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:

\[ \cos^2(\pi) = (-1)^2 = 1 \]
Thus, we have:

\[ \tan(\pi \cos^2(\pi)) = \tan(\pi \cdot 1) = \tan(\pi) = 0 \]

Now, we also need to evaluate \( \sin^2(\pi) \):

\[ \sin(\pi) = 0 \quad \Rightarrow \quad \sin^2(\pi) = 0 \]

So, we have:

\[ \frac{\tan(\pi \cos^2 x)}{\sin^2 x} = \frac{0}{0} \]

This is an indeterminate form, so we will use L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

- The numerator is \( \tan(\pi \cos^2 x) \).
- The denominator is \( \sin^2 x \).

We need to differentiate both:

1. **Differentiate the numerator**:
\[ \frac{d}{dx} \tan(\pi \cos^2 x) = \sec^2(\pi \cos^2 x) \cdot \frac{d}{dx}(\pi \cos^2 x) = \sec^2(\pi \cos^2 x) \cdot (-2\pi \cos x \sin x) \]

2. **Differentiate the denominator**:
\[ \frac{d}{dx} \sin^2 x = 2 \sin x \cos x \]

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to \pi} \frac{\tan(\pi \cos^2 x)}{\sin^2 x} = \lim_{x \to \pi} \frac{-2\pi \cos x \sin x \sec^2(\pi \cos^2 x)}{2 \sin x \cos x} \]

### Step 3: Simplify the expression
The \( 2 \sin x \cos x \) cancels out:

\[ = \lim_{x \to \pi} -\pi \sec^2(\pi \cos^2 x) \]

### Step 4: Evaluate the limit again
As \( x \to \pi \):

\[ \cos^2(\pi) = 1 \quad \Rightarrow \quad \sec^2(\pi \cdot 1) = \sec^2(\pi) = \frac{1}{\cos^2(\pi)} = \frac{1}{(-1)^2} = 1 \]

Thus, we have:

\[ -\pi \cdot 1 = -\pi \]

### Final Answer
Therefore, the value of the limit is:

\[ \lim_{x \to \pi} \frac{\tan(\pi \cos^2 x)}{\sin^2 x} = -\pi \]

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