The value of `lim_(x to pi) (tan(picos^2 x))/(sin^2 x)` is equal to

Options

A. 1
B. `pi`
C. `-pi`
D. `pi/2`

Answer

To solve the limit \( \lim_{x \to \pi} \frac{\tan(\pi \cos^2 x)}{\sin^2 x} \), we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \( x = \pi \) into the expression:
\[ \sin(\pi) = 0 \quad \text{and} \quad \cos(\pi) = -1 \quad \Rightarrow \quad \cos^2(\pi) = 1 \]
Thus, we have:
\[ \tan(\pi \cos^2(\pi)) = \tan(\pi \cdot 1) = \tan(\pi) = 0 \]
So, both the numerator and denominator approach 0 as \( x \to \pi \):
\[ \frac{0}{0} \text{ form} \]
This indicates that we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:
- The derivative of the numerator \( \tan(\pi \cos^2 x) \):
\[ \frac{d}{dx}[\tan(\pi \cos^2 x)] = \sec^2(\pi \cos^2 x) \cdot \frac{d}{dx}[\pi \cos^2 x] = \sec^2(\pi \cos^2 x) \cdot (-2\pi \cos x \sin x) \]
- The derivative of the denominator \( \sin^2 x \):
\[ \frac{d}{dx}[\sin^2 x] = 2 \sin x \cos x \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using these derivatives:
\[ \lim_{x \to \pi} \frac{-2\pi \cos x \sin x \sec^2(\pi \cos^2 x)}{2 \sin x \cos x} \]
We can simplify this:
\[ = \lim_{x \to \pi} \frac{-\pi \sec^2(\pi \cos^2 x)}{1} \]

### Step 4: Evaluate the limit again
Now we substitute \( x = \pi \) into the simplified expression:
\[ \sec^2(\pi \cos^2(\pi)) = \sec^2(\pi) = 1 \]
Thus, we have:
\[ \lim_{x \to \pi} -\pi \cdot 1 = -\pi \]

### Final Answer
The value of the limit is:
\[ \boxed{-\pi} \]