The value of `lim_(x to (pi)/4) (sin 2x )^(sec^2 2x)` is equal to

Options

A. `-1/2`
B. `1/2`
C. `e^(-1//2)`
D. `e^(1//2)`

Answer

To solve the limit \( \lim_{x \to \frac{\pi}{4}} (\sin 2x)^{\sec^2 2x} \), we can follow these steps:

### Step 1: Identify the Indeterminate Form
First, we substitute \( x = \frac{\pi}{4} \):
- \( \sin 2x = \sin \left(2 \cdot \frac{\pi}{4}\right) = \sin \left(\frac{\pi}{2}\right) = 1 \)
- \( \sec^2 2x = \sec^2 \left(2 \cdot \frac{\pi}{4}\right) = \sec^2 \left(\frac{\pi}{2}\right) = \infty \)

Thus, we have the form \( 1^\infty \), which is an indeterminate form.

### Step 2: Rewrite the Limit
We can rewrite the limit in a form that is easier to evaluate:
Let \( y = (\sin 2x)^{\sec^2 2x} \).
Taking the natural logarithm on both sides:
\[ \ln y = \sec^2 2x \cdot \ln(\sin 2x) \]
Now, we need to evaluate:
\[ \lim_{x \to \frac{\pi}{4}} \ln y = \lim_{x \to \frac{\pi}{4}} \sec^2 2x \cdot \ln(\sin 2x) \]

### Step 3: Evaluate the Limit
Substituting \( x = \frac{\pi}{4} \) gives us:
- \( \ln(\sin 2x) = \ln(1) = 0 \)
- \( \sec^2 2x \) approaches \( \infty \)

Thus, we have the form \( \infty \cdot 0 \), which is still indeterminate. We can rewrite it as:
\[ \lim_{x \to \frac{\pi}{4}} \frac{\ln(\sin 2x)}{\cos^2 2x} \]

### Step 4: Apply L'Hôpital's Rule
Since we have the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:
Differentiate the numerator and denominator:
- The derivative of \( \ln(\sin 2x) \) is \( \frac{2 \cos 2x}{\sin 2x} \)
- The derivative of \( \cos^2 2x \) is \( -4 \cos 2x \sin 2x \)

Now we have:
\[ \lim_{x \to \frac{\pi}{4}} \frac{\frac{2 \cos 2x}{\sin 2x}}{-4 \cos 2x \sin 2x} \]
This simplifies to:
\[ \lim_{x \to \frac{\pi}{4}} \frac{-\frac{1}{2 \sin^2 2x}}{1} \]

### Step 5: Evaluate the Limit
Substituting \( x = \frac{\pi}{4} \):
\[ \sin 2x = \sin \left(\frac{\pi}{2}\right) = 1 \quad \Rightarrow \quad \sin^2 2x = 1 \]
Thus, we have:
\[ \lim_{x \to \frac{\pi}{4}} -\frac{1}{2} = -\frac{1}{2} \]

### Step 6: Exponentiate to Find \( y \)
Since \( \ln y = -\frac{1}{2} \), we exponentiate to find \( y \):
\[ y = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

### Final Answer
Thus, the value of the limit is:
\[ \lim_{x \to \frac{\pi}{4}} (\sin 2x)^{\sec^2 2x} = \frac{1}{\sqrt{e}} \]