Let's solve the given limit step by step:

The problem is:

$\lim_{x \to \frac{\pi}{4}} \frac{\int_{2}^{\sec^2x} f(t) dt}{x^2 - \frac{\pi^2}{16}}$

Step 1: Check the form of the limit

As $x \to \frac{\pi}{4}$:

• The numerator becomes $\int_{2}^{\sec^2(\pi/4)} f(t) dt$, which is $\int_{2}^{2} f(t) dt = 0$.
• The denominator becomes $x^2 - \frac{\pi^2}{16} = \left(\frac{\pi}{4}\right)^2 - \frac{\pi^2}{16} = 0$.

Thus, this is an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

To use L'Hôpital's Rule, differentiate the numerator and denominator with respect to $x$.

Differentiating the numerator

$\frac{d}{dx} \int_{2}^{\sec^2 x} f(t) dt = f(\sec^2 x) \cdot \frac{d}{dx}(\sec^2 x) = f(\sec^2 x) \cdot 2\sec^2 x \tan x$

Differentiating the denominator

$\frac{d}{dx} \left(x^2 - \frac{\pi^2}{16}\right) = 2x$

So, the limit becomes: $\lim_{x \to \frac{\pi}{4}} \frac{f(\sec^2 x) \cdot 2\sec^2 x \tan x}{2x}$

Step 3: Evaluate the limit as $x \to \frac{\pi}{4}$

At $x = \frac{\pi}{4}$:

• $\sec^2 x = \sec^2\left(\frac{\pi}{4}\right) = 2$
• $\tan x = \tan\left(\frac{\pi}{4}\right) = 1$
• $x = \frac{\pi}{4}$

Substitute these values into the expression: