The problem asks to calculate the limit:

$\lim_{x \to 0} \left[\frac{\left(\frac{2}{x} + 3\right)}{\left(\frac{3}{x} - 1\right)}\right]$

Step-by-step solution:

1. Analyze the behavior as $x \to 0$:

   • The term $\frac{2}{x}$ grows large as $x \to 0$ (tends towards $\infty$ or $-\infty$ depending on the direction of approach).
   • Similarly, $\frac{3}{x}$ grows large as $x \to 0$.

2. Factor out $\frac{1}{x}$ from both the numerator and denominator:

   In the numerator, we have: $\frac{2}{x} + 3 = \frac{1}{x} (2 + 3x)$

   In the denominator: [ \frac{3}{x} - 1 = \frac{1}{x} (3 - x) \