If the function $f$ defined on $\left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)$ by $f(x) = \begin{cases} \dfrac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x \ne \dfrac{\pi}{4} \ k, & x = \dfrac{\pi}{4} \end{cases}$ is continuous, then $k$ is equal to