Limit of a sequence defined by recursive relation : $ a_n = \sqrt{a_{n-1}a_{n-2}}$
Using the relation $a_{n} = \sqrt{a_{n-1}a_{n-2}}$ for $n \geqslant 2$, we find that
$$a_{n+1}^2a_n = (\sqrt{a_na_{n-1}})^2a_n = a_na_{n-1}a_n = a_n^2a_{n-1}$$
is independent of $n$, so $a_{n+1}^2a_n = a_2^2a_1$ for all $n$, and hence $\lambda^3 = a_2^2a_1$ for $\lambda = \lim\limits_{n\to\infty} a_n$.