To solve the limit:

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

Recall that $\sec(x) = \frac{1}{\cos(x)}$. Therefore, the limit becomes:

$\lim_{x \to \frac{\pi}{2}} \frac{1}{\cos(x)}$

Now, we know that:

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

Thus, we are trying to evaluate:

$\frac{1}{0}$

Since division by zero is undefined, the secant function approaches infinity as $x$ approaches $\frac{\pi}{2}$. More precisely:

$\lim_{x \to \frac{\pi}{2}} \sec(x) = \infty$

Therefore, the limit is:

$\lim_{x \to \frac{\pi}{2}} \sec(x) = \infty$

Would you like a more detailed breakdown or have any further questions? Here are some related questions to expand on this topic:

1. What is the behavior of $\sec(x)$ as $x$ approaches other points where $\cos(x) = 0$?
2. How does the secant function behave around the vertical asymptotes?
3. Can you find the limit of $\sec(x)$ as $x \to 0$?
4. What is the limit of $\csc(x)$ as $x \to 0$?
5. How do the sine and cosine functions behave as $x$ approaches $\frac{\pi}{2}$?

Tip: When dealing with trigonometric limits involving secant, it's helpful to remember that $\sec(x)$ is undefined wherever $\cos(x) = 0$.