1. **State the problem:** Find the limit $$\lim_{x \to -\frac{3}{2}} \frac{1}{(2x + 3)^2}$$. 2. **Recall the formula and rules:** The limit of a function as $x$ approaches a value is the value the function approaches. Here, the denominator is squared, so it is always non-negative. 3. **Evaluate the expression inside the denominator:** Substitute $x = -\frac{3}{2}$ into $2x + 3$: $$2 \left(-\frac{3}{2}\right) + 3 = -3 + 3 = 0$$ 4. **Analyze the denominator:** Since the denominator is $(2x + 3)^2$, it becomes $0^2 = 0$ at $x = -\frac{3}{2}$. 5. **Check the behavior near $x = -\frac{3}{2}$:** For values of $x$ close to $-\frac{3}{2}$ but not equal, $(2x + 3)^2$ is a very small positive number because squaring makes it positive. 6. **Determine the limit:** As the denominator approaches $0$ from positive values, the fraction $\frac{1}{(2x + 3)^2}$ grows without bound. 7. **Conclusion:** The limit is $$\lim_{x \to -\frac{3}{2}} \frac{1}{(2x + 3)^2} = +\infty$$ This means the function grows arbitrarily large near $x = -\frac{3}{2}$.