$$ \lim_{x\rarr\dfrac{\pi}{4}} $$$$ \frac{\sqrt2-2\cos x}{1-\sqrt2\sin x} $$

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.

Learn how to evaluate the limit lim x->pi/4 (sqrt(2) - 2cos(x)) / (1 - sqrt(2)sin(x)) using L'Hôpital's Rule. Step-by-step solution provided. Suitable for advanced high school students and anyone interested in calculus and trigonometry.

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