Answer
$0$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin x - \sin x\cos x}}{{{x^2}}} \cr
& {\text{Factor the numerator of the function}}{\text{, the GCF is }}\sin x \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x\left( {1 - \cos x} \right)}}{{{x^2}}} \cr
& {\text{Split }}{x^2} \cr
& = \mathop {\lim }\limits_{x \to 0} \left[ {\left( {\frac{{\sin x}}{x}} \right)\left( {\frac{{1 - \cos x}}{{{x}}}} \right)} \right] \cr
& {\text{Use the product law of limits}}{\text{, the limit of a product is the }} \cr
& {\text{product of the limits}}{\text{, then}} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{x}} \right) \cr
& {\text{Use the special limits }}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right) = 1{\text{ and }}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{x}} \right) = 0 \cr
& = \left( 1 \right)\left( 0 \right) \cr
& = 0 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin x - \sin x\cos x}}{{{x^2}}} = 0 \cr} $$
