The problem asks us to calculate the limits of certain sums using the Riemann sum approach. Here's a breakdown of the exercise:

Problem 1:

We are tasked to calculate the following limits:

1. $\lim_{n \to \infty} \left( \sum_{k=1}^n \frac{n}{n^2 + k^2} \right)$
2. $\lim_{n \to \infty} \left( \sum_{k=1}^n \frac{k}{n^2 + k^2} \right)$
3. $\lim_{n \to \infty} \left( \frac{1}{n} \sum_{k=1}^{n-1} \frac{k}{\sqrt{n^2 - k^2}} \right)$

Let me solve these step by step:

1.

The sum can be interpreted as a Riemann sum. Rewrite the term inside: $\frac{n}{n^2 + k^2} = \frac{1}{n} \cdot \frac{1}{1 + \left(\frac{k}{n}\right)^2}$

Using the Riemann sum definition, where $x_k = \frac{k}{n}$, the sum becomes: $\sum_{k=1}^n \frac{n}{n^2 + k^2} \to \int_0^1 \frac{1}{1 + x^2} \, dx \quad \text{as } n \to \infty$

The integral evaluates to: $\int_0^1 \frac{1}{1 + x^2} \, dx = \left[ \arctan(x) \right]_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4}$

Thus: $\lim_{n \to \infty} \sum_{k=1}^n \frac{n}{n^2 + k^2} = \frac{\pi}{4}$

2.

Similarly, rewrite the term inside: $\frac{k}{n^2 + k^2}$ This requires careful treatment. Rewrite it as: $\frac{k}{n^2 + k^2} = \frac{\frac{k}{n}}{n + \left(\frac{k}{n}\right)^2}$

Using the substitution $x_k = \frac{k}{n}$, the sum approximates: $\sum_{k=1}^n \frac{k}{n^2 + k^2} \to \int_0^1 \frac{x}{1 + x^2} \, dx \quad \text{as } n \to \infty$

The integral evaluates to: $\int_0^1 \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln(1 + x^2) \bigg|_0^1 = \frac{1}{2} \ln(2) - 0 = \frac{\ln(2)}{2}$

Thus: $\lim_{n \to \infty} \sum_{k=1}^n \frac{k}{n^2 + k^2} = \frac{\ln(2)}{2}$

3.

Here, we have: $\frac{1}{n} \sum_{k=1}^{n-1} \frac{k}{\sqrt{n^2 - k^2}}$

The term inside the sum becomes: $\frac{k}{\sqrt{n^2 - k^2}} = \frac{\frac{k}{n}}{\sqrt{1 - \left(\frac{k}{n}\right)^2}}$

Using $x_k = \frac{k}{n}$, the sum approximates: [ \frac