Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$ 
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\rightarrow \infty\;,$ Then $x^2<x^2+x+1<x^2+2x+1$
So $x<\sqrt{x^2+x+1}<(x+1)\;,$ So $\lfloor x^2+x+1 \rfloor = x\;,$ bcz it lies between two integers.
So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right) = \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right)\cdot \frac{\left(\sqrt{x^2+x+1}+x\right)}{\left(\sqrt{x^2+x+1}+x\right)}$ 
So $\displaystyle \lim_{x\rightarrow \infty} \frac{x+1}{\sqrt{x^2+x+1}+x} = \frac{1}{2}$
$\bullet\; $ If $x\notin \mathbb{Z}$ and $x\rightarrow \infty\;,$ Then How can i solve the above limit in that case,
Help me, Thanks

I would say
$\displaystyle \lim_{x\to \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor=\lim_{x\to \infty} \lfloor \sqrt{x^2+x+1 }\rfloor+\{ \sqrt{x^2+x+1 } \}-\lfloor \sqrt{x^2+x+1 }\rfloor=$
$\displaystyle=\lim_{x\to \infty} \{ \sqrt{x^2+x+1 } \}= (0 \sim  1) \Rightarrow\, $limit not exist
