Evaluate, `lim_(xto(pi//4)) (sinx-cosx)/(x-pi/4)`

Options

A. `sqrt(2)`
B. `2`
C. `0`
D. `1`

Explanation

To evaluate the limit

\[ \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{x - \frac{\pi}{4}}, \]

we first substitute \( x = \frac{\pi}{4} \) into the expression:

\[ \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0, \]

and

\[ x - \frac{\pi}{4} = \frac{\pi}{4} - \frac{\pi}{4} = 0. \]

This gives us the indeterminate form \( \frac{0}{0} \). To resolve this, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form of type \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator.

1. **Differentiate the numerator**:
The derivative of \( \sin x - \cos x \) is \( \cos x + \sin x \).

2. **Differentiate the denominator**:
The derivative of \( x - \frac{\pi}{4} \) is \( 1 \).

Now we can rewrite the limit using L'Hôpital's Rule:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{x - \frac{\pi}{4}} = \lim_{x \to \frac{\pi}{4}} \frac{\cos x + \sin x}{1}. \]

Next, we substitute \( x = \frac{\pi}{4} \) into the new limit:

\[ \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}. \]

Thus, the limit evaluates to:

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

**Final Answer:**
\[ \sqrt{2}. \]