<s>[INST] Find $\lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-\big\lfloor \sqrt{n^2+n+1} \big\rfloor \right)$ I need to find the limit
$$
\displaystyle \lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-\big\lfloor  \sqrt{n^2+n+1} \big\rfloor \right),$$ where $n\in \mathbb{N}$.
My attempt. As $\displaystyle \lim_{n\rightarrow \infty} (n^2+n+1)\approx n^2$, then $\displaystyle \lim_{n\rightarrow \infty}\sqrt{n^2+n+1}\approx \displaystyle \lim_{n\rightarrow \infty}\sqrt{n^2} = n$.
So $$\displaystyle \lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-n\right) = \displaystyle \lim_{n\rightarrow \infty}\frac{\left(\sqrt{n^2+n+1}-n\right).\left(\sqrt{n^2+n+1}+n\right)}{\left(\sqrt{n^2+n+1}+n\right)}.$$
So $$\displaystyle \lim_{n\rightarrow \infty}\frac{n\cdot\left(1+\frac{1}{n}\right)}{n \left(\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1\right)} = \frac{1}{2}.$$
My Question is , Is my Process is Right OR Not ,OR Is there is any error .
If Not Then How can I Solve it
Help Required
Thanks
 [/INST] 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 $n\to \infty$.
</s>