Answer
$$ \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\tan x-1} =\frac{1}{\sqrt 2}$$
Work Step by Step
Given $$ \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\tan x-1} $$
let $$ f(x) = \ \frac{\sin x-\cos x}{\tan x-1} $$
Since, we have
$$ f(\frac{\pi}{4})= \ \frac{\sin \frac{\pi}{4}-\cos \frac{\pi}{4}}{\tan \frac{\pi}{4}-1} =\frac{\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}}{1-1}=0$$
So, we get
\begin{aligned}
L&= \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\tan x-1}\\
&= \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\frac{\sin x}{\cos x}-1} \\
&= \lim _{x \rightarrow \frac{\pi}{4}}
\frac{\sin x-\cos x}{\frac{\sin x-\cos x}{\cos x}} \\
&
= \lim _{x \rightarrow \frac{\pi}{4}}\frac{\sin x-\cos x}{\sin x-\cos x} \times \cos x \\
&= \lim _{x \rightarrow \frac{\pi}{4}} \cos x\\
&=\cos \frac{\pi}{4}\\
&=\frac{1}{\sqrt 2}
\end{aligned}
