To evaluate the limit

$\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - 2\cos x}{1 - \sqrt{2}\sin x},$

we'll first substitute $x = \frac{\pi}{4}$ into the expression to check if we get an indeterminate form:

1. Substitute $x = \frac{\pi}{4}$:

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

   Plugging these into the expression, we have:

   $\text{Numerator: } \sqrt{2} - 2\cos\left(\frac{\pi}{4}\right) = \sqrt{2} - 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} - \sqrt{2} = 0$

   $\text{Denominator: } 1 - \sqrt{2}\sin\left(\frac{\pi}{4}\right) = 1 - \sqrt{2} \cdot \frac{\sqrt{2}}{2} = 1 - 1 = 0$

   Since both the numerator and the denominator are $0$, we have an indeterminate form $\frac{0}{0}$.

2. Apply L'Hôpital's Rule:

   Because the limit is an indeterminate form of $\frac{0}{0}$, we can apply L'Hôpital's rule, which involves differentiating the numerator and the denominator with respect to $x$:

   $\text{Numerator: } \frac{d}{dx} \left(\sqrt{2} - 2\cos x\right) = 2\sin x$

   $\text{Denominator: } \frac{d}{dx} \left(1 - \sqrt{2}\sin x\right) = -\sqrt{2}\cos x$

   Therefore, applying L'Hôpital's rule, the limit becomes:

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

3. Evaluate the limit:

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

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

   Substituting these values in, we get:

   $-\lim_{x \to \frac{\pi}{4}} \frac{2 \cdot \frac{\sqrt{2}}{2}}{\sqrt{2} \cdot \frac{\sqrt{2}}{2}} = -\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}}{\frac{2}{2}} = -\frac{\sqrt{2}}{1} = -\sqrt{2}$

So, the value of the limit is

$\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - 2\cos x}{1 - \sqrt{2}\sin x} = -\sqrt{2}.$

Would you like more details on this solution, or do you have any questions?

Here are some related questions that might interest you:

1. What is L'Hôpital's rule and how is it applied in different scenarios?
2. How do you determine if a function has an indeterminate form?
3. What are the common trigonometric limits and how are they evaluated?
4. Can you use L'Hôpital's rule for limits involving higher-order indeterminate forms?
5. How can you simplify expressions involving trigonometric functions?

Tip: When applying L'Hôpital's rule, make sure that both the numerator and the denominator are differentiable and that the limit remains indeterminate after substitution.