\[{a_1} = 2,{\text{ }}{a_2} = 3,{\text{ }}{a_n} = \sqrt {{a_{n - 1}} \cdot {a_{n - 2}}} {\text{ if }}n \geqslant 3.{\text{ Find }}\mathop {\lim }\limits_{n \to \infty } {a_n}.\]
\[{b_n} = \ln {a_n} \Rightarrow {b_n} = \frac{{{b_{n - 1}} + {b_{n - 2}}}}{2}\]
\[\sqrt[3]{{18}}\]
