The given limit is:

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

Step-by-Step Solution:

1. Identify the degrees of the polynomial in the numerator and the denominator:

   • The highest degree term in the numerator is $4x^2$, so the degree of the numerator is 2.
   • The highest degree term in the denominator is $3x^3$, so the degree of the denominator is 3.

2. Divide both the numerator and denominator by the highest power of $x$, which is $x^3$, since the degree of the denominator is 3:

   $\frac{\frac{4x^2}{x^3}}{\frac{3x^3}{x^3} - \frac{x^2}{x^3} + \frac{2}{x^3}} = \frac{\frac{4}{x}}{3 - \frac{1}{x} + \frac{2}{x^3}}$

3. Simplify the expression as $x \to \infty$:

   • As $x \to \infty$, $\frac{4}{x} \to 0$, $\frac{1}{x} \to 0$, and $\frac{2}{x^3} \to 0$.
   • The expression simplifies to: $\frac{0}{3 - 0 + 0} = \frac{0}{3} = 0$

Thus, the limit is:

$\lim_{x \to \infty} \frac{4x^2}{3x^3 - x^2 + 2} = 0$

Would you like more details, or do you have any questions?

Here are five related questions to further your understanding:

1. How would the result change if the degree of the numerator was higher than the denominator?
2. What happens if you approach $x \to -\infty$ instead of $x \to \infty$?
3. How do we compute limits when both numerator and denominator have the same degree?
4. Can L'Hôpital's rule be applied to this limit? How would it work?
5. How do we handle limits involving exponential or logarithmic functions?

Tip: When the degrees of the polynomials in the numerator and denominator differ, the result is often determined by the degree of the highest power of $x$.