Find $\lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-\big\lfloor \sqrt{n^2+n+1} \big\rfloor \right)$
Note that $$ n^2<n^2+n+1<n^2+2n+1=(n+1)^2, $$ and hence $$ n<\sqrt{n^2+n+1}<n+1, $$ and therefore $\lfloor\sqrt{n^2+n+1}\rfloor=n$. Therefore \begin{align} \sqrt{n^2+n+1}-\lfloor\sqrt{n^2+n+1}\rfloor&=\sqrt{n^2+n+1}-n =\frac{\big(\sqrt{n^2+n+1}-n\big)\big(\sqrt{n^2+n+1}+n\big)}{\big(\sqrt{n^2+n+1}+n\big)}\\ &= \frac{n+1}{\sqrt{n^2+n+1}+n}=\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1}\to\frac{1}{2}, \end{align} as $n\to \infty$.