8. From the set $\{1,2, \cdots, 63\}$, seven numbers are randomly drawn, and the smallest number among them is denoted as $x$. Then the mathematical expectation $\mathrm{E}(x)=$ $\qquad$
2. 2 .

It is known that, $\frac{a_{n}}{n!}=\frac{1}{\mathrm{C}_{n}^{0}}+\frac{1}{\mathrm{C}_{n}^{1}}+\cdots+\frac{1}{\mathrm{C}_{n}^{n}}$.
When $n>3$, by the monotonicity of binomial coefficients, we have
$$
\begin{array}{l}
2+\frac{2}{n}<\frac{a_{n}}{n!}<2+\frac{2}{n}+\frac{n-3}{\mathrm{C}_{n}^{2}} \\
=2+\frac{2}{n}+\frac{2(n-3)}{n(n+1)}<2+\frac{4}{n} \\
\Rightarrow 2+\frac{2}{n}<\frac{a_{n}}{n!}<2+\frac{4}{n} .
\end{array}
$$

When $n \rightarrow+\infty$, taking the limit on both sides of the above inequality, we get 2. By the squeeze theorem, we have
$$
\lim _{n \rightarrow \infty} \frac{a_{n}}{n!}=2
$$