30. Let $f(x)=4 x-x^{2}$. Given $x_{0}$, consider how many real numbers $x_{0}$, make the sequence $x_{0}, x_{1}, x_{2}, \cdots H$ take infinitely many different values.
(A)0.(B)1 or 2.(C)3, 1, or 6.
(D)More than 6 but finite. (E)Infinitely many.
30
E
30. Notice that when $x_{0}=0$, from $f(0)=4 \cdot 0-0^{2}=0$ we get the constant sequence $0,0, \cdots, 0$.
Solving $4 x-x^{2}=0$ yields $x=0, x=4$.
When $x_{0}=4$, from $f(4)=4 \times 4-4^{2}=0$ we get the sequence $4,0,0, \cdots$.
Solving $4 x-x^{2}=4$ yields $x=2$.
When $x_{0}=2$, from $f(2)=4 \times 2-2^{2}=4$ we get the sequence $2,4,0,0, \cdots$

Solving $4 x-x^{2}=2$ yields $x=2 \pm \sqrt{2}$, resulting in the sequences $2+\sqrt{2}, 2,4,0,0, \cdots$,
$2-\sqrt{2}, 2,4,0,0, \cdots$.
Assuming $x=a_{n}$ results in a sequence with only $n$ values $a_{n}, a_{n-1}, \cdots, a_{2}, a_{1}, a_{1}, \cdots$
(This sequence can also be written as $a_{n}, a_{n-1}, \cdots, 2,4,0,0, \cdots$).
Then let $f\left(a_{n+1}\right)=a_{n}$, i.e., $4 a_{n+1}^{2}-a_{n+1}=a_{n}$,
Solving yields $a_{n+1}=2 \pm \sqrt{4-a}$.
Therefore, when $0 \leqslant a_{n} \leqslant 4$, $a_{n+1}$ is a real number and satisfies
$$
0 \leqslant a_{n+1} \leqslant 4 .
$$

Thus, we can obtain a sequence with only $n+1$ values, hence there are infinitely many $x_{0}$ such that the sequence $x_{0}, x_{1}, \cdots$ has only a finite number of distinct values, so the answer should be (E).