sequences and series - Show $lim_{n->infty} (3sqrt{n})^{frac{1}{2n}}=1$ (no L'hopital).

I need show $\lim_{n->\infty} (3\sqrt{n})^{\frac{1}{2n}}=1$ in a course of Real Analysis, but I can't use derivative. Can you give me a hit?

I can use:

• Squeeze theorem


• Convergence with εε-δδ


• Sequential convergence


• The notion of a monotone increasing/decreasing function

Limit laws no. Thanks in advance.