The limit problem you've uploaded is:

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

Step-by-Step Solution:

1. Expand the numerator: $(3x^2 - 2)^2 = (3x^2)^2 - 2 \cdot 3x^2 \cdot 2 + (-2)^2 = 9x^4 - 12x^2 + 4$ So, the expression becomes: $\lim_{{x \to \infty}} \frac{9x^4 - 12x^2 + 4}{4x^3 + 3x^2 - 2}$

2. Factor out the highest powers of $x$: In the numerator, the highest degree term is $9x^4$, and in the denominator, the highest degree term is $4x^3$.

   Divide both the numerator and the denominator by $x^3$, the highest degree in the denominator:

   $\lim_{{x \to \infty}} \frac{x^4(9 - \frac{12}{x^2} + \frac{4}{x^4})}{x^3(4 + \frac{3}{x} - \frac{2}{x^3})}$

   Simplifying the $x^4/x^3 = x$, we get:

   $\lim_{{x \to \infty}} x \cdot \frac{9 - \frac{12}{x^2} + \frac{4}{x^4}}{4 + \frac{3}{x} - \frac{2}{x^3}}$

3. Evaluate the limit as $x \to \infty$: As $x \to \infty$, the terms involving $\frac{1}{x^n}$ approach zero. So we are left with:

   $x \cdot \frac{9}{4}$

   As $x \to \infty$, this expression grows without bound. Therefore, the limit is:

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

Conclusion:

The limit as $x \to \infty$ is $\infty$.

Do you want further details or have any questions about this problem?

Here are five related questions to explore further:

1. How would the result change if the highest power of $x$ in the denominator was higher than that in the numerator?
2. Can this limit approach a finite value for different functions?
3. What happens to limits if we introduce logarithmic functions in the numerator or denominator?
4. How does L'Hôpital's Rule apply in cases where both numerator and denominator approach infinity?
5. How does polynomial degree comparison help in finding limits at infinity?

Tip: When dealing with limits approaching infinity, always look at the highest degree terms in both the numerator and denominator for a quick estimate of the behavior.