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

Your proof is good. Alternatively, you can use the fact that 
$$\lim_{x\to 0}\frac{\sin(x)}{x} = 1$$
And also use the following fact: if $f$ is continuous and $\lim_{x\to c}f(x) = a$ and $x_n\to c$, then $f(x_n)\to a$. 
