Lim x->0 1+sinx-cosx/1-sinx-cosx

Решите задачу:

$\frac{1+\sin x - \cos x}{1-\sin x -\cos x}=\frac{(1+\sin x - \cos x)(1+\sin x +\cos x)}{(1-\sin x -\cos x)(1+\sin x +\cos x)}=\\ \frac{1+\sin x-\cos x+\sin x+\sin^2 x - \sin x \cos x+\cos x + \sin x \cos x - \cos^2 x}{1-\sin x - \cos x + \sin x - \sin^2 x - \sin x \cos x + \cos x - \sin x \cos x - \cos^2 x}=\\ =\frac{2\sin x + 2\sin^2 x}{-2\sin x \cos x}=\frac{1+\sin x}{-\cos x}\\ \\ \lim_{x \to 0}\frac{1+\sin x - \cos x}{1-\sin x -\cos x} = \lim_{x \to 0}\frac{1+\sin x}{-\cos x}=-1$