Consider the function `f(x)=(1-sinx)/((pi-2x)^(2))` where `xne(pi)/(2)andf((pi)/(2))=lambda`
What is `underset(xrarr(pi)/(2))limf(x)` equal to?

Options

A. A) 1
B. B) `(1)/(2)`
C. C) `(1)/(4)`
D. D) `(1)/(8)`

Answer

To find the limit of the function \( f(x) = \frac{1 - \sin x}{(\pi - 2x)^2} \) as \( x \) approaches \( \frac{\pi}{2} \), we can follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{2} - h \)
We want to evaluate the limit as \( x \) approaches \( \frac{\pi}{2} \). To do this, we can let \( h \) approach \( 0 \) such that \( x = \frac{\pi}{2} - h \). Therefore, we rewrite the limit:

\[ \lim_{x \to \frac{\pi}{2}} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2} - h\right) \]

### Step 2: Substitute into the function
Now, we substitute \( x = \frac{\pi}{2} - h \) into the function:

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

Using the identity \( \sin\left(\frac{\pi}{2} - h\right) = \cos h \):

\[ = \frac{1 - \cos h}{(2h)^2} \]

### Step 3: Simplify the expression
Now we simplify the expression:

\[ = \frac{1 - \cos h}{4h^2} \]

### Step 4: Evaluate the limit
As \( h \) approaches \( 0 \), both the numerator and denominator approach \( 0 \), creating a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule:

\[ \lim_{h \to 0} \frac{1 - \cos h}{4h^2} = \lim_{h \to 0} \frac{\sin h}{8h} \]

### Step 5: Apply L'Hôpital's Rule
Now we differentiate the numerator and denominator:

- The derivative of \( 1 - \cos h \) is \( \sin h \).
- The derivative of \( 4h^2 \) is \( 8h \).

Thus, we have:

\[ = \lim_{h \to 0} \frac{\sin h}{8h} \]

### Step 6: Use the known limit
We know that:

\[ \lim_{h \to 0} \frac{\sin h}{h} = 1 \]

Therefore:

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

### Final Result
Thus, the limit is:

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