Answer
$\mathop {\lim }\limits_{n \to \infty } {a_n} = 3$
Work Step by Step
For $n \ge 1$, we have ${a_n} = {\left( {{2^n} + {3^n}} \right)^{1/n}} \ge {\left( {{3^n}} \right)^{1/n}}$. So,
${a_n} \ge 3$
Since ${a_n} = {\left( {{2^n} + {3^n}} \right)^{1/n}} \le {\left( {{3^n} + {3^n}} \right)^{1/n}}$, so
${a_n} \le {\left( {2\cdot{3^n}} \right)^{1/n}}$
Hence,
$3 \le {a_n} \le {\left( {2\cdot{3^n}} \right)^{1/n}} = {2^{1/n}}\cdot3$
Next, we evaluate $\mathop {\lim }\limits_{n \to \infty } {2^{1/n}}\cdot3$
$\mathop {\lim }\limits_{n \to \infty } {2^{1/n}}\cdot3 = 3\cdot\mathop {\lim }\limits_{n \to \infty } {2^{1/n}} = 3\cdot{2^0} = 3$
Since $\mathop {\lim }\limits_{n \to \infty } 3 = 3$ and $\mathop {\lim }\limits_{n \to \infty } {2^{1/n}}\cdot3 = 3$, by Squeeze Theorem we conclude that $\mathop {\lim }\limits_{n \to \infty } {a_n} = 3$
