To solve the given limit:

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

we analyze the behavior of the numerator and denominator as $x \to \pi/2$.

Step 1: Rewrite $\csc x$

$\csc x = \frac{1}{\sin x}.$ Thus, $\ln(\csc x) = \ln\left(\frac{1}{\sin x}\right) = -\ln(\sin x)$.

The limit becomes: $\lim_{x \to \pi/2} \frac{-\ln(\sin x)}{(x - \pi/2)^2}.$

Step 2: Approximation of $\sin x$ near $x = \pi/2$

Using the Taylor expansion of $\sin x$ near $x = \pi/2$: $\sin x \approx 1 - \frac{(x - \pi/2)^2}{2}, \quad \text{for } x \to \pi/2.$ Thus: $\ln(\sin x) \approx \ln\left(1 - \frac{(x - \pi/2)^2}{2}\right).$

For small $z = (x - \pi/2)^2$, $\ln(1 - z) \approx -z$. Therefore: $\ln(\sin x) \approx -\frac{(x - \pi/2)^2}{2}.$

Step 3: Substitute the approximation

Substitute $\ln(\sin x) \approx -\frac{(x - \pi/2)^2}{2}$ into the numerator: $\ln(\csc x) = -\ln(\sin x) \approx \frac{(x - \pi/2)^2}{2}.$

The limit now becomes: $\lim_{x \to \pi/2} \frac{\frac{(x - \pi/2)^2}{2}}{(x - \pi/2)^2}.$

Step 4: Simplify the fraction

Simplify the fraction: $\frac{\frac{(x - \pi/2)^2}{2}}{(x - \pi/2)^2} = \frac{1}{2}.$

Final Answer:

$\lim_{x \to \pi/2} \frac{\ln(\csc x)}{(x - \pi/2)^2} = \frac{1}{2}.$

Would you like further clarification or a breakdown of specific steps? Here are five related questions to deepen understanding:

1. Why is $\ln(\sin x) \approx -z$ for small $z$?
2. How can Taylor expansions be used to approximate trigonometric functions?
3. What happens to the limit if $(x - \pi/2)^3$ appears in the denominator instead of $(x - \pi/2)^2$?
4. How is $\csc x$ related to $\sin x$ and its logarithmic behavior near asymptotes?
5. Why is L'Hôpital's Rule not