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