We are given the function:

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

and we are asked to find:

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

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

First, substitute $x = \frac{\pi}{4}$ directly into the function to see if we encounter any indeterminate forms:

$g\left(\frac{\pi}{4}\right) = \frac{\cos{\frac{\pi}{4}} - \sin{\frac{\pi}{4}}}{1 - 2\sin^2{\frac{\pi}{4}}}$

We know: $\cos{\frac{\pi}{4}} = \sin{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}$

Substituting this into the expression:

$g\left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{1 - 2\left(\frac{\sqrt{2}}{2}\right)^2}$

This simplifies to:

$g\left(\frac{\pi}{4}\right) = \frac{0}{1 - 2 \cdot \frac{1}{2}} = \frac{0}{1 - 1} = \frac{0}{0}$

So, we have an indeterminate form $\frac{0}{0}$, which suggests we need to apply L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule

Since we have the indeterminate form $\frac{0}{0}$, we can differentiate the numerator and the denominator separately and then evaluate the limit.

Differentiate the numerator:

$\frac{d}{dx} \left(\cos{x} - \sin{x}\right) = -\sin{x} - \cos{x}$

Differentiate the denominator:

$\frac{d}{dx} \left(1 - 2\sin^2{x}\right) = -4\sin{x}\cos{x}$

Now, the limit becomes:

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

Step 3: Substitute $x = \frac{\pi}{4}$ again

At $x = \frac{\pi}{4}$, we know:

$\sin{\frac{\pi}{4}} = \cos{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}$

Substitute these values into the new expression:

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

Simplifying:

$= \frac{-\sqrt{2}}{-4 \cdot \frac{1}{2}} = \frac{-\sqrt{2}}{-2} = \frac{\sqrt{2}}{2}$

Final Answer:

$\lim_{x \to \frac{\pi}{4}} g(x) = \frac{\sqrt{2}}{2}$

Would you like any further clarification or details on this solution?

Here are five related questions you might find useful:

1. What are the conditions for applying L'Hopital's Rule?
2. How do you differentiate trigonometric functions like $\sin(x)$ and $\cos(x)$?
3. What other methods can be used to evaluate limits that result in indeterminate forms?
4. How can we simplify trigonometric expressions before applying limits?
5. What are some real-world applications of limits and indeterminate forms?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying L'Hopital's Rule to avoid incorrect assumptions about the limit's value.