Answer is: option3

Solution:

The given limit is:

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

This can be rewritten as:

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left( \frac{k}{n} \right)^2$$

The general form of a Riemann sum is:

$$\lim_{n \to \infty} \sum_{k=1}^n \frac{b-a}{n} f\left( a + \frac{(b-a)k}{n} \right)$$

Comparing with the given limit:

1. The coefficient \( \frac{1}{n} \) corresponds to \( \frac{b-a}{n} \), so \( b-a = 1 \)
2. The term \( \left( \frac{k}{n} \right)^2 \) corresponds to \( f\left( a + \frac{(b-a)k}{n} \right) \)

Let’s choose \( a = 0 \), then:

$$b - 0 = 1 \Rightarrow b = 1$$

Now, the function argument becomes:

$$f\left( 0 + \frac{1 \cdot k}{n} \right) = \left( \frac{k}{n} \right)^2$$

This suggests that \( f(x) = x^2 \)

Substituting \( f(x) = x^2 \), \( a = 0 \), and \( b = 1 \) into the Riemann sum, the limit becomes:

$$\int_0^1 x^2 \, dx$$

This matches option (C).