$\lim \limits_{n \to \infty} \frac{\sin x_n}{x_n}$ if $\lim \limits_{n \to \infty} x_n =0$ How to easily prove that 
$$\lim \limits_{n \to \infty} \frac{\sin x_n}{x_n}=1,$$ 
if $\lim \limits_{n \to \infty} x_n =0$?
I proved it using inequality 
$$ 1-\frac{x^2}{2}<\frac{\sin x}{x}<1$$
therefore,
$$1\xleftarrow[\text{$x_n \to 0$}]{}1-\frac{x_n^2}{2}<\frac{\sin x_n}{x_n}<1 \longrightarrow 1$$

$$
\lim \limits_{x \to 0} \frac{\sin x}{x}=\lim \limits_{x\to0} \frac{\sin x-\sin 0}{x-0}=(\sin x)'|_{x=0}=\cos 0=1.
$$ 
