`lim_(xrarrpi//2)(cos((pi)/(2)sin^(2)x))/(cot((pi)/(2)sin^(2)x))` equals

Options

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

Answer

To solve the limit problem, we need to evaluate:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos\left(\frac{\pi}{2} \sin^2 x\right)}{\cot\left(\frac{\pi}{2} \sin^2 x\right)} \]

### Step 1: Analyze the limit
First, we substitute \( x = \frac{\pi}{2} \):

- \(\sin^2\left(\frac{\pi}{2}\right) = 1\)
- Therefore, \(\frac{\pi}{2} \sin^2 x = \frac{\pi}{2}\)

Now, we evaluate:

- \(\cos\left(\frac{\pi}{2}\right) = 0\)
- \(\cot\left(\frac{\pi}{2}\right) = 0\)

This gives us the indeterminate form \(\frac{0}{0}\).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \(\lim_{x \to c} \frac{f(x)}{g(x)}\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivatives of the numerator and denominator:

\[ \lim_{x \to \frac{\pi}{2}} \frac{f'(x)}{g'(x)} \]

Let:

- \(f(x) = \cos\left(\frac{\pi}{2} \sin^2 x\right)\)
- \(g(x) = \cot\left(\frac{\pi}{2} \sin^2 x\right)\)

### Step 3: Differentiate the numerator and denominator
1. Differentiate \(f(x)\):
\[ f'(x) = -\sin\left(\frac{\pi}{2} \sin^2 x\right) \cdot \frac{\pi}{2} \cdot 2 \sin x \cos x = -\pi \sin\left(\frac{\pi}{2} \sin^2 x\right) \sin x \cos x \]

2. Differentiate \(g(x)\):
\[ g'(x) = -\csc^2\left(\frac{\pi}{2} \sin^2 x\right) \cdot \frac{\pi}{2} \cdot 2 \sin x \cos x = -\pi \sin x \cos x \csc^2\left(\frac{\pi}{2} \sin^2 x\right) \]

### Step 4: Substitute back into the limit
Now we substitute \(f'(x)\) and \(g'(x)\) back into the limit:

\[ \lim_{x \to \frac{\pi}{2}} \frac{-\pi \sin\left(\frac{\pi}{2} \sin^2 x\right) \sin x \cos x}{-\pi \sin x \cos x \csc^2\left(\frac{\pi}{2} \sin^2 x\right)} \]

### Step 5: Simplify the expression
The \(-\pi \sin x \cos x\) terms cancel out:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\sin\left(\frac{\pi}{2} \sin^2 x\right)}{\csc^2\left(\frac{\pi}{2} \sin^2 x\right)} = \lim_{x \to \frac{\pi}{2}} \sin\left(\frac{\pi}{2} \sin^2 x\right) \cdot \sin^2\left(\frac{\pi}{2} \sin^2 x\right) \]

### Step 6: Evaluate the limit
Substituting \(x = \frac{\pi}{2}\):

- \(\sin\left(\frac{\pi}{2}\right) = 1\)
- Therefore, the limit evaluates to \(1\).

### Final Answer
Thus, the limit is:

\[ \boxed{1} \]