Answer
$$\lim\limits_{x\to\pi/4}\frac{1-\tan x}{\sin x-\cos x}=-\sqrt 2$$
Work Step by Step
$$A=\lim\limits_{x\to\pi/4}\frac{1-\tan x}{\sin x-\cos x}$$ $$A=\lim\limits_{x\to\pi/4}\frac{1-\frac{\sin x}{\cos x}}{\sin x-\cos x}$$ $$A=\lim\limits_{x\to\pi/4}\frac{\frac{\cos x-\sin x}{\cos x}}{\sin x-\cos x}$$ $$A=\lim\limits_{x\to\pi/4}\frac{\cos x-\sin x}{\cos x(\sin x-\cos x)}$$ $$A=\lim\limits_{x\to\pi/4}\frac{-(\sin x-\cos x)}{\cos x(\sin x-\cos x)}$$ $$A=\lim\limits_{x\to\pi/4}\frac{-1}{\cos x}$$ $$A=\frac{-1}{\cos(\pi/4)}$$ $$A=\frac{-1}{\frac{\sqrt 2}{2}}$$ $$A=\frac{-2}{\sqrt 2}=-\sqrt 2$$
