Answer
$$1$$
Work Step by Step
In order to simplify the above expression, we will use the following rules.
$(a) \lim\limits_{x \to a} [p(x) \cdot q(x)]=\lim\limits_{x \to a} p(x) \lim\limits_{x \to a} q(x) \\ (b) \lim\limits_{x \to a} k(x)=k(a)$
where $a$ as a constant.
Thus, we have:
$\lim\limits_{x\to 0}\dfrac{\sin^2 x+\sin x(\cos x-1)}{x^2}\\=\lim\limits_{x\to 0}\dfrac{\sin^2 x}{x^2} +\lim\limits_{x\to 0} \dfrac{\sin x(\cos x-1)}{x^2} \\=\lim\limits_{x\to 0} (\dfrac{\sin{x}}{x})^2+ [\lim\limits_{x\to 0} \dfrac{\sin x}{x} \cdot \lim\limits_{x\to 0} \dfrac{\cos x-1}{x}]\\=1+1(0)\\=1$
