The following is basically Martin's argument but without messing with $\,\lim\sup\,$:
take any $\,\epsilon>0\Longrightarrow\,\exists\,N_\epsilon\in\Bbb N\,\,s.t.\,\,|x_n-x|<\epsilon\,\,,\,\forall\,n>N_\epsilon\,$ . Now, take $\,n>N_\epsilon\,$:
$\left|\frac{x_1+\ldots+x_n}{n}-x\right|=\left|\frac{(x_1-x)+(x_2-x)+\ldots+(x_n-x)}{n}\right|\leq\left|\frac{(x_1-x)+\ldots+(x_{N_\epsilon}-x)}{n}\right|+\left|\frac{(x_{N_\epsilon+1}-x)+\ldots (x_n-x)}{n}\right|\leq$
$\leq \frac{k}{n}+\frac{n-N_\epsilon}{n}\epsilon$
with $\,k\,$ a fixed positive constant. Well, now just make $\,n\to\infty\,$ ,and you'll get
$0\leq\left|\frac{x_1+\ldots+x_n}{n}-x\right|\leq 0+\epsilon=\epsilon$
and since $\,\epsilon\,$ was arbitrary we're done
Acclaration: As the above is a very common exercise at the start of limits of sequences, it may be it is given before $\lim\sup\,,\,\lim\inf\,$ and other beasts are studied, so perhaps the above approach is slightly more elementary.