Find $\lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-\big\lfloor \sqrt{n^2+n+1} \big\rfloor \right)$

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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