The value of `lim_(x to 0) ("sin" (pi cos^(2) x))/(x^(2))` equals

Options

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

Answer

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 sine function
We know that \( \sin(\theta) \) can be rewritten using the identity \( \sin(\pi - \theta) = \sin(\theta) \). Here, we will express \( \sin(\pi \cos^2 x) \) in a different form.

### Step 2: Use the identity
We can express \( \sin(\pi \cos^2 x) \) as:
\[ \sin(\pi \cos^2 x) = \sin(\pi (1 - \sin^2 x)) \]
This simplifies to:
\[ \sin(\pi \cos^2 x) = \sin(\pi - \pi \sin^2 x) = \sin(\pi \sin^2 x) \]

### Step 3: Substitute in the limit
Now, we substitute this back into our limit:
\[ \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{x^2} \]

### Step 4: Use the small angle approximation
For small values of \( x \), we can use the approximation \( \sin y \approx y \) when \( y \) is close to 0. Thus, we can write:
\[ \sin(\pi \sin^2 x) \approx \pi \sin^2 x \]

### Step 5: Substitute the approximation
Now we substitute this approximation into our limit:
\[ \lim_{x \to 0} \frac{\pi \sin^2 x}{x^2} \]

### Step 6: Use the limit property
Using the property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we can rewrite \( \sin^2 x \) as:
\[ \sin^2 x = \left(\frac{\sin x}{x}\right)^2 x^2 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{\pi \sin^2 x}{x^2} = \lim_{x \to 0} \pi \left(\frac{\sin x}{x}\right)^2 \]

### Step 7: Evaluate the limit
As \( x \to 0 \), \( \left(\frac{\sin x}{x}\right)^2 \to 1 \). Therefore, we have:
\[ \lim_{x \to 0} \pi \left(\frac{\sin x}{x}\right)^2 = \pi \cdot 1 = \pi \]

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

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