`lim_(xrarr0) (sin(picos^2x))/(x^2)` equals

Options

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

Explanation

To solve the limit \( \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2} \), we will follow these steps:

### Step 1: Substitute \(\cos^2 x\)
We know that \( \cos^2 x = 1 - \sin^2 x \). So we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\sin(\pi (1 - \sin^2 x))}{x^2} \]

### Step 2: Simplify the sine expression
Using the identity \( \sin(a - b) = \sin a \cos b - \cos a \sin b \), we can express:
\[ \sin(\pi (1 - \sin^2 x)) = \sin(\pi - \pi \sin^2 x) = \sin(\pi \sin^2 x) \]
Thus, our limit becomes:
\[ \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{x^2} \]

### Step 3: Rewrite the limit
Now we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{x^2} \]

### Step 4: Analyze the first part of the limit
The first part, \( \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \), approaches 1 as \( x \to 0 \) because \( \sin(h) \approx h \) when \( h \) is small. Therefore:
\[ \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} = 1 \]

### Step 5: Analyze the second part of the limit
Next, we need to evaluate \( \frac{\pi \sin^2 x}{x^2} \). Using the limit property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we have:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)^2 = 1^2 = 1 \]
Thus,
\[ \lim_{x \to 0} \frac{\pi \sin^2 x}{x^2} = \pi \]

### Step 6: Combine the results
Combining both parts, we get:
\[ \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{x^2} = 1 \cdot \pi = \pi \]

### Final Answer
Therefore, the value of the limit is:
\[ \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2} = \pi \]

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