To evaluate the given limit:

$\lim_{x \to \frac{\pi}{2}} \frac{\sec^2(x)}{\sec^2(3x)}$

Step 1: Understanding the behavior of secant

The secant function is defined as: $\sec(x) = \frac{1}{\cos(x)}.$ When $x \to \frac{\pi}{2}$, $\cos(x) \to 0$, so $\sec(x) \to \infty$. Similarly, $3x \to \frac{3\pi}{2}$, and $\cos(3x) \to 0$, making $\sec(3x) \to \infty$.

Both $\sec^2(x)$ and $\sec^2(3x)$ diverge to infinity as $x \to \frac{\pi}{2}$. This requires further analysis to determine their relative growth.

Step 2: Rewrite the ratio

We use the definition: $\frac{\sec^2(x)}{\sec^2(3x)} = \frac{\frac{1}{\cos^2(x)}}{\frac{1}{\cos^2(3x)}} = \frac{\cos^2(3x)}{\cos^2(x)}.$

Step 3: Analyze $\cos(x)$ and $\cos(3x)$

Near $x = \frac{\pi}{2}$:

1. $\cos(x) \to 0$, but $\cos(x)$ can be approximated as $\cos(x) \approx -\sin(x - \frac{\pi}{2}) = -(x - \frac{\pi}{2})$.
2. $\cos(3x) \to 0$, and $\cos(3x) \approx -\sin(3x - \frac{3\pi}{2}) = -3(x - \frac{\pi}{2})$ because $3x$ is a linear scaling.

Thus: $\cos^2(x) \approx (x - \frac{\pi}{2})^2 \quad \text{and} \quad \cos^2(3x) \approx [3(x - \frac{\pi}{2})]^2 = 9(x - \frac{\pi}{2})^2.$

Step 4: Simplify the ratio

Substituting into the ratio: $\frac{\cos^2(3x)}{\cos^2(x)} \approx \frac{9(x - \frac{\pi}{2})^2}{(x - \frac{\pi}{2})^2} = 9.$

Final Answer:

$\lim_{x \to \frac{\pi}{2}} \frac{\sec^2(x)}{\sec^2(3x)} = 9.$

Would you like a detailed explanation of a specific step or have further questions?

Follow-up questions:

1. Why does $\cos(x)$ approach $0$ at $\frac{\pi}{2}$?
2. How is $\sec(x)$ related to $\cos(x)$?
3. What is the role of Taylor series or approximations in solving limits like this?
4. Can the limit be solved using L'Hopital's Rule instead of approximation?
5. What happens if $x \to 3\pi/2$ instead of $\pi/2$?

Tip:

Always analyze the behavior of trigonometric functions near critical points like $\pi/2$ using approximations or expansions to simplify complex ratios.