We are asked to find the limit of the following expression as $x \to \infty$:

$\lim_{x \to \infty} \frac{5x^3 + 2x}{\sqrt{4x^6 - 2x - 3x^3}}$

Step 1: Analyze the behavior of the numerator and denominator

• Numerator: The numerator is $5x^3 + 2x$. As $x \to \infty$, the term $5x^3$ will dominate because it grows much faster than the $2x$ term. Thus, for large $x$, the numerator behaves like $5x^3$.

• Denominator: The denominator is $\sqrt{4x^6 - 2x - 3x^3}$. As $x \to \infty$, the term $4x^6$ dominates because it grows faster than $-2x$ and $-3x^3$. So, the denominator behaves like $\sqrt{4x^6} = 2x^3$.

Step 2: Approximate the expression for large $x$

For large $x$, the expression simplifies to: $\frac{5x^3 + 2x}{\sqrt{4x^6 - 2x - 3x^3}} \approx \frac{5x^3}{2x^3} = \frac{5}{2}$

Step 3: Conclusion

The limit of the expression as $x \to \infty$ is:

$\lim_{x \to \infty} \frac{5x^3 + 2x}{\sqrt{4x^6 - 2x - 3x^3}} = \frac{5}{2}$

If you have any further questions or need more details, feel free to ask!

Here are 5 related questions to deepen understanding:

1. What happens if the highest degree terms in the numerator and denominator cancel each other out?
2. How would the result change if the dominant terms in the numerator and denominator were of different degrees?
3. What techniques can be used to handle limits involving square roots at infinity?
4. Can L'Hopital's Rule be applied to this type of problem? Why or why not?
5. How would this limit change if we approached $x \to -\infty$ instead of $x \to \infty$?

Tip: When calculating limits at infinity, always identify the highest degree terms to simplify the expression.