Question

Find :
$\underset{x\to 0}{lim}\frac{\sin x(cosx-sinx-1)+cosx(sinx+cosx-1)}{x}$

Answer 1

$\underset{x\to 0}{lim}\frac{\sin x(\cos x-\sin x-1)+\cos x(\sin x+\cos x-1)}{x}$

$=\underset{x\to 0}{lim}\frac{2\sin x\cos x+{\cos }^{2}x-{\sin }^{2}x-\sin x-\cos x}{x}$

$=\underset{x\to 0}{lim}\frac{\sin 2x+\cos 2x-\sin x-\cos x}{x}$ $|\frac{0}{0}$ form$

Applying $L^{\prime }$ Hospital's Rule

$=\underset{x\to 0}{lim}\frac{2\cos 2x-2\sin 2x-\cos x+\sin x}{1}$

$=\frac{2-0-1+0}{1}$

$=1$