Solution: Note that \begin{align*} & \lim_{n \to \infty} \left( \sqrt{n^{2}+n} - \sqrt{n^{2}+1 } \right) \\ = & \lim_{n \to \infty} \left( \left( \sqrt{n^{2}+n} - \sqrt{n^{2}+1 } \right) \times \frac{\sqrt{n^{2}+n} + \sqrt{n^{2}+1 } }{\sqrt{n^{2}+n} + \sqrt{n^{2}+1 }} \right) \\ = & \lim_{n \to \infty} \left( \frac{n^{2}+n-n^2 -1 }{\sqrt{n^2 +n} + \sqrt{n^2 +1} } \right) \\ = & \lim_{n \to \infty} \frac{n-1}{\sqrt{n^2 +n} + \sqrt{n^2 +1}} \\ = & \lim_{n \to \infty} \frac{1-\frac{1}{n}}{\sqrt{1+\frac{1}{n}} + \sqrt{1+\frac{1}{n^2}} } \quad \boxed{\text{divide by $n$}}\\ = & \frac{1}{1+1} \\ = & \frac{1}{2}. \end{align*} Therefore, \[ \textcolor{blue}{\boxed{\lim_{n \to \infty} \left( \sqrt{n^{2}+n} - \sqrt{n^{2}+1 } \right) = \frac{1}{2}}} \]