This limit problem involves evaluating:

$\lim_{x \to \pi/2} \frac{\ln(\csc x)}{(x - \pi/2)^2}.$

Solution Steps:

1. Expand $\ln(\csc x)$: Recall $\csc x = \frac{1}{\sin x}$, so: $\ln(\csc x) = \ln\left(\frac{1}{\sin x}\right) = -\ln(\sin x).$ Thus, the expression becomes: $\lim_{x \to \pi/2} \frac{-\ln(\sin x)}{(x - \pi/2)^2}.$

2. Behavior of $\sin x$ as $x \to \pi/2$: Near $x = \pi/2$, $\sin x \approx 1 - \frac{1}{2}(x - \pi/2)^2$ by Taylor expansion. Substituting this approximation: $\ln(\sin x) \approx \ln\left(1 - \frac{1}{2}(x - \pi/2)^2\right).$

3. Simplify $\ln(1 - u)$: Using $\ln(1 - u) \approx -u$ for small $u$, we have: $\ln\left(1 - \frac{1}{2}(x - \pi/2)^2\right) \approx -\frac{1}{2}(x - \pi/2)^2.$

4. Substitute back: The numerator $-\ln(\sin x)$ becomes: $-\left(-\frac{1}{2}(x - \pi/2)^2\right) = \frac{1}{2}(x - \pi/2)^2.$

5. Simplify the limit: The expression is now: $\lim_{x \to \pi/2} \frac{\frac{1}{2}(x - \pi/2)^2}{(x - \pi/2)^2}.$ Cancel $(x - \pi/2)^2$ from the numerator and denominator: $\lim_{x \to \pi/2} \frac{1}{2} = \frac{1}{2}.$

Final Answer:

$\boxed{\frac{1}{2}}$

Let me know if you'd like further clarification or a breakdown of any step.

Related Questions:

1. What is the Taylor expansion of $\sin x$ near $x = \pi/2$?
2. Why does $\ln(1 - u) \approx -u$ for small $u$?
3. How do we handle indeterminate forms in limits like this one?
4. Can this limit be evaluated using L'Hôpital's Rule?
5. What are the implications of Taylor expansions in evaluating limits?

Tip: Always simplify expressions by approximating small terms when approaching indeterminate forms in limits.