The limit problem you're asking to solve is:

$\lim_{x \to \left(\frac{\pi}{2}\right)^+} \left( \frac{1}{x} \sec x \right)$

Step-by-Step Solution:

1. Examine the behavior as $x \to \frac{\pi}{2}^+$:

   • The secant function, $\sec x = \frac{1}{\cos x}$, becomes undefined at $x = \frac{\pi}{2}$, because $\cos \left( \frac{\pi}{2} \right) = 0$.
   • As $x \to \frac{\pi}{2}^+$ (from the right-hand side), $\cos x$ approaches 0 from the negative side (since cosine is negative in the second quadrant). Therefore, $\sec x$ tends to $-\infty$.

2. Analyze $\frac{1}{x}$ as $x \to \frac{\pi}{2}^+$:

   • As $x$ approaches $\frac{\pi}{2}$ from the right, $\frac{1}{x}$ approaches some finite positive number (because $x$ is still positive).

3. Combine the behavior:

   • As $x \to \frac{\pi}{2}^+$, $\frac{1}{x}$ stays finite and positive, but $\sec x$ tends to $-\infty$.
   • Multiplying a positive number by $-\infty$ gives $-\infty$.

Final Result:

$\lim_{x \to \left( \frac{\pi}{2} \right)^+} \left( \frac{1}{x} \sec x \right) = -\infty$

Would you like more details or explanations on any step?

Related Questions:

1. What happens to $\frac{1}{x} \sec x$ if $x \to \frac{\pi}{2}^-$ (from the left)?
2. How does the behavior of secant change in different quadrants of the unit circle?
3. Can you use L'Hopital's Rule for this type of limit?
4. What are the typical discontinuities of the secant function?
5. What is the behavior of $\sec x$ for values of $x$ near $0$?

Tip: Always check if the function inside a limit has discontinuities or singularities at the point you're evaluating the limit for accurate results.