4. Given that $\triangle A B C$ is an equilateral triangle with side length 2, and points $M_{1}, M_{2}, \cdots, M_{n}$ are taken sequentially on segment $B C$, dividing $B C$ into $n+1$ equal parts. Then
$$
\lim _{n \rightarrow \infty} \frac{1}{n}\left(A M_{1}^{2}+A M_{2}^{2}+\cdots+A M_{n}^{2}\right)=(\quad) \text {. }
$$
(A) $\frac{8}{3}$
(B) 3
(C) $\frac{10}{3}$
(D) 4
4. C.

As shown in Figure 4, draw $B D \perp A C$, with the foot of the perpendicular being $D$. Draw $M_{k} N_{k} \perp B D$ and $M_{k} P_{k} \perp A C$, with the feet of the perpendiculars being $N_{k}$ and $P_{k}$, respectively.
From $\frac{B M_{k}}{B C}=\frac{k}{n+1}$, we get
$$
\begin{array}{c}
M_{k} N_{k}=\frac{k}{n+1}, \\
M_{k} P_{k}=\frac{\sqrt{3}(n+1-k)}{n+1} .
\end{array}
$$

Thus, $A P_{k}=A D+M_{k} N_{k}=1+\frac{k}{n+1}$.
Therefore, $A M_{k}^{2}=A P_{k}^{2}+M_{k} P_{k}^{2}$
$$
\begin{array}{l}
=\left(1+\frac{k}{n+1}\right)^{2}+\left[\frac{\sqrt{3}(n+1-k)}{n+1}\right]^{2} \\
=1+\frac{2 k}{n+1}+\frac{k^{2}}{(n+1)^{2}}+\frac{3(n+1-k)^{2}}{(n+1)^{2}} .
\end{array}
$$

Then, $\frac{1}{n}\left(A M_{1}^{2}+A M_{2}^{2}+\cdots+A M_{n}^{2}\right)$
$$
\begin{array}{l}
=\frac{1}{n}\left\{n+\frac{2(1+2+\cdots+n)}{n+1}+\frac{1^{2}+2^{2}+\cdots+n^{2}}{(n+1)^{2}}+\right. \\
\left.\frac{3\left[n^{2}+(n-1)^{2}+\cdots+1^{2}\right]}{(n+1)^{2}}\right\} \\
=\frac{1}{n}\left[2 n+\frac{4}{(n+1)^{2}} \cdot \frac{1}{6} n(n+1)(2 n+1)\right] \\
=2+\frac{2(2 n+1)}{3(n+1)} .
\end{array}
$$

Therefore, $\lim _{n \rightarrow \infty} \frac{1}{n}\left(A M_{1}^{2}+A M_{2}^{2}+\cdots+A M_{n}^{2}\right)$
$$
=\lim _{n \rightarrow \infty}\left[2+\frac{2(2 n+1)}{3(n+1)}\right]=2+\frac{4}{3}=\frac{10}{3} \text {. }
$$