Answer
$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$
Work Step by Step
For $n \ge 1$, we write
${a_n} = \frac{1}{{\sqrt {{n^4} + {n^8}} }} = \frac{1}{{\sqrt {{n^8}\left( {\frac{1}{{{n^4}}} + 1} \right)} }} = \frac{1}{{{n^4}\sqrt {\frac{1}{{{n^4}}} + 1} }}$
Since $\frac{1}{{{n^4}}} \le 1$, therefore ${a_n} \ge \frac{1}{{{n^4}\sqrt 2 }}$.
Next, we write
${a_n} = \frac{1}{{\sqrt {{n^4} + {n^8}} }} = \frac{1}{{\sqrt {{n^4}\left( {1 + {n^4}} \right)} }} = \frac{1}{{{n^2}\sqrt {1 + {n^4}} }}$
Since ${n^4} \ge 1$, therefore ${a_n} \le \frac{1}{{{n^2}\sqrt 2 }}$.
Hence,
$\frac{1}{{\sqrt 2 {n^4}}} \le {a_n} \le \frac{1}{{\sqrt 2 {n^2}}}$.
Since $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{\sqrt 2 {n^4}}} = 0$ and $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{\sqrt 2 {n^2}}} = 0$, by Squeeze Theorem we conclude that $\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$.
