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The given mathematical problem is as follows:

\[
\lim_{n \to \infty} \left( \frac{1}{n^2} + \frac{2}{n^2} + \dots + \frac{n-1}{n^2} \right)
\]

### Solution:
The expression inside the limit is a sum of terms of the form:

\[
S_n = \sum_{k=1}^{n-1} \frac{k}{n^2}
\]

This can be written as:

\[
S_n = \frac{1}{n^2} \sum_{k=1}^{n-1} k
\]

We know that the sum of the first \(n-1\) integers is given by the formula:

\[
\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}
\]

Thus, the sum \(S_n\) becomes:

\[
S_n = \frac{1}{n^2} \cdot \frac{(n-1)n}{2} = \frac{n(n-1)}{2n^2}
\]

Simplifying this expression:

\[
S_n = \frac{n-1}{2n}
\]

Now, we take the limit as \(n \to \infty\):

\[
\lim_{n \to \infty} \frac{n-1}{2n} = \lim_{n \to \infty} \frac{n}{2n} - \frac{1}{2n} = \frac{1}{2} - 0 = \frac{1}{2}
\]

Thus, the value of the limit is:

\[
\boxed{\frac{1}{2}}
\]

Would you like more details on any specific step or clarification?

Here are some related questions you might consider:
1. What is the general formula for the sum of an arithmetic series?
2. How does this method relate to Riemann sums and integrals?
3. Can this type of sum be generalized for different exponents in the denominator?
4. What if the upper limit of the sum was extended to \(n\) instead of \(n-1\)?
5. How would this problem change if the terms inside the sum were squared?

**Tip:** Simplifying sums using known formulas is a powerful technique in evaluating limits.

Evaluate the limit: lim (n → ∞) (1/n² + 2/n² + ... + (n-1)/n²).

Solve the limit problem involving the summation of an arithmetic series: lim (n → ∞) (1/n² + 2/n² + ... + (n-1)/n²). Learn how to simplify using summation formulas and take limits as n approaches infinity. This step-by-step solution is suitable for high school calculus students.

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