`lim_(theta to pi//2) (1- sin theta)/(pi-2theta)^(2) = 1/8`

Answer

To solve the limit problem \( \lim_{\theta \to \frac{\pi}{2}} \frac{1 - \sin \theta}{(\pi - 2\theta)^2} \), we will follow these steps:

### Step 1: Identify the limit
We start by substituting \( \theta = \frac{\pi}{2} \) into the expression:

\[ \lim_{\theta \to \frac{\pi}{2}} \frac{1 - \sin \theta}{(\pi - 2\theta)^2} \]

Calculating the numerator:

\[ 1 - \sin\left(\frac{\pi}{2}\right) = 1 - 1 = 0 \]

Calculating the denominator:

\[ (\pi - 2\cdot\frac{\pi}{2})^2 = (\pi - \pi)^2 = 0^2 = 0 \]

Since both the numerator and denominator approach 0, we have an 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 \( \frac{0}{0} \), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

Here, let \( f(\theta) = 1 - \sin \theta \) and \( g(\theta) = (\pi - 2\theta)^2 \).

### Step 3: Differentiate the numerator and denominator
Now we compute the derivatives:

1. Derivative of the numerator \( f(\theta) = 1 - \sin \theta \):
\[ f'(\theta) = -\cos \theta \]

2. Derivative of the denominator \( g(\theta) = (\pi - 2\theta)^2 \):
Using the chain rule:
\[ g'(\theta) = 2(\pi - 2\theta)(-2) = -4(\pi - 2\theta) \]

### Step 4: Rewrite the limit using the derivatives
Now we can rewrite the limit using these derivatives:

\[ L = \lim_{\theta \to \frac{\pi}{2}} \frac{-\cos \theta}{-4(\pi - 2\theta)} = \lim_{\theta \to \frac{\pi}{2}} \frac{\cos \theta}{4(\pi - 2\theta)} \]

### Step 5: Substitute again
Substituting \( \theta = \frac{\pi}{2} \):

The numerator becomes:

\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

The denominator becomes:

\[ 4(\pi - 2\cdot\frac{\pi}{2}) = 4(\pi - \pi) = 0 \]

Again, we have the indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again.

### Step 6: Differentiate again
1. Derivative of the numerator \( \cos \theta \):
\[ -\sin \theta \]

2. Derivative of the denominator \( 4(\pi - 2\theta) \):
\[ -8 \]

### Step 7: Rewrite the limit again
Now we rewrite the limit:

\[ L = \lim_{\theta \to \frac{\pi}{2}} \frac{-\sin \theta}{-8} = \lim_{\theta \to \frac{\pi}{2}} \frac{\sin \theta}{8} \]

### Step 8: Substitute the limit
Substituting \( \theta = \frac{\pi}{2} \):

\[ L = \frac{\sin\left(\frac{\pi}{2}\right)}{8} = \frac{1}{8} \]

### Conclusion
Thus, we have shown that:

\[ \lim_{\theta \to \frac{\pi}{2}} \frac{1 - \sin \theta}{(\pi - 2\theta)^2} = \frac{1}{8} \]