`Lim_( x to pi/2) (1-sin x)/((pi-2x)^(2))`

Answer

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

### Step 1: Substitute \( x = \frac{\pi}{2} \)

First, we substitute \( x = \frac{\pi}{2} \) into the limit expression:

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

And for the denominator:

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

Thus, we have 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:

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

if the limit on the right exists. Here, \( f(x) = 1 - \sin x \) and \( g(x) = (\pi - 2x)^2 \).

### Step 3: Differentiate the Numerator and Denominator

Now we differentiate both the numerator and the denominator:

- The derivative of the numerator \( f(x) = 1 - \sin x \) is:

\[ f'(x) = -\cos x \]

- The derivative of the denominator \( g(x) = (\pi - 2x)^2 \) using the chain rule is:

\[ g'(x) = 2(\pi - 2x)(-2) = -4(\pi - 2x) \]

### Step 4: Rewrite the Limit

Now we can rewrite the limit using the derivatives:

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

### Step 5: Substitute \( x = \frac{\pi}{2} \) Again

Now we substitute \( x = \frac{\pi}{2} \) again:

- The numerator becomes:

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

- The denominator becomes:

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

This is again an indeterminate form \( \frac{0}{0} \).

### Step 6: Apply L'Hôpital's Rule Again

We apply L'Hôpital's Rule a second time:

- The derivative of the numerator \( \cos x \) is:

\[ -\sin x \]

- The derivative of the denominator \( 4(\pi - 2x) \) is:

\[ -8 \]

### Step 7: Rewrite the Limit Again

Now we can rewrite the limit again:

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

### Step 8: Substitute \( x = \frac{\pi}{2} \) One Last Time

Now we substitute \( x = \frac{\pi}{2} \):

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

Thus, the limit becomes:

\[ \frac{1}{8} \]

### Final Answer

The limit is:

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