The problem asks to evaluate the limit:

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

Step 1: Simplify $\frac{4x - 2}{4x + 3}$ as $x \to \infty$

The numerator and denominator are polynomials of the same degree. Divide through by $x$ in both the numerator and denominator: $\frac{4x - 2}{4x + 3} = \frac{4 - \frac{2}{x}}{4 + \frac{3}{x}}.$ As $x \to \infty$, the terms $\frac{2}{x} \to 0$ and $\frac{3}{x} \to 0$, leaving: $\frac{4x - 2}{4x + 3} \to \frac{4}{4} = 1.$

Step 2: Simplify the second factor, $x^2 - 2x + 3$

The term $x^2 - 2x + 3$ is dominated by the $x^2$ term as $x \to \infty$, so: $x^2 - 2x + 3 \to x^2.$

Step 3: Multiply the simplified expressions

As $x \to \infty$, the expression simplifies to: $\lim_{x \to \infty} \left( \frac{4x - 2}{4x + 3} \right) \cdot \left( x^2 - 2x + 3 \right) \to 1 \cdot x^2.$

Thus, the limit is: $\boxed{\infty}.$

This indicates the expression grows unbounded as $x$ increases.

Would you like a deeper explanation of any step, or further assistance?

Here are 5 related questions for further practice:

1. How would the result change if the numerator and denominator of $\frac{4x - 2}{4x + 3}$ were both squared?
2. What happens to $\frac{4x - 2}{4x + 3}$ if we evaluate the limit as $x \to -\infty$?
3. How does the behavior of $x^2 - 2x + 3$ differ as $x \to -\infty$?
4. Can you rewrite $x^2 - 2x + 3$ in vertex form to analyze its minimum value?
5. How would the limit change if $x^2 - 2x + 3$ were replaced with $x^2 + 2x - 3$?

Tip: For rational expressions, always check the highest degree terms for dominant behavior as $x \to \infty$ or $x \to -\infty$.