3. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=$ $\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$, then $x_{1999}-x_{601}=$ $\qquad$ .
3.0 .

From $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$ we get $x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$.
Let $x_{n}=\tan \alpha_{n}$, then
$$
x_{n+1}=\tan \alpha_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right) \text {. }
$$

Therefore, $x_{n+6}=x_{n}$,
which means the sequence $\left\{x_{n}\right\}$ is a periodic sequence.
From $1999=333 \times 6+1,601=100 \times 6+1$ we get
$$
x_{1999}-x_{601}=0 \text {. }
$$