`lim_(xto0) (sin(picos^(2)x))/(x^(2))` is equal to

Explanation

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

### Step 1: Rewrite the limit
Let \( L = \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2} \).

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

### Step 3: Use the sine subtraction identity
Using the identity \( \sin(\pi - \theta) = \sin(\theta) \), we can rewrite the limit:
\[ L = \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{x^2} \]

### Step 4: Recognize the form
As \( x \to 0 \), \( \sin^2 x \to 0 \), so we have a \( \frac{0}{0} \) form. We can apply L'Hôpital's Rule or use the limit property:
\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]
We rewrite the limit:
\[ L = \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{x^2} \]

### Step 5: Evaluate the first limit
The first part \( \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \) approaches 1 as \( x \to 0 \).

### Step 6: Evaluate the second limit
Now we need to evaluate \( \lim_{x \to 0} \frac{\sin^2 x}{x^2} \):
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = \left( \lim_{x \to 0} \frac{\sin x}{x} \right)^2 = 1^2 = 1 \]

### Step 7: Combine the results
Putting it all together:
\[ L = 1 \cdot \pi \cdot 1 = \pi \]

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

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