2. Define the sequence $\left\{a_{n}\right\}$ :
$$
a_{i} \in\left(0, \frac{1}{2}\right](i=1,2, \cdots) \text {, }
$$

and. $a_{1}=\frac{1}{2}, \frac{a_{n-1}}{a_{n}}+4 a_{n}^{2}=3$.
Then $\lim _{n \rightarrow+\infty} a_{n}=(：$.
(A) $\frac{\sqrt{3}}{4}$
(B) $\frac{1}{12}$
(C) $\frac{\sqrt{2}}{4}$
(D) 0
2.1).

Let $a_{n}=\sin \theta$. Since $a_{n}$ satisfies $a_{n-1}=3 a_{n}-4 a_{n}^{3}>0$, and $a_{n}<1$, it follows from the double angle formula that $a_{n-1}=\sin 3 \theta$.

Considering that $f(x)=3 x-4 x^{3}$ is monotonically increasing on $\left(0, \frac{1}{2}\right]$, for a given $\dot{a}_{n-1}, a_{n}$ is uniquely determined. Therefore,
$$
a_{2}=\sin \frac{\pi}{18}, \cdots, a_{n}=\sin \frac{\pi}{2 \times 3^{n}} .
$$

As $n^{\rightarrow} \rightarrow+\infty$, we have $\lim _{n \rightarrow+} a_{n}=0$.