شرط پیوستگی تابع $f$ در $x=\frac{\pi }{4}$ آن است که $\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,f(x)=f(\frac{\pi }{4})$ باشد.

$\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,f(x)=\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{\cot x-1}{\operatorname{Sin}x-\operatorname{Cos}x}=\frac{0}{0}$

$\Rightarrow \underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{\frac{\operatorname{Cos}x}{\operatorname{Sin}x}-1}{\operatorname{Sin}x-\operatorname{Cos}x}=\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{\frac{\operatorname{Cos}x-\operatorname{Sin}x}{\operatorname{Sin}x}}{\operatorname{Sin}x-\operatorname{Cos}x}$

$=\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{-(\operatorname{Sin}x-\operatorname{Cos}x)}{\operatorname{Sin}x(\operatorname{Sin}x-\operatorname{Cos}x)}=\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{-1}{\operatorname{Sin}x}$

$=\frac{-1}{\frac{\sqrt{2}}{2}}=\frac{-2}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{-2\sqrt{2}}{2}=-\sqrt{2}\Rightarrow f(\frac{\pi }{4})=k=-\sqrt{2}$