Ta có
$\sin\left( x - \dfrac{\pi}{4} \right) = \dfrac{1}{\sqrt{2}} (\sin x - \cos x)$
Thay vào ta có
$\underset{x \to \frac{\pi}{4}}{\lim} \dfrac{\sin x - \cos x}{\sqrt{2} - 2\sin x} = \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{(\sin x - \cos x)(\sqrt{2} + 2\sin x)}{2 - 4\sin^2x}$
$= \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{(\sin x - \cos x)(\sqrt{2} + 2\sin x)}{2(1 - 2\sin^2x)}$
Lại có
$1 - 2\sin^2x = \cos(2x) = \cos^2x - \sin^2x$
Thay vào ta có
$\underset{x \to \frac{\pi}{4}}{\lim} \dfrac{\sin \left( x - \frac{\pi}{4} \right)}{1 - \sin x \sqrt{2}} = \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{-\sqrt{2} - 2\sin x}{2(\sin x + \cos x)}$
Thay $x = \dfrac{\pi}{4}$, suy ra
$\sin x = \cos x = \dfrac{\sqrt{2}}{2}$
vào ta có
$\underset{x \to \frac{\pi}{4}}{\lim} \dfrac{\sin \left( x - \frac{\pi}{4} \right)}{1 - \sin x \sqrt{2}} = -1$
