Answer
$\mathop {\lim }\limits_{n \to \infty } {a_n} = 10$
Work Step by Step
For $n \ge 1$, we have ${a_n} = {\left( {n + {{10}^n}} \right)^{1/n}} \ge {\left( {{{10}^n}} \right)^{1/n}}$. So,
${a_n} \ge 10$
Since ${a_n} = {\left( {n + {{10}^n}} \right)^{1/n}} \le {\left( {{{10}^n} + {{10}^n}} \right)^{1/n}}$, so
${a_n} \le {\left( {2\cdot{{10}^n}} \right)^{1/n}}$
Hence,
$10 \le {a_n} \le {\left( {2\cdot{{10}^n}} \right)^{1/n}}$
Since ${\left( {2\cdot{{10}^n}} \right)^{1/n}} = {2^{1/n}}\cdot10$, we evaluate $\mathop {\lim }\limits_{n \to \infty } {2^{1/n}}\cdot10$.
$\mathop {\lim }\limits_{n \to \infty } {2^{1/n}}\cdot10 = 10\cdot\mathop {\lim }\limits_{n \to \infty } {2^{1/n}} = 10\cdot{2^0} = 10$
Since $\mathop {\lim }\limits_{n \to \infty } 10 = 10$ and $\mathop {\lim }\limits_{n \to \infty } {2^{1/n}}\cdot10 = 10$, by Squeeze Theorem we conclude that $\mathop {\lim }\limits_{n \to \infty } {a_n} = 10$
