Answer
$$\lim _{x \rightarrow \frac{\pi}{ 4}} \frac{1-\tan x}{\sin x-\cos x}=-\sqrt{2}$$
Work Step by Step
Given $$\lim _{x \rightarrow \frac{\pi}{ 4}} \frac{1-\tan x}{\sin x-\cos x}$$
so,
\begin{align}
\lim _{x \rightarrow \frac{\pi}{ 4}} \frac{1-\tan x}{\sin x-\cos x}&=\lim _{x \rightarrow\frac{\pi}{ 4}} \frac{\left(1-\frac{\sin x}{\cos x}\right) \cdot \cos x}{\sin x-\cos x}\\
&=\lim _{x \rightarrow \frac{\pi}{ 4}} \frac{\cos x-\sin x}{(\sin x-\cos x) \cos x}\\
&=\lim _{x \rightarrow \frac{\pi}{ 4}} \frac{-1}{\cos x}=\frac{-1}{1 / \sqrt{2}}=-\sqrt{2}
\end{align}
