If we set f(x)=xcosx-sinx and g(x)=x-sinx, f'(x)=-xsinx+cosx-cosx=-xsinx and g'(x)=1-cosx. Applying L'Hôpital's rule isn't going to help so far, so let's differentiate again: f''(x)=-xcosx-sinx and g''(x)=sinx. Using sinx=x for small x, we have f''(x)=-x-x=-2x and g''(x)=x. So f''(x)/g''(x)=-2x/x=-2 as x approaches zero.