To solve this problem, let us analyze the given limit:

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

The sum has $n$ terms, and each term is of the form $\frac{n}{n^2+k}$, where $k$ ranges from $1$ to $n$.

Simplify each term

For each term $\frac{n}{n^2 + k}$, we factor $n^2$ from the denominator:

$\frac{n}{n^2 + k} = \frac{n}{n^2(1 + \frac{k}{n^2})} = \frac{1}{n(1 + \frac{k}{n^2})}.$

Approximation for large $n$

As $n \to \infty$, the term $\frac{k}{n^2} \to 0$, so $\frac{1}{n(1 + \frac{k}{n^2})} \approx \frac{1}{n}$. This approximation holds for all $k$ between $1$ and $n$.

Thus, the entire sum can be approximated as: $\sum_{k=1}^n \frac{1}{n}.$

Evaluate the sum

Since there are $n$ terms, each approximately equal to $\frac{1}{n}$, the sum simplifies to: $\sum_{k=1}^n \frac{1}{n} = n \cdot \frac{1}{n} = 1.$

Conclusion

The value of the limit is:

$\boxed{1}.$

Would you like additional details on any step or a different explanation? Here are five questions that build on this problem:

1. What are other approaches to evaluate this type of sum using integral approximations?
2. How would the result change if the numerator was a function of $k$ instead of $n$?
3. Can this result be generalized to similar sums with different denominator structures?
4. What is the role of the approximation in simplifying the sum?
5. How does the behavior of the sum change for finite values of $n$?

Tip: Always check if the denominator grows significantly faster than the numerator in limit problems—it often leads to simplifications!