Answer
By the Squeeze Theorem (Theorem 3):
$\mathop {\lim }\limits_{n \to \infty } \frac{{{8^{2n}}}}{{n!}} = 0$.
Work Step by Step
Write ${a_n} = \frac{{{8^{2n}}}}{{n!}} = \frac{{{{64}^n}}}{{n!}}$.
Similar method to Exercise 41, for $n>64$, we write $\frac{{{{64}^n}}}{{n!}}$ as a product of $n$ factors:
$\frac{{{{64}^n}}}{{n!}} = \left( {\frac{{64}}{1}\cdot\frac{{64}}{2}\cdot...\cdot\frac{{64}}{{64}}} \right)\left( {\frac{{64}}{{65}}\cdot\frac{{64}}{{66}}\cdot...\cdot\frac{{64}}{n}} \right)$.
Since $\frac{{{{64}^n}}}{{n!}} \ge 0$ and $\frac{{64}}{1}\cdot\frac{{64}}{2}\cdot...\cdot\frac{{64}}{{64}} = \frac{{{{64}^{64}}}}{{64!}}$, we have
$0 \le \frac{{{{64}^n}}}{{n!}} = \frac{{{{64}^{64}}}}{{64!}}\left( {\frac{{64}}{{65}}\cdot\frac{{64}}{{66}}\cdot...\cdot\frac{{64}}{n}} \right)$.
Since $\frac{{64}}{{65}},\frac{{64}}{{66}},...$ are all less than 1, so
$\frac{{{{64}^{64}}}}{{64!}}\left( {\frac{{64}}{{65}}\cdot\frac{{64}}{{66}}\cdot...\cdot\frac{{64}}{n}} \right) \le \frac{{{{64}^{64}}}}{{64!}}\left( {\frac{{64}}{n}} \right)$.
Therefore,
$0 \le \frac{{{{64}^n}}}{{n!}} = \frac{{{{64}^{64}}}}{{64!}}\left( {\frac{{64}}{{65}}\cdot\frac{{64}}{{66}}\cdot...\cdot\frac{{64}}{n}} \right) \le \frac{{{{64}^{64}}}}{{64!}}\frac{{64}}{n}$.
Since $\frac{{{{64}^{64}}}}{{64!}}$ is a constant and $\mathop {\lim }\limits_{n \to \infty } \frac{{64}}{n} = 0$, the Squeeze Theorem (Theorem 3) gives us $\mathop {\lim }\limits_{n \to \infty } \frac{{{8^{2n}}}}{{n!}} = 0$.
