Let $\alpha > 0$ be a real number. Compute the limit of the sequence $\{x_n\}_{n\geq 1}$ defined by $$x_n=\begin{cases} \sum \limits_{k=1}^n \sinh \left(\frac{k}{n^2}\right),& \text{when}\ n>\frac{1}{\alpha}\\ 0,& \text{when}\ n\leq \frac{1}{\alpha}\end{cases}$$
1. Given the sequence \( \{x_n\}_{n \geq 1} \) defined by:
   \[
   x_n = \begin{cases} 
   \sum \limits_{k=1}^n \sinh \left(\frac{k}{n^2}\right), & \text{when}\ n > \frac{1}{\alpha} \\
   0, & \text{when}\ n \leq \frac{1}{\alpha}
   \end{cases}
   \]
   we need to compute the limit of \( x_n \) as \( n \to \infty \).

2. For \( n > \frac{1}{\alpha} \), we use the Taylor series expansion of \( \sinh x \):
   \[
   \sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots
   \]
   Therefore,
   \[
   \sinh \left(\frac{k}{n^2}\right) = \frac{k}{n^2} + \frac{1}{3!} \left(\frac{k}{n^2}\right)^3 + \cdots
   \]

3. Summing these terms for \( k = 1 \) to \( n \):
   \[
   x_n = \sum_{k=1}^n \sinh \left(\frac{k}{n^2}\right) = \sum_{k=1}^n \left( \frac{k}{n^2} + \frac{1}{3!} \left(\frac{k}{n^2}\right)^3 + \cdots \right)
   \]

4. Considering the leading term \( \frac{k}{n^2} \):
   \[
   \sum_{k=1}^n \frac{k}{n^2} = \frac{1}{n^2} \sum_{k=1}^n k = \frac{1}{n^2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{2n^2} = \frac{n+1}{2n}
   \]
   As \( n \to \infty \), \( \frac{n+1}{2n} \to \frac{1}{2} \).

5. Higher-order terms:
   \[
   \sum_{k=1}^n \frac{1}{3!} \left(\frac{k}{n^2}\right)^3 = \frac{1}{6n^6} \sum_{k=1}^n k^3
   \]
   Using the formula for the sum of cubes:
   \[
   \sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2 = \frac{n^2(n+1)^2}{4}
   \]
   Thus,
   \[
   \frac{1}{6n^6} \cdot \frac{n^2(n+1)^2}{4} = \frac{(n+1)^2}{24n^4}
   \]
   As \( n \to \infty \), \( \frac{(n+1)^2}{24n^4} \to 0 \).

6. Therefore, the higher-order terms vanish as \( n \to \infty \), and the dominant term is \( \frac{1}{2} \).

7. For \( n \leq \frac{1}{\alpha} \), \( x_n = 0 \), but this is irrelevant for large \( n \) due to the Archimedean property.

8. Hence, the limit of \( x_n \) as \( n \to \infty \) is:
   \[
   \lim_{n \to \infty} x_n = \frac{1}{2}
   \]

The final answer is \( \boxed{\frac{1}{2}} \).