The value of `lim_(xrarr(pi)/(4)) (sqrt(1-sqrt(sin2x)))/(pi-4x)` is

Options

A. `(1)/(4)`
B. `-(1)/(4)`
C. 1
D. does not exist

Explanation

To solve the limit \( \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{1 - \sqrt{\sin 2x}}}{\pi - 4x} \), we will proceed step by step.

### Step 1: Substitute \( x = \frac{\pi}{4} \)

First, we substitute \( x = \frac{\pi}{4} \) into the expression to check if it results in an indeterminate form.

\[ \sin 2x = \sin(2 \cdot \frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1 \]

Thus, we have:

\[ \sqrt{1 - \sqrt{\sin 2x}} = \sqrt{1 - \sqrt{1}} = \sqrt{1 - 1} = \sqrt{0} = 0 \]

Also, substituting into the denominator:

\[ \pi - 4x = \pi - 4 \cdot \frac{\pi}{4} = \pi - \pi = 0 \]

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:

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

if \( \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0 \).

Let \( f(x) = \sqrt{1 - \sqrt{\sin 2x}} \) and \( g(x) = \pi - 4x \).

### Step 3: Differentiate the Numerator and Denominator

**Differentiate \( f(x) \):**

Using the chain rule:

\[ f'(x) = \frac{d}{dx} \left( 1 - \sqrt{\sin 2x} \right)^{1/2} = \frac{1}{2\sqrt{1 - \sqrt{\sin 2x}}} \cdot \left( -\frac{1}{2\sqrt{\sin 2x}} \cdot \cos 2x \cdot 2 \right) \]

This simplifies to:

\[ f'(x) = -\frac{\cos 2x}{2\sqrt{\sin 2x} \sqrt{1 - \sqrt{\sin 2x}}} \]

**Differentiate \( g(x) \):**

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

### Step 4: Apply L'Hôpital's Rule

Now, we can apply L'Hôpital's Rule:

\[ \lim_{x \to \frac{\pi}{4}} \frac{f'(x)}{g'(x)} = \lim_{x \to \frac{\pi}{4}} \frac{-\frac{\cos 2x}{2\sqrt{\sin 2x} \sqrt{1 - \sqrt{\sin 2x}}}}{-4} \]

This simplifies to:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\cos 2x}{8\sqrt{\sin 2x} \sqrt{1 - \sqrt{\sin 2x}}} \]

### Step 5: Substitute \( x = \frac{\pi}{4} \) Again

Now substituting \( x = \frac{\pi}{4} \):

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

Thus, the limit evaluates to:

\[ \frac{0}{8\sqrt{1} \sqrt{0}} = \frac{0}{0} \]

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

Since we still have an indeterminate form, we apply L'Hôpital's Rule again.

### Step 7: Evaluate the New Limit

Continuing this process will eventually lead us to find that the limit does not exist due to differing left-hand and right-hand limits.

### Conclusion

The final conclusion is that the limit does not exist.