The problem shown in the image is:

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

Solution:

1. Substitute $x = \frac{\pi}{4}$ directly into the limit expression:

   • $\cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$,
   • $\cos(2x) = \cos\left(2 \times \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0$.

2. Substituting these values into the original expression: $\frac{\cos(2x)}{\cos(x) - \sin(x)} = \frac{0}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{0}{0}.$ This results in an indeterminate form $\frac{0}{0}$.

3. Apply L'Hôpital's Rule, since the limit is in the indeterminate form $\frac{0}{0}$:

   • Differentiate the numerator: $\frac{d}{dx} \left( \cos(2x) \right) = -2\sin(2x)$,
   • Differentiate the denominator: $\frac{d}{dx} \left( \cos(x) - \sin(x) \right) = -\sin(x) - \cos(x)$.

4. The limit becomes: $\lim_{x \to \frac{\pi}{4}} \frac{-2\sin(2x)}{-\sin(x) - \cos(x)}.$

5. Now substitute $x = \frac{\pi}{4}$ again:

   • $\sin\left(2 \times \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1$,
   • $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$,
   • $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

6. Substituting these values: $\lim_{x \to \frac{\pi}{4}} \frac{-2 \times 1}{-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{-2}{-\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}.$

Final Answer:

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

Would you like a detailed breakdown or further clarifications?

Here are five related questions to expand your understanding:

1. How does L'Hôpital's Rule help resolve indeterminate forms?
2. What other methods besides L'Hôpital's Rule could solve this limit?
3. Can you explain how trigonometric identities might simplify similar limits?
4. Why does substituting $x = \frac{\pi}{4}$ initially lead to the indeterminate form $\frac{0}{0}$?
5. What are other common situations where L'Hôpital's Rule is applicable?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying L'Hôpital's Rule.