【解析】

$\lim\limits_{n \to \infty} \frac{1}{n^2} \left[ \ln \frac{1}{n} + 2\ln \frac{2}{n} + \cdots + (n - 1)\ln \frac{n - 1}{n} \right]$

$= \lim\limits_{n \to \infty} \left( \frac{1}{n}\ln \frac{1}{n} + \frac{2}{n}\ln \frac{2}{n} + \cdots + \frac{n - 1}{n}\ln \frac{n - 1}{n} \right) \cdot \frac{1}{n}$

$= \lim\limits_{n \to \infty} \left( \frac{1}{n}\ln \frac{1}{n} + \frac{2}{n}\ln \frac{2}{n} + \cdots + \frac{n - 1}{n}\ln \frac{n - 1}{n} + \frac{n}{n}\ln \frac{n}{n} \right) \cdot \frac{1}{n}$

$= \lim\limits_{n \to \infty} \sum_{i = 1}^{n} \frac{i}{n}\ln \frac{i}{n} \cdot \frac{1}{n}$

$= \int_{0}^{1} x\ln x dx$

$= \frac{1}{2} \int_{0}^{1} \ln x dx^2 = \frac{1}{2} \left( x^2\ln x \big|_{0}^{1} - \int_{0}^{1} x^2 \cdot \frac{1}{x} dx \right)$

$= -\frac{1}{2} \int_{0}^{1} x dx = -\frac{1}{4}$