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To evaluate the limit, we are given the function:

\[
g(x) = \frac{\cos x - \sin x}{1 - 2 \sin^2 x}
\]

and need to find:

\[
\lim_{x \to \frac{\pi}{4}} g(x)
\]

### Step 1: Simplify the Expression
We start by substituting \( x = \frac{\pi}{4} \) directly to check for any indeterminate form:

1. **For the numerator**: \( \cos \frac{\pi}{4} - \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0 \).
2. **For the denominator**: \( 1 - 2 \sin^2 \frac{\pi}{4} = 1 - 2 \cdot \left( \frac{\sqrt{2}}{2} \right)^2 = 1 - 2 \cdot \frac{1}{2} = 1 - 1 = 0 \).

This yields a \( \frac{0}{0} \) indeterminate form, so we can apply **L'Hôpital's Rule**.

### Step 2: Apply L'Hôpital's Rule
To apply L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to \( x \).

1. **Differentiate the numerator**: 
   \[
   \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x
   \]

2. **Differentiate the denominator**:
   \[
   \frac{d}{dx}(1 - 2 \sin^2 x) = -4 \sin x \cos x = -2 \sin(2x)
   \]

After applying L'Hôpital's Rule, we have:

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

### Step 3: Substitute \( x = \frac{\pi}{4} \) Again
Now, substitute \( x = \frac{\pi}{4} \):

1. **For the numerator**: \( -\sin \frac{\pi}{4} - \cos \frac{\pi}{4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} \).
2. **For the denominator**: \( -2 \sin\left(2 \cdot \frac{\pi}{4}\right) = -2 \sin\left(\frac{\pi}{2}\right) = -2 \cdot 1 = -2 \).

Thus,

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

### Final Answer
The correct answer is:

\[
\boxed{\frac{1}{\sqrt{2}}}
\]

or, in decimal form, this is equivalent to answer choice **B**.

Would you like more details or have any questions?

### Related Questions:
1. How would we approach the limit if L'Hôpital's Rule didn't apply?
2. What are common indeterminate forms that lead to using L'Hôpital's Rule?
3. How do trigonometric identities help in simplifying limits?
4. How can we use Taylor series expansions for similar trigonometric limits?
5. What if we needed to evaluate \(\lim_{x \to \frac{\pi}{2}} g(x)\)?

#### Tip:
When dealing with trigonometric limits, checking for direct substitution and possible simplifications using trigonometric identities can often save time before jumping to L'Hôpital's Rule.

If g is the function defined by g(x) = \frac{\cos x - \sin x}{1 - 2\sin^2 x}, what is \lim_{x \to \frac{\pi}{4}} g(x)?

Learn how to evaluate the limit \lim_{x \to \frac{\pi}{4}} g(x) for the function g(x) = (cos x - sin x) / (1 - 2sin^2 x). This solution involves concepts of limits, indeterminate forms, trigonometric identities, and L'Hôpital's Rule. Suitable for advanced high school and college calculus students.

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