The value of `lim _( x to (pi)/(4))(int_(2 ) ^(cosec ^(2)x)tg (t )dt)/(x ^(2)-(pi^(2))/(16))` is:

Options

A. `(2)/(pi) g (2)`
B. `-(4)/(pi) g (2)`
C. `-(16)/(pi) g (2)`
D. `(16)/(pi) g (2)`

Answer

To solve the limit

\[ \lim_{x \to \frac{\pi}{4}} \frac{\int_{2}^{\csc^2 x} \tan(t) \, dt}{x^2 - \frac{\pi^2}{16}}, \]

we start by evaluating the expression as \( x \) approaches \( \frac{\pi}{4} \).

### Step 1: Evaluate the denominator

First, we calculate the denominator:

\[ x^2 - \frac{\pi^2}{16}. \]

Substituting \( x = \frac{\pi}{4} \):

\[ \left(\frac{\pi}{4}\right)^2 - \frac{\pi^2}{16} = \frac{\pi^2}{16} - \frac{\pi^2}{16} = 0. \]

### Step 2: Evaluate the numerator

Next, we evaluate the numerator:

\[ \int_{2}^{\csc^2\left(\frac{\pi}{4}\right)} \tan(t) \, dt. \]

Since \( \csc\left(\frac{\pi}{4}\right) = \sqrt{2} \), we have:

\[ \csc^2\left(\frac{\pi}{4}\right) = 2. \]

Thus, the integral becomes:

\[ \int_{2}^{2} \tan(t) \, dt = 0. \]

### Step 3: Form of the limit

Now we have both the numerator and denominator approaching 0:

\[ \lim_{x \to \frac{\pi}{4}} \frac{0}{0}. \]

This indicates that we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule

We differentiate the numerator and the denominator with respect to \( x \).

**Numerator:**

Using the Fundamental Theorem of Calculus, we differentiate:

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

Now, we find \( \frac{d}{dx}(\csc^2 x) \):

\[ \frac{d}{dx}(\csc^2 x) = 2 \csc^2 x \cdot (-\csc x \cot x) = -2 \csc^3 x \cot x. \]

Thus, the derivative of the numerator is:

\[ \tan(\csc^2 x) \cdot (-2 \csc^3 x \cot x). \]

**Denominator:**

The derivative of the denominator is:

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

### Step 5: Evaluate the limit again

Now we apply the limit again:

\[ \lim_{x \to \frac{\pi}{4}} \frac{-2 \tan(\csc^2 x) \csc^3 x \cot x}{2x}. \]

Substituting \( x = \frac{\pi}{4} \):

- \( \csc^2\left(\frac{\pi}{4}\right) = 2 \)
- \( \tan(2) \) and \( \cot\left(\frac{\pi}{4}\right) = 1 \)
- \( \csc\left(\frac{\pi}{4}\right) = \sqrt{2} \)

Thus, we have:

\[ \lim_{x \to \frac{\pi}{4}} \frac{-2 \tan(2) \cdot (\sqrt{2})^3 \cdot 1}{2 \cdot \frac{\pi}{4}} = \frac{-8 \tan(2)}{\frac{\pi}{2}} = \frac{-16 \tan(2)}{\pi}. \]

### Final Result

Thus, the value of the limit is:

\[ \frac{-16 \tan(2)}{\pi}. \]