4. $\left(0, \frac{\sqrt{6}+\sqrt{2}}{2}\right)$

Analysis: There are 4 edges of length 1 and 2 edges of length $x$, which can be divided into two cases. The first case considers the unfolded diagram of the tetrahedron $V-ABC$, as shown in Figure 1. The second case involves the two edges of length $x$ being skew, i.e., opposite edges, as shown in Figure 2.

First, consider the first case. To form a tetrahedron, it must satisfy $\angle A V B + \angle B V C + \angle C V A^{\prime} \angle B V C$.

For the case in Figure 1: $\angle A V B = \angle C V A^{\prime} \in \left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$, by the cosine rule: $\cos \angle A V B = \frac{2 - x^{2}}{2}$, hence
$$
-\frac{\sqrt{3}}{2} < 1 - \frac{x^{2}}{2} < \frac{\sqrt{3}}{2} \Leftrightarrow x \in \left(\frac{\sqrt{6}-\sqrt{2}}{2}, \frac{\sqrt{6}+\sqrt{2}}{2}\right)
$$

Next, consider the second case. Here, $A B = B C = C D = D A = 1, B D = A C = x$. It is easy to see that $0 < x < \sqrt{2}$. Combining both cases, we get the answer.
4. There are four line segments of length 1 and two line segments of length $x$. If these six line segments can form a tetrahedron, then the range of values for $x$ is . $\qquad$