We are tasked with finding the limit of the following expression:

$\lim_{{x \to \frac{\pi}{4}}} \frac{1 - \sin(2x)}{\cos^2(2x)}$

Let's break down the problem:

Step 1: Simplify the trigonometric functions

• First, recall that $\sin(2x)$ and $\cos(2x)$ are standard double-angle trigonometric functions.

We need to evaluate the limit as $x$ approaches $\frac{\pi}{4}$.

Step 2: Substitute $x = \frac{\pi}{4}$

At $x = \frac{\pi}{4}$:

$\sin(2x) = \sin\left(2 \times \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1$ $\cos(2x) = \cos\left(2 \times \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0$

Now, substituting into the original expression:

$\frac{1 - \sin(2x)}{\cos^2(2x)} = \frac{1 - 1}{\cos^2(2x)} = \frac{0}{0}$

This yields an indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule

To apply L'Hôpital's Rule, differentiate the numerator and the denominator with respect to $x$.

• The derivative of $1 - \sin(2x)$ is: $\frac{d}{dx}[1 - \sin(2x)] = -2\cos(2x)$
• The derivative of $\cos^2(2x)$ is: $\frac{d}{dx}[\cos^2(2x)] = 2\cos(2x) \cdot (-2\sin(2x)) = -4\cos(2x)\sin(2x)$

Step 4: Evaluate the limit again

Now, the limit becomes:

$\lim_{{x \to \frac{\pi}{4}}} \frac{-2\cos(2x)}{-4\cos(2x)\sin(2x)}$

We can simplify this expression by canceling out $\cos(2x)$ (since it's non-zero near $x = \frac{\pi}{4}$):

$\lim_{{x \to \frac{\pi}{4}}} \frac{2}{4\sin(2x)} = \frac{1}{2\sin\left(\frac{\pi}{2}\right)} = \frac{1}{2 \times 1} = \frac{1}{2}$

Thus, the value of the limit is $\boxed{\frac{1}{2}}$.

Final Answer:

The correct option is C. $\frac{1}{2}$.

Do you want more details or have any questions?

Here are 5 related questions for further practice:

1. What is L'Hôpital's Rule, and when can it be applied?
2. How do you find the limit of $\lim_{x \to 0} \frac{\sin(x)}{x}$?
3. What are the double-angle formulas for sine and cosine?
4. How would the answer change if the limit were approaching a different value?
5. Can you explain the importance of recognizing indeterminate forms in calculus?

Tip: Always check for indeterminate forms before applying L'Hôpital's Rule!