The problem asks to determine the infinite limit:

$\lim_{x \to -\frac{3}{2}^{-}} x \sec(\pi x)$

Step-by-step solution:

1. Understanding the behavior of $\sec(\pi x)$:

   • Recall that $\sec(\theta) = \frac{1}{\cos(\theta)}$, so $\sec(\pi x) = \frac{1}{\cos(\pi x)}$.
   • The cosine function, $\cos(\pi x)$, is periodic with a period of $2$. Let's check what happens when $x = -\frac{3}{2}$: $\cos\left(\pi \times \left(-\frac{3}{2}\right)\right) = \cos\left(-\frac{3\pi}{2}\right) = 0.$ Since $\cos\left(-\frac{3\pi}{2}\right) = 0$, it implies that $\sec\left(\pi x\right)$ approaches infinity as $x \to -\frac{3}{2}^{-}$, because $\sec(\pi x) = \frac{1}{0}$ leads to vertical asymptotes.

2. Limit behavior near $x = -\frac{3}{2}$:

   • As $x$ approaches $-\frac{3}{2}$ from the left ($x \to -\frac{3}{2}^{-}$), $\sec(\pi x)$ approaches $-\infty$ because the cosine value goes to $0$ from the negative side.
   • Therefore, we have: $\lim_{x \to -\frac{3}{2}^{-}} \sec(\pi x) = -\infty.$
   • Also, $x$ approaches $-\frac{3}{2}$ from the left, meaning $x < -\frac{3}{2}$, so $x$ is a negative number near this value.

3. Multiplying by $x$:

   • The term $x \sec(\pi x)$ now behaves as: $x \times (-\infty) \quad \text{as} \quad x \to -\frac{3}{2}^{-}.$
   • Since $x$ is negative and $\sec(\pi x)$ tends to $-\infty$, multiplying a negative number by $-\infty$ gives $+\infty$.

4. Conclusion: $\lim_{x \to -\frac{3}{2}^{-}} x \sec(\pi x) = +\infty.$

Let me know if you'd like further details or explanations on this.

Here are five related questions to extend your understanding:

1. What is the general behavior of $\sec(\theta)$ as $\theta$ approaches values where $\cos(\theta) = 0$?
2. How does the sign of $x$ affect the result in infinite limits when multiplying by a divergent function?
3. What happens to $\sec(\pi x)$ when $x$ approaches other values like $0$ or $1$?
4. Can you determine the limit of $x \tan(\pi x)$ as $x \to -\frac{3}{2}^{-}$?
5. How would the limit change if the cosine term was replaced by a sine or tangent function?

Tip: When evaluating limits involving trigonometric functions, always check for points where the denominator might become zero, leading to asymptotes or undefined behavior.