Find the value of $k$ for which the function $f(x) = \begin{cases} \frac{\sin x - \cos x}{4x - \pi} &, x \neq \frac{\pi}{4} \\ k &, x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$.

$\frac{1}{2\sqrt{2}}$

The correct answer is Option (2) → $\frac{1}{2\sqrt{2}}$ ##

$\lim\limits_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{4x - \pi} = f\left(\frac{\pi}{4}\right)$

$\lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x \right)}{4x - \pi} = k$

$\lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \left( \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} \right)}{4x - \pi} = k$

$\lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \sin \left( x - \frac{\pi}{4} \right)}{4 \left( x - \frac{\pi}{4} \right)} = k$

$\frac{\sqrt{2}}{4} = k$

$k = \frac{1}{2\sqrt{2}}$