Answer
By the Squeeze Theorem (Theorem 3):
$\mathop {\lim }\limits_{n \to \infty } \frac{{{9^n}}}{{n!}} = 0$.
Work Step by Step
For $n>9$, we write $\frac{{{9^n}}}{{n!}}$ as a product of $n$ factors:
$\frac{{{9^n}}}{{n!}} = \left( {\frac{9}{1}\cdot\frac{9}{2}\cdot...\cdot\frac{9}{9}} \right)\left( {\frac{9}{{10}}\cdot\frac{9}{{11}}\cdot...\cdot\frac{9}{n}} \right)$.
Since $\frac{{{9^n}}}{{n!}} \ge 0$ and $\frac{9}{1}\cdot\frac{9}{2}\cdot...\cdot\frac{9}{9} = \frac{{{9^9}}}{{9!}}$, we have
$0 \le \frac{{{9^n}}}{{n!}} = \frac{{{9^9}}}{{9!}}\left( {\frac{9}{{10}}\cdot\frac{9}{{11}}\cdot...\cdot\frac{9}{n}} \right)$.
Since $\frac{9}{{10}},\frac{9}{{11}},...$ are all less than 1, so
$\frac{{{9^9}}}{{9!}}\left( {\frac{9}{{10}}\cdot\frac{9}{{11}}\cdot...\cdot\frac{9}{n}} \right) \le \frac{{{9^9}}}{{9!}}\left( {\frac{9}{n}} \right)$.
Therefore,
$0 \le \frac{{{9^n}}}{{n!}} = \frac{{{9^9}}}{{9!}}\left( {\frac{9}{{10}}\cdot\frac{9}{{11}}\cdot...\cdot\frac{9}{n}} \right) \le \frac{{{9^9}}}{{9!}}\cdot\frac{9}{n}$.
Since $\frac{{{9^9}}}{{9!}}$ is a constant and $\mathop {\lim }\limits_{n \to \infty } \frac{9}{n} = 0$, the Squeeze Theorem (Theorem 3) gives us $\mathop {\lim }\limits_{n \to \infty } \frac{{{9^n}}}{{n!}} = 0$.
