No. Note, however, two things:
1. You missed a corner case allowing L = 0 as an additional solution.
2. You've only proven that *if* the sequence converges, then it must converge to 2 or 0. You still need to show that it does indeed converge (in this case for any initial value ). In this case that's easy, thanks to some useful inequalities satisfied by the geometric mean.
For example, a similar argument shows that the sequence formally converges to 1/2, but in fact that's only true for . The sequence fails to converge for all other initial values.