Combine fractions: \(\dfrac{x-\sin x}{x\sin x}\). As \(x\to0\), numerator \(x-\sin x\approx\dfrac{x^3}{6}\) and denominator \(x\sin x\approx x\cdot x=x^2\). So the limit \(\approx\dfrac{x^3/6}{x^2}=\dfrac{x}{6}\to0\). L'Hôpital or Taylor series confirms the limit is \(0\).