}}}$

Now, cancel out the $x^2$ terms:

$= \lim_{x \to +\infty} \frac{1 - \frac{8}{x^2}}{\sqrt{3 + \frac{1}{x^3}}}$

Step 3: Apply the limit

As $x \to +\infty$, both $\frac{8}{x^2} \to 0$ and $\frac{1}{x^3} \to 0$. Therefore, the limit simplifies to:

$= \frac{1 - 0}{\sqrt{3 + 0}} = \frac{1}{\sqrt{3}}$

Final Answer:

$\boxed{\frac{1}{\sqrt{3}}}$

Would you like more details or explanations on any of the steps?

Here are some related questions to explore this topic further:

1. How would the result change if the numerator were $x^3$ instead of $x^2$?
2. What happens if the limit tends to negative infinity instead of positive infinity?
3. How would we approach this problem if we had $x^5$ in the denominator instead of $x^4$?
4. Can this limit be solved using L'Hôpital's Rule? If so, how?
5. How does the inclusion of small terms (like $\frac{8}{x^2}$) affect the behavior of the limit?

Tip: When solving limits as $x \to \infty$, focusing on dominant terms helps simplify the process significantly.