Limit of Definite Integral with Secant Upper Limit

To evaluate the given limit:
$L = \lim_{x \to \frac{\pi}{4}} \frac{\int_{2}^{\sec^2 x} f(t) \, dt}{x^2 - \frac{\pi^2}{16}}$

First, check the form of the limit as $x \to \frac{\pi}{4}$:

• The upper limit of the integral approaches $\sec^2\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2$, so the numerator approaches $\int_{2}^{2} f(t) \, dt = 0$.
• The denominator approaches $\left(\frac{\pi}{4}\right)^2 - \frac{\pi^2}{16} = 0$.

Since the limit is in the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Differentiating the numerator with respect to $x$ using the Leibniz Rule:
$\frac{d}{dx} \left( \int_{2}^{\sec^2 x} f(t) \, dt \right) = f(\sec^2 x) \cdot \frac{d}{dx}(\sec^2 x) = f(\sec^2 x) \cdot (2\sec^2 x \tan x)$

Differentiating the denominator with respect to $x$:
$\frac{d}{dx} \left( x^2 - \frac{\pi^2}{16} \right) = 2x$

Now, substituting the derivatives back into the limit:
$L = \lim_{x \to \frac{\pi}{4}} \frac{f(\sec^2 x) \cdot 2\sec^2 x \tan x}{2x} = \lim_{x \to \frac{\pi}{4}} \frac{f(\sec^2 x) \cdot \sec^2 x \tan x}{x}$

Evaluating the limit as $x \to \frac{\pi}{4}$:

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

Thus,
$L = \frac{f(2) \cdot 2 \cdot 1}{\frac{\pi}{4}} = \frac{8}{\pi} f(2)$

Therefore, the correct option is A.