why $\lim_{x \to \frac{\pi}{4}} \frac{\cos 2x}{\sin x-\cos x}=-\sqrt{2}$?

Question

I have this very simple limit to find $$\lim_{x \to \frac{\pi}{4}} \frac{\cos (2x)}{\sin x-\cos x}$$ which is equal to $-\sqrt{2}$. However I can get the outcome as mentioned, or $\sqrt{2}$ in the following way: $$\lim_{x \to \frac{\pi}{4}} \frac{\cos (2x)}{\sin x-\cos x}=\lim_{x \to \frac{\pi}{4}} \frac{\cos ^2x-\sin^2x}{\sin x-\cos x}=\lim_{x \to \frac{\pi}{4}} \frac{(\cos x-\sin x)(\cos x+\sin x)}{\sin x-\cos x}$$ $$=\lim_{x \to \frac{\pi}{4}} (\sin x+\cos x)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$$ Why the mentioned solution method is wrong? Is limit outcome dependent on the expression rearrangement?

Answer 1

Note that $$\cos x-\sin x\not=\sin x-\cos x.$$ We have $$\frac{(\cos x-\sin x)(\cos x+\sin x)}{\sin x-\cos x}=\frac{-(\color{red}{\sin x-\cos x})(\cos x+\sin x)}{\color{red}{\sin x-\cos x}}$$

Answer 2

The limit never depend from the expression rearrangements: it is unique (because you are working in $\Bbb R$ with euclidean metric, which makes it an Hausdorff space). For example you can solve your limit using de l'Hopital rule as well: $$ \lim_{x\to\frac{\pi}4}\frac{\cos(2x)}{\sin x-\cos x}= \lim_{x\to\frac{\pi}4}\frac{-2\sin(2x)}{\cos x+\sin x}=\frac{-2}{\frac{\sqrt2}{2}+\frac{\sqrt2}2}=-\sqrt2\;\;. $$