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The given expression is:

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

### Step-by-Step Solution:

1. **Simplify the numerator**:
   Using the double-angle identity for sine, \( 2 \sin x \cos x = \sin(2x) \), we can rewrite the numerator:
   \[
   1 - 2 \sin x \cos x = 1 - \sin(2x)
   \]
   So, the expression becomes:
   \[
   \lim_{x \to \frac{\pi}{4}} \frac{1 - \sin(2x)}{\sin x - \cos x}
   \]

2. **Evaluate at \( x = \frac{\pi}{4} \)**:
   - \( \sin \left( \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \)
   - Substituting into the denominator:
     \[
     \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0
     \]
   - Substituting into the numerator:
     \[
     1 - \sin\left(2 \cdot \frac{\pi}{4}\right) = 1 - \sin\left(\frac{\pi}{2}\right) = 1 - 1 = 0
     \]
   So, both the numerator and denominator evaluate to 0, giving us an indeterminate form \( \frac{0}{0} \). This suggests we should apply **L'Hopital's Rule**.

3. **Apply L'Hopital's Rule**:
   To resolve the indeterminate form, we differentiate the numerator and the denominator with respect to \( x \):
   
   - The derivative of the numerator \( 1 - \sin(2x) \) is:
     \[
     -2 \cos(2x)
     \]
   - The derivative of the denominator \( \sin x - \cos x \) is:
     \[
     \cos x + \sin x
     \]

   So, applying L'Hopital's Rule, we now compute the limit of:
   \[
   \lim_{x \to \frac{\pi}{4}} \frac{-2 \cos(2x)}{\cos x + \sin x}
   \]

4. **Evaluate the new expression at \( x = \frac{\pi}{4} \)**:
   - \( \cos\left(2 \cdot \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \)
   - \( \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \)

   Therefore, the expression becomes:
   \[
   \frac{-2 \cdot 0}{\sqrt{2}} = 0
   \]

### Final Answer:
\[
\lim_{x \to \frac{\pi}{4}} \frac{1 - 2 \sin x \cos x}{\sin x - \cos x} = 0
\]

Would you like more details on any step or further clarification?

Here are 5 related questions you might be interested in:
1. What is L'Hopital's Rule, and when is it used?
2. How can trigonometric identities simplify limits involving sine and cosine?
3. What is the double-angle identity, and how does it help in solving limits?
4. How would this problem change if we were approaching a different value for \(x\)?
5. What are some common forms of indeterminate limits?

**Tip:** Always check for indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying L'Hopital's Rule!

lim (x→1/4π) (1 - 2sinxcosx) / (sinx - cosx)

Understand how to calculate the limit lim (x→1/4π) (1 - 2sinxcosx) / (sinx - cosx) using L'Hopital's Rule and trigonometric identities. This detailed solution is suitable for high school students in Grades 10-12.

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