`lim_(x to pi/4)(int_(2)^(sec^(2)x) xx f(t) dt)/(x^(2) - pi^(2)/16)` equals

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 problem

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

we start by substituting \( x = \frac{\pi}{4} \).

### Step 1: Evaluate the limit directly
When we substitute \( x = \frac{\pi}{4} \):

- The upper limit of the integral becomes \( \sec^2\left(\frac{\pi}{4}\right) = 2 \).
- Thus, the integral becomes \( \int_{2}^{2} x f(t) \, dt = 0 \) (since the limits of integration are the same).
- The denominator becomes \( \left(\frac{\pi}{4}\right)^2 - \frac{\pi^2}{16} = 0 \).

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator.

#### Numerator:
The numerator is \( \int_{2}^{\sec^2 x} x f(t) \, dt \). We will differentiate this with respect to \( x \):

Using the Fundamental Theorem of Calculus and the Chain Rule, we have:

\[ \frac{d}{dx} \left( \int_{2}^{\sec^2 x} x f(t) \, dt \right) = f(\sec^2 x) \cdot \frac{d}{dx}(\sec^2 x) \cdot x - 0, \]

where we evaluate the upper limit at \( \sec^2 x \) and multiply by the derivative of the upper limit \( \sec^2 x \).

The derivative of \( \sec^2 x \) is \( 2 \sec^2 x \tan x \). Thus, we have:

\[ \frac{d}{dx} \left( \int_{2}^{\sec^2 x} x f(t) \, dt \right) = x f(\sec^2 x) \cdot 2 \sec^2 x \tan x. \]

#### Denominator:
The denominator is \( x^2 - \frac{\pi^2}{16} \). The derivative is:

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

### Step 3: Rewrite the limit
Now, we can rewrite the limit using L'Hôpital's Rule:

\[ L = \lim_{x \to \frac{\pi}{4}} \frac{x f(\sec^2 x) \cdot 2 \sec^2 x \tan x}{2x}. \]

### Step 4: Simplify the expression
We can simplify the expression:

\[ L = \lim_{x \to \frac{\pi}{4}} f(\sec^2 x) \cdot \sec^2 x \tan x. \]

### Step 5: Substitute \( x = \frac{\pi}{4} \)
Now substituting \( x = \frac{\pi}{4} \):

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

Thus, we have:

\[ L = f(2) \cdot 2 \cdot 1 = 2 f(2). \]

### Final Result
Therefore, the limit is:

\[ L = 2 f(2). \]