If $f\left(x\right)=\begin{cases} \dfrac{\sin x - \cos x}{x - \frac{\pi}{4}}, & x \neq \dfrac{\pi}{4} \\ k, & x = \dfrac{\pi}{4} \end{cases}$ $\;\;$ is continuous at $x=\dfrac{\pi}{4}$, find k.
Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{4}$ $\implies$ $f\left(\dfrac{\pi}{4}\right) = \lim\limits_{x \to \frac{\pi}{4}} f\left(x\right)$
i.e. $k = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin x - \cos x}{x - \frac{\pi}{4}}$
Let $x = \dfrac{\pi}{4}+h$
As $x \to \dfrac{\pi}{4}, \; h \to 0$
$\begin{aligned}
\therefore \; k & = \lim\limits_{h \to 0} \; \dfrac{\sin \left(\dfrac{\pi}{4}+h\right)-\cos \left(\dfrac{\pi}{4}+h\right)}{h} \\
& \\
& \left[\begin{aligned}
\text{Note: } & \sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\
& \cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
\end{aligned}\right] \\
& \\
& = \lim\limits_{h \to 0} \; \dfrac{\sin \left(\dfrac{\pi}{4}\right) \cos h + \cos \left(\dfrac{\pi}{4}\right)\sin h- \left[\cos \left(\dfrac{\pi}{4}\right) \cos h - \sin \left(\dfrac{\pi}{4} \right) \sin h\right]}{h} \\
& \\
& = \lim\limits_{h \to 0} \; \dfrac{\dfrac{\cos h}{\sqrt{2}} + \dfrac{\sin h}{\sqrt{2}} - \dfrac{\cos h}{\sqrt{2}}+ \dfrac{\sin h}{\sqrt{2}}}{h} \\
& \\
& = \dfrac{2}{\sqrt{2}} \; \lim\limits_{h \to 0} \; \dfrac{\sin h}{h} \\
& \\
& = \sqrt{2} \times 1 = \sqrt{2}
\end{aligned}$