Use LH rule:
First Derivative$:\dfrac{ [x(e^{x}) + (e^{x}-1) - 2(\sin x)]}{[x \sin x + (1 - \cos x)]}$
Second Derivative$: \dfrac{[xe^{x} + e^{x} + e^{x} - 2 \cos x]}{[x \cos x + \sin x + \sin x]}$
Third Derivative$: \dfrac {[x e^{x} + e^{x} + e^{x} + e^{x} + 2 \sin x]}{[-x \sin x + \cos x + \cos x + \cos x]}$
Put $x = 0: \dfrac{[0+1+1+1+0]}{[0+1+1+1]} =\dfrac {3}{3} = 1.$