If $$f : R \to R$$ is given by $$f(x) = x + 1$$, then the value of
$$\lim_{n \to \infty} \frac{1}{n}\left[f(0) + f\left(\frac{5}{n}\right) + f\left(\frac{10}{n}\right) + \ldots + f\left(\frac{5(n-1)}{n}\right)\right]$$ is:

The given limit is $$L = \lim_{n \to \infty} \frac{1}{n}\left[f(0) + f\!\left(\frac{5}{n}\right) + f\!\left(\frac{10}{n}\right) + \cdots + f\!\left(\frac{5(n-1)}{n}\right)\right] = \lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^{n-1} f\!\left(\frac{5k}{n}\right).$$

Multiplying and dividing by 5: $$L = \frac{1}{5} \lim_{n\to\infty} \frac{5}{n}\sum_{k=0}^{n-1} f\!\left(\frac{5k}{n}\right) = \frac{1}{5}\int_0^5 f(x)\,dx.$$

Since $$f(x) = x + 1$$: $$\int_0^5 f(x)\,dx = \int_0^5 (x+1)\,dx = \left[\frac{x^2}{2} + x\right]_0^5 = \frac{25}{2} + 5 = \frac{35}{2}.$$

Therefore, $$L = \frac{1}{5} \cdot \frac{35}{2} = \frac{7}{2}$$.