To solve the limit:

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

we observe that direct substitution of $x = \frac{\pi}{4}$ leads to an indeterminate form $\frac{0}{0}$. To resolve this, we can apply L'Hôpital's Rule, which states that if a limit is in the form $\frac{0}{0}$, then

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

if the limit on the right exists.

Step 1: Differentiate the numerator and denominator

1. Numerator: $f(x) = \sin x - \cos x$
   • $f'(x) = \cos x + \sin x$
2. Denominator: $g(x) = x - \frac{\pi}{4}$
   • $g'(x) = 1$

Step 2: Apply L'Hôpital's Rule

Now, substitute these derivatives into the limit:

$\lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{x - \frac{\pi}{4}} = \lim_{x \to \frac{\pi}{4}} \frac{\cos x + \sin x}{1}$

This simplifies to:

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

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

Since $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ and $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, we get:

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

Final Answer:

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

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

Here are 5 related questions to explore this topic further:

1. What is L'Hôpital's Rule and when can it be applied?
2. How do we determine when to use trigonometric identities instead of L'Hôpital's Rule for limits?
3. Can this limit be solved without L'Hôpital's Rule, and if so, how?
4. What are common indeterminate forms encountered in calculus, and how are they resolved?
5. Why does direct substitution sometimes lead to indeterminate forms in limit problems?

Tip: L'Hôpital's Rule is especially useful for limits that result in indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.