Answer
$$2$$
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:
$\lim\limits_{x\to 0}\dfrac{\sin 2x}{x}\\=\lim\limits_{x\to 0}\dfrac{2 \sin x \cos x}{x}\\=\lim\limits_{x\to 0}[(\dfrac{\sin{x}}{x}) \cdot (2 \cos x)]\\=2 (\lim\limits_{x\to 0} \dfrac{\sin x}{x}) (\lim\limits_{x\to 0} \cos x) \\=(2) (1) (\cos 0)\\=(2) (1) \\=2$
