`lim_(xrarr(pi)/(4)) (int_(2)^(sec^(2)x)f(t)dt)/(x^(2-)(pi^(2))/(16))` is equal to

Options

A. `(8)/(pi)f(2)`
B. `(2)/(pi)f(2)`
C. `(2)/(pi)f((1)/(2))`
D. `4f(2)`

Answer

To solve the limit

\[ L = \lim_{x \to \frac{\pi}{4}} \frac{\int_{2}^{\sec^2 x} f(t) \, dt}{x^2 - \frac{\pi^2}{16}}, \]

we will apply L'Hôpital's rule since both the numerator and denominator approach 0 as \( x \to \frac{\pi}{4} \).

### Step 1: Check the limit of the numerator and denominator
As \( x \to \frac{\pi}{4} \):
- The upper limit of the integral, \( \sec^2\left(\frac{\pi}{4}\right) = 2 \).
- Thus, the integral becomes \( \int_{2}^{2} f(t) \, dt = 0 \).
- The denominator \( x^2 - \frac{\pi^2}{16} \) also approaches \( \left(\frac{\pi}{4}\right)^2 - \frac{\pi^2}{16} = 0 \).

Since both the numerator and denominator approach 0, we can apply L'Hôpital's rule.

### Step 2: Differentiate the numerator and denominator
Using L'Hôpital's rule, we differentiate the numerator and the denominator with respect to \( x \).

**Numerator:**
Using the Fundamental Theorem of Calculus and the chain rule:
\[ \frac{d}{dx} \left( \int_{2}^{\sec^2 x} f(t) \, dt \right) = f(\sec^2 x) \cdot \frac{d}{dx}(\sec^2 x). \]
The derivative of \( \sec^2 x \) is \( 2 \sec^2 x \tan x \).

Thus, the derivative of the numerator is:
\[ f(\sec^2 x) \cdot 2 \sec^2 x \tan x. \]

**Denominator:**
The derivative of \( x^2 - \frac{\pi^2}{16} \) is:
\[ \frac{d}{dx}(x^2) = 2x. \]

### Step 3: Apply L'Hôpital's Rule
Now we can rewrite the limit:
\[ L = \lim_{x \to \frac{\pi}{4}} \frac{f(\sec^2 x) \cdot 2 \sec^2 x \tan x}{2x}. \]

### Step 4: Substitute \( x = \frac{\pi}{4} \)
Now, we substitute \( x = \frac{\pi}{4} \):
- \( \sec^2\left(\frac{\pi}{4}\right) = 2 \).
- \( \tan\left(\frac{\pi}{4}\right) = 1 \).
- \( 2x = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \).

Thus, we have:
\[ L = \frac{f(2) \cdot 2 \cdot 2 \cdot 1}{\frac{\pi}{2}} = \frac{4f(2)}{\frac{\pi}{2}} = \frac{8f(2)}{\pi}. \]

### Final Answer
The limit is:
\[ L = \frac{8f(2)}{\pi}. \]