Trigonometric Limits - solution needed: $\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$ how to solve this problem? (without using l'Hopital rule)
$$\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$$
thanks for helping.

take $x=\frac\pi2-t$, since $\cos x=1-2\sin^2\frac x2, \sin 2x=2\sin x\cos x$, we have 
$$\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}=\lim_{x\to 0} \frac{1-\cos x+\sin x}{\sin 2x -\sin x}= \lim_{x\to 0} \frac{2\sin^2\frac x2+\sin x}{2\sin x\cos x -\sin x} $$
notice $\sin x=2\sin \frac x2\cos \frac x2$, so
$$\frac{2\sin^2\frac x2+\sin x}{2\sin x\cos x -\sin x} =\frac{2\sin^2\frac x2+2\sin \frac x2\cos \frac x2}{4\sin \frac x2\cos \frac x2\cos x -2\sin \frac x2\cos \frac x2}\\=\frac{2\sin\frac x2+2\cos \frac x2}{4\cos \frac x2\cos x -2\cos \frac x2} \to \frac2{4-2}=1(x\to 0)$$
