We need to evaluate the limit using L'Hôpital's Rule:

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

Step 1: Check Indeterminate Form

As $x \to \frac{\pi}{2}$:

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

• Since $\sin\left(\frac{\pi}{2}\right) = 1$, we get: $\ln(\csc x) = -\ln(1) = 0$

• The denominator is also $x - \frac{\pi}{2}$, which approaches $0$.

Thus, we have a 0/0 indeterminate form, so we apply L'Hôpital’s Rule, which states:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

if the right-hand limit exists.

Step 2: Differentiate Numerator and Denominator

We differentiate the numerator:

$f(x) = \ln(\csc x) = -\ln(\sin x)$

Using the chain rule:

$f'(x) = -\frac{1}{\sin x} \cdot \cos x = -\cot x$

For the denominator:

$g(x) = x - \frac{\pi}{2}, \quad g'(x) = 1$

Step 3: Compute the Limit

Applying L’Hôpital’s Rule:

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

Since $\cot x = \frac{\cos x}{\sin x}$, we evaluate at $x = \frac{\pi}{2}$:

$\cot\left(\frac{\pi}{2}\right) = \frac{\cos\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)} = \frac{0}{1} = 0$

Thus,

$\lim_{x \to \frac{\pi}{2}} \frac{\ln(\csc x)}{x - \frac{\pi}{2}} = 0$

Final Answer:

$\lim_{x \to \frac{\pi}{2}} \frac{\ln(\csc x)}{x - \frac{\pi}{2}} = 0$

Would you like additional explanations or a graph to visualize this behavior?

Related Questions:

1. How do you apply L'Hôpital’s Rule when both numerator and denominator tend to infinity?
2. What happens if higher-order derivatives are needed in L'Hôpital’s Rule?
3. Can you apply L'Hôpital’s Rule to limits involving exponentials?
4. What is the graphical interpretation of this limit?
5. How can Taylor series help in evaluating similar limits?

Tip: Always check the form of the limit before applying L'Hôpital’s Rule to avoid unnecessary differentiation! 🚀