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 start by substituting \( x = \frac{\pi}{4} \) into the expression to check for indeterminate forms.

1. **Substituting \( x = \frac{\pi}{4} \)**:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}. \]
Thus, the numerator becomes:
\[ \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0. \]
The denominator is:
\[ \frac{\pi}{4} - \frac{\pi}{4} = 0. \]
Therefore, we have the indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit on the right side exists.

We need to find the derivatives of the numerator and the denominator:
- Derivative of the numerator \( \sin x - \cos x \):
\[ f'(x) = \cos x + \sin x. \]
- Derivative of the denominator \( x - \frac{\pi}{4} \):
\[ g'(x) = 1. \]

3. **Re-evaluating the limit**:
Now we can rewrite the limit:
\[ \lim_{x \to \frac{\pi}{4}} \frac{\cos x + \sin x}{1}. \]
Substituting \( x = \frac{\pi}{4} \):
\[ \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}. \]

4. **Final Result**:
Thus, the limit evaluates to:
\[ \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{x - \frac{\pi}{4}} = \sqrt{2}. \]