I think I have a solution (thanks to Guido De Philippis).

First I want to show that if $x>0$ is sufficiently small for all positive integers $n$ one has:
$$
 \sin x \le n \sin \frac x n \le \tan x.
$$
To prove this let $y=\frac x n$ so that our thesis becomes
$$
\sin ny \le n \sin y \le \tan ny.
$$
We are going to use induction on $n$.

For the first inequality we have 
$$
\sin (n+1)y = \sin ny\cdot \cos y + \cos ny \cdot \sin y \le \sin ny + \sin y.
$$
Using inductive assumption we obtain
$$
\sin(n+1)y \le n\sin y + \sin y = (n+1)\sin y
$$
which is what we wanted to prove.

For the second inequality we have:
$$
 \tan (n+1)y = \frac{\tan ny + \tan y}{1 - \tan ny \cdot \tan y} \ge \tan ny + \tan y
\ge \tan ny + \sin y
$$
supposing that $0\le \tan y$, $0\le \tan ny$, $0\le\tan ny \cdot \tan y < 1$. 
So by inductive assumption we have 
$$
\tan (n+1)y \le n \sin y + \sin y = (n+1)\sin y
$$
as we wanted to prove.

So we have proved that fixed a sufficiently small positive $x$
one has
$$
\frac{\sin x}{x} \le \frac{\sin \frac x n}{\frac x n } \le \frac{\tan x}{x}.
$$
It is also easy to show that the sequence 
$$
  a_n = \frac{\sin \frac x n }{\frac x n}
$$
is increasing. Hence $a_n\to \ell$ with $0 < \ell < +\infty$.

Now given any $y<x$ one can find $n=n(y)$ such that $\frac{x}{n+1} \le y \le \frac{x}{n}$
whence
$$
a_{n+1} \cdot \frac{n}{n+1}
= \frac{\sin \frac x {n+1}}{\frac x n }
\le \frac{\sin y}{y}
\le \frac{\sin \frac x n}{\frac x {n+1}}
= a_n \cdot \frac{n+1}{n}
$$
and now if $y\to 0^+$ one has $n\to +\infty$ hence $\frac{\sin y}{y}\to \ell$.