The problem you uploaded involves evaluating the following limit as $x \to \infty$:

$\lim_{x \to \infty} \frac{3x^3 - x\sqrt{x+1} - 2}{0.2x - 4x^3 - 3}$

To solve this limit, we will analyze the behavior of both the numerator and the denominator as $x \to \infty$.

Step-by-step breakdown:

1. Numerator:

   • The highest degree term in the numerator is $3x^3$, which dominates as $x \to \infty$.
   • The term involving $x\sqrt{x+1}$ needs to be simplified. For large $x$, $\sqrt{x+1} \approx \sqrt{x}$, so $x\sqrt{x+1} \approx x \cdot \sqrt{x} = x^{3/2}$.
   • As a result, the numerator behaves like $3x^3 - x^{3/2} - 2$.

2. Denominator:

   • The highest degree term in the denominator is $-4x^3$, which dominates as $x \to \infty$.
   • The other terms like $0.2x$ and $-3$ are negligible compared to $x^3$ for large $x$.

Thus, for large $x$, the expression behaves like:

$\frac{3x^3}{-4x^3}$

This simplifies to:

$-\frac{3}{4}$

Conclusion:

The limit is:

$\lim_{x \to \infty} \frac{3x^3 - x\sqrt{x+1} - 2}{0.2x - 4x^3 - 3} = -\frac{3}{4}$

Would you like any further details or clarification?

Here are some additional questions that might interest you:

1. How do you find limits of indeterminate forms?
2. What happens when the highest degree terms in both the numerator and denominator are the same?
3. How do you deal with radical terms in limits as $x \to \infty$?
4. Can you apply L'Hopital's Rule to this problem?
5. How does the behavior of the function change if the limit approaches negative infinity?

Tip: When evaluating limits involving polynomials and radicals as $x \to \infty$, always focus on the highest degree terms, as these dominate the behavior of the function.