8. In square $A B C D$, $A B=2, E$ is the midpoint of $A B$, $F$ is a point on side $B C$, let $B F=x$, to make $\triangle A E D, \triangle D C F$ fold up (so that $A, C$ coincide at point $A^{\prime}$) to form a tetrahedron $A^{\prime}-E F D$, then the range of values for $x$ is . $\qquad$
$$
\begin{array}{l}
0 1 } \\
{ \sqrt { 1 + x ^ { 2 } } + 1 > 2 - x } \\
{ 1 + 2 - x > \sqrt { 1 + x ^ { 2 } } }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ 0 < x - 1 } \\
{ \sqrt { 1 + x ^ { 2 } } > 1 - x } \\
{ \sqrt { 1 + x ^ { 2 } } < 3 - x }
\end{array} \Rightarrow \left\{\begin{array}{l}
0 < x < 2 \\
\sqrt{1 + x^{2}} < 3 - x
\end{array} \text {. Therefore } 0 < x < \frac{4}{3}\right.\right.\right. \text {. }
$$