$\lim \limits_{n \to \infty} \frac{\sin x_n}{x_n}$ 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,
$$1\xleftarrow[\text{$x_n \to 0$}]{}1-\frac{x_n^2}{2}<\frac{\sin x_n}{x_n}<1 \longrightarrow 1$$

Thanks everyone for the help! I've came up with the proper estimate, which could be obtained without 'magic logic circles':
$$\sin x < x < \tan x$$
$$\frac{1}{\sin x} > \frac{1}{x} > \frac{1}{\tan x}$$
$$1\leftarrow1=\frac{\sin x_n}{\sin x_n} > \frac{\sin x_n}{x_n} > \frac{\sin x_n}{\tan x_n} = \cos x_n \xrightarrow[\text{$x_n \to 0$}]{} 1$$
